All language subtitles for Calculus-2-Full-College-Course_en

Hi, I'm Beau with Free Code Camp. This course\n

at the University of North Carolina at Chapel\n

for many years to undergraduate students,\n

can follow along at home. Let's start.

Hi, I'm Beau with Free Code Camp. This course\n

at the University of North Carolina at Chapel\n

for many years to undergraduate students,\n

can follow along at home. Let's start.

This video introduces the idea of finding\n

some review. in calculus one, you approximate\n

dividing it up into tall skinny rectangles.\n

rectangles using the symbols delta x, here\n

axis or a small change in x values. You picked\n

each of these little sub intervals along the\n

The sample point on the x axis for rectangle\n

sample points and the function f to figure\n

of rectangle number i is given by the functions\n

this, you calculate the area of each rectangle,\n

So the area of the first rectangle is delta\n

rectangle is delta x times f of x two star.\n

to be its base delta x times its height, f\n

than the last rectangle will have base delta\n

area under the curve is given by adding up\n

sigma notation, this can be written as sigma,\n

of rectangles of the area of the rectangle,\n

is then given by the limit of these approximating\n

infinity. That's the limit as n goes to infinity\nof this

This video introduces the idea of finding\n

some review. in calculus one, you approximate\n

dividing it up into tall skinny rectangles.\n

rectangles using the symbols delta x, here\n

axis or a small change in x values. You picked\n

each of these little sub intervals along the\n

The sample point on the x axis for rectangle\n

sample points and the function f to figure\n

of rectangle number i is given by the functions\n

this, you calculate the area of each rectangle,\n

So the area of the first rectangle is delta\n

rectangle is delta x times f of x two star.\n

to be its base delta x times its height, f\n

than the last rectangle will have base delta\n

area under the curve is given by adding up\n

sigma notation, this can be written as sigma,\n

of rectangles of the area of the rectangle,\n

is then given by the limit of these approximating\n

infinity. That's the limit as n goes to infinity\n

A limit of a Riemann sum like this is by definition,\n

the integral sign as the integral of f of\n

based on our picture are from x equals a 2x\n

limit of a Riemann sum to integral notation,\n

variable x, and the delta x becomes our dx.\n

the area under a curve, we're going to use\n

two curves. To compute the area between two\n

x, in between the x values of a and b, we\n

skinny rectangles as shown in this picture.\n

of each rectangle. And let's let x i star\n

So x sub i star is a point on the x axis that\n

want to compute the area of one of these tall\n

a rectangle is the base times the height.\n

by delta x. But the height is different for\n

number, I assume this is rectangle number\n

to F of the sample point f of x by star and\n

that x, y star. So the height of that rectangle\n

sample point. In other words, it's the difference,\n

that we have an expression for the area of\n

those areas as before, to get an expression\n

it's just the sum of these areas. And as before,\n

rectangles skinnier and skinnier. By taking\n

of rectangles goes to infinity of this Riemann\n

sum is given by the integral where the exabyte\n

x, and the delta x becomes our dx, let's put\n

x goes between A and B, and we have an expression\n

this formula only works if f of x is greater\n

a to b. That inequality guarantees that this\n

will be a positive number, we want a positive\n

a positive area. If instead, f of x is less\n

to switch around your subtraction and take\n

in order to get a positive area. One way to\n

as the integral from a to b of the top y value\n

that you'll need to replace the top y value\n

A limit of a Riemann sum like this is by definition,\n

the integral sign as the integral of f of\n

based on our picture are from x equals a 2x\n

limit of a Riemann sum to integral notation,\n

variable x, and the delta x becomes our dx.\n

the area under a curve, we're going to use\n

two curves. To compute the area between two\n

x, in between the x values of a and b, we\n

skinny rectangles as shown in this picture.\n

of each rectangle. And let's let x i star\n

So x sub i star is a point on the x axis that\n

want to compute the area of one of these tall\n

a rectangle is the base times the height.\n

by delta x. But the height is different for\n

number, I assume this is rectangle number\n

to F of the sample point f of x by star and\n

that x, y star. So the height of that rectangle\n

sample point. In other words, it's the difference,\n

that we have an expression for the area of\n

those areas as before, to get an expression\n

it's just the sum of these areas. And as before,\n

rectangles skinnier and skinnier. By taking\n

of rectangles goes to infinity of this Riemann\n

sum is given by the integral where the exabyte\n

x, and the delta x becomes our dx, let's put\n

x goes between A and B, and we have an expression\n

this formula only works if f of x is greater\n

a to b. That inequality guarantees that this\n

will be a positive number, we want a positive\n

a positive area. If instead, f of x is less\n

to switch around your subtraction and take\n

in order to get a positive area. One way to\n

as the integral from a to b of the top y value\n

that you'll need to replace the top y value\n

Let's look at an example. We want to find\n

plus x and y equals three minus x squared.\n

So that must be the red curve here, while\n

downwards. So that must be this blue curve.\n

from our starting x value to our ending x\n

y values, all integrated with respect to x,\n

three minus x squared, and our bottom y values\n

Now we still need to figure out the values\n

from the graph, it looks like a should be\n

should be about one, since that's where the\n

direction. But to find exact values of a and\n

two equations equal to each other and solve\n

to x squared plus x, I can add x squared to\n

three equals zero. This factors into 2x plus\n

to equal negative three halves and or x equals\n

So we can finish doing our setup, our bounds\n

to one, and I can simplify my integrand here.\n

minus x. Or in other words, the integral from\n

minus x plus three dx. This integrates to\n

over two plus 3x, evaluated between one and\n

plug in my bounds of integration and then\n

Let's look at an example. We want to find\n

plus x and y equals three minus x squared.\n

So that must be the red curve here, while\n

downwards. So that must be this blue curve.\n

from our starting x value to our ending x\n

y values, all integrated with respect to x,\n

three minus x squared, and our bottom y values\n

Now we still need to figure out the values\n

from the graph, it looks like a should be\n

should be about one, since that's where the\n

direction. But to find exact values of a and\n

two equations equal to each other and solve\n

to x squared plus x, I can add x squared to\n

three equals zero. This factors into 2x plus\n

to equal negative three halves and or x equals\n

So we can finish doing our setup, our bounds\n

to one, and I can simplify my integrand here.\n

minus x. Or in other words, the integral from\n

minus x plus three dx. This integrates to\n

over two plus 3x, evaluated between one and\n

plug in my bounds of integration and then\n

In this video, we saw that the area between\n

b can be given by the integral from a to b\n

dx, were the top y values and bottom y values\n

the top curve is given by y equals f of x,\n

this integral is the integral from a to b\n

we'll calculate the volumes of solids of revolution.\n

object that can be formed by rotating a region\n

by rotating a region of the plane around a\n

figure on the left, this three dimensional\n

plane shaped like this, around the x axis,\n

a crescent shaped region of the plane like\n

these two solids of revolution, using planes\n

our cross sections look like disks.

In this video, we saw that the area between\n

b can be given by the integral from a to b\n

dx, were the top y values and bottom y values\n

the top curve is given by y equals f of x,\n

this integral is the integral from a to b\n

we'll calculate the volumes of solids of revolution.\n

object that can be formed by rotating a region\n

by rotating a region of the plane around a\n

figure on the left, this three dimensional\n

plane shaped like this, around the x axis,\n

a crescent shaped region of the plane like\n

these two solids of revolution, using planes\n

our cross sections look like disks.

A disc here means the inside of a circle.\n

it's hollow inside. And when we slice it by\n

cross sections that are shaped like washers.\n

two concentric circles. So for solids of revolution,\n

disk, or the shape of a washer. The area of\n

r squared, where r is the radius. And the\n

r outer squared minus pi times our inner squared,\n

and our inner is the radius of the little\n

of the washer is just the area of the larger\n

circle. Now we know that the volume of any\n

using planes perpendicular to the x axis is\n

to x equals B of the area of the cross section\n

are disks, this formula becomes the integral\n

a function of x. If instead, the cross sections\n

becomes the integral of pi r outer squared\n

and our inner are functions of x here. These\n

is formed by rotating a region around the\n

around a horizontal line, then our cross sectional\n

x axis and are thin in the x direction. So\n

we want to rotate around the y axis or a vertical\n

are going to be perpendicular to the y axis\n

So when rotating around the y axis or a vertical\n

to y. Our cross sectional area will be a function\n

integration will have to be Y values, our\n

just have to calculate our Radia and bounds\n

As our first example, let's consider the region\n

of x, the x axis and the line x equals eight.\n

revolution found by rotating this region around\n

to be discs. And these discs are thin in the\n

dx, our smallest x value is zero, and our\n

of integration. And we want to integrate pi\n

of our disks is given by the y coordinate\n

to y, which is equal to the cube root of x\n

the volume as the integral from zero to eight\n

I can pull out the PI and rewrite this integral\n

using bounds of integration to get three fifths\n

Now, eight to the five thirds means eight\n

to the 1/3 is two and two to the fifth is\n

A disc here means the inside of a circle.\n

it's hollow inside. And when we slice it by\n

cross sections that are shaped like washers.\n

two concentric circles. So for solids of revolution,\n

disk, or the shape of a washer. The area of\n

r squared, where r is the radius. And the\n

r outer squared minus pi times our inner squared,\n

and our inner is the radius of the little\n

of the washer is just the area of the larger\n

circle. Now we know that the volume of any\n

using planes perpendicular to the x axis is\n

to x equals B of the area of the cross section\n

are disks, this formula becomes the integral\n

a function of x. If instead, the cross sections\n

becomes the integral of pi r outer squared\n

and our inner are functions of x here. These\n

is formed by rotating a region around the\n

around a horizontal line, then our cross sectional\n

x axis and are thin in \n

dx. If instead, we want to rotate around the\n

wash our cross sections are going to be perpendicular\n

the y direction. So when rotating around the\n

our integral with respect to y. Our cross\n

integrate d y and our bounds of integration\n

look pretty much the same. We'll just have\n

in terms of y instead of x. As our first example,\n

y equals a cube root of x, the x axis and\n

volume of the solid of revolution found by\n

cross sections here are going to be discs.\n

So we're going to be integrating dx, our smallest\n

So those are our bounds of integration. And\n

dx. Now the radius of our disks is given by\n

write r is equal to y, which is equal to the\n

So we can rewrite the volume as the integral\n

of x squared dx, I can pull out the PI and\n

integrate and then evaluate using bounds of\n

to the five thirds minus zero. Now, eight\n

raised to the fifth power, eight to the 1/3\n

expression simplifies to three fifths pi times\n32, or 96/5

As our next example, let's consider the region\n

the curve y equals the cube root of x and\n

this region around the x axis to get this\n

cross sections this time are shaped like washers,\n

out by the curve y equals cube root of x,\n

y equals 1/4 X. We know our volume is given\n

minus pi times our inner squared. And since\n

know we'll need to integrate dx, our bounds\n

of zero, and our largest x value, which is\n

an x value of eight. We can confirm that the\n

setting them equal to each other. And solving\n

And multiplying both sides by four gives us\n

both sides to the three halves power gives\n

As our next example, let's consider the region\n

the curve y equals the cube root of x and\n

this region around the x axis to get this\n

cross sections this time are shaped like washers,\n

out by the curve y equals cube root of x,\n

y equals 1/4 X. We know our volume is given\n

minus pi times our inner squared. And since\n

know we'll need to integrate dx, our bounds\n

of zero, and our largest x value, which is\n

an x value of eight. We can confirm that the\n

setting them equal to each other. And solving\n

And multiplying both sides by four gives us\n

both sides to the three halves power gives\n

This confirms we have the correct bound of\n

a formula for the outer radius as a function\n

This confirms we have the correct bound of\n

a formula for the outer radius as a function\n

since the outer circle is swept out by the\n

is just given by the y coordinate of this\n

root of x. Now the inner radius is given by\n

of this line as a function of x is 1/4 x.\n

our solid and I routine computation gives\n

let's switch gears and rotate this region\n

are still washers, but this time the washers\n

to be integrating with respect to y. Our bounds\n

at the minimum y value of zero and the maximum\n

where x equals eight, and y, which is the\n

to two. For this problem, we need our our\n

From the picture, we see that our outer is\n

line has the equation y equals 1/4 x. And\n

our outer as a function of y. defined our\n

the x coordinate of the curve, y equals cube\n

terms of y, we have y equals the cube root\n

so our inner is equal to y cubed as a function\n

equation for volume. We can simplify this\n

21. In this video, we calculated the volumes\n

the washer methods, and the following formulas.\n

of bread is to slice it into slices, and calculate\n

the idea to keep in mind as we cover this\n

lot of three dimensional objects are kind\n

sliced into slabs or slices like this. Let's\n

we'll call them s one s two through sn in\n

here's s two, here's s3, and so on, we'll\n

slice is the same thickness, we'll call that\n

solid is just the sum of the volumes of the\n

as the sum from i equals one to n, the number\n

volume of the slab is approximately its cross\n

area of the slab, I mean the area of the front\n

spread peanut butter. Well, now, as you've\n

the area of the front of the slice might be\n

of the slice. So what we'll do is for each\n

point x vystar, that's in that I've interval,\n

be on the right endpoint, or it could be in\n

at the cross sectional area, I'll call it\n

I were to go karate chop right at that x y\n

since the outer circle is swept out by the\n

is just given by the y coordinate of this\n

root of x. Now the inner radius is given by\n

of this line as a function of x is 1/4 x.\n

our solid and I routine computation gives\n

let's switch gears and rotate this region\n

are still washers, but this time the washers\n

to be integrating with respect to y. Our bounds\n

at the minimum y value of zero and the maximum\n

where x equals eight, and y, which is the\n

to two. For this problem, we need our our\n

From the picture, we see that our outer is\n

line has the equation y equals 1/4 x. And\n

our outer as a function of y. defined our\n

the x coordinate of the curve, y equals cube\n

terms of y, we have y equals the cube root\n

so our inner is equal to y cubed as a function\n

equation for volume. We can simplify this\n

21. In this video, we calculated the volumes\n

the washer methods, and the following formulas.\n

of bread is to slice it into slices, and calculate\n

the idea to keep in mind as we cover this\n

lot of three dimensional objects are kind\n

sliced into slabs or slices like this. Let's\n

we'll call them s one s two through sn in\n

here's s two, here's s3, and so on, we'll\n

slice is the same thickness, we'll call that\n

solid is just the sum of the volumes of the\n

as the sum from i equals one to n, the number\n

volume of the slab is approximately its cross\n

area of the slab, I mean the area of the front\n

spread peanut butter. Well, now, as you've\n

the area of the front of the slice might be\n

of the slice. So what we'll do is for each\n

point x vystar, that's in that I've interval,\n

be on the right endpoint, or it could be in\n

at the cross sectional area, I'll call it\n

I were to go karate chop right at that x y\n

Now we can go back and write our volume as\n

x the thickness of the slice. Now this is\n

expression here gives you the volume as if\n

same area from one side to the other. But\n

thin. And in fact, we can calculate the exact\n

thinner and thinner. Or in other words, as\n

we have here is the limit of a Riemann sum.\n

where the x i star becomes our variable x\n

of integration, we'll just use the abstract\n

would fill these in based on In the context\n

expression for the volume of a three dimensional\n

volumes like this, we'll first need a formula\n

function of x. As an example, let's try to\n

ellipse given by this equation, and whose\n

are squares. First, let me graph the base,\n

the extraction than in the y direction. So\n

base, are a bunch of squares. And the squares\n

to the x axis. So they're oriented sort of\n

that's supposed to be coming out of the picture\n

out of the picture. Here's a slightly better\n

tilted. So we're looking at it from below,\n

can see the square cross sections, the x axis\n

y axis is in that direction. This picture\n

where they're only about eight or 10 slices,\n

on the front and the back. better picture\n

are infinitely thin, but they're still square\n

a way that they're perpendicular to the x\n

the integral from a to b of area as a function\n

is negative two, and the maximum x value is\n

of integration. Also, I know that the area\n

So I can write my cross sectional area as\n

of the square as a function of x. Notice that\n

different. But my side length is always twice\n

axis to the y value on the ellipse. So I'll\n

of x squared. I can simplify this a little\n

dx. Now all I need to do is find a formula\n

this equation up here, relating y and x, all\n

In fact, I can get by solving for y squared,\n

solving for y squared, I have y squared over\n

four, which means that y squared is equal\n

Now we can go back and write our volume as\n

x the thickness of the slice. Now this is\n

expression here gives you the volume as if\n

same area from one side to the other. But\n

thin. And in fact, we can calculate the exact\n

thinner and thinner. Or in other words, as\n

we have here is the limit of a Riemann sum.\n

where the x i star becomes our variable x\n

of integration, we'll just use the abstract\n

would fill these in based on In the context\n

expression for the volume of a three dimensional\n

volumes like this, we'll first need a formula\n

function of x. As an example, let's try to\n

ellipse given by this equation, and whose\n

are squares. First, let me graph the base,\n

the extraction than in the y direction. So\n

base, are a bunch of squares. And the squares\n

to the x axis. So they're oriented sort of\n

that's supposed to be coming out of the picture\n

out of the picture. Here's a slightly better\n

tilted. So we're looking at it from below,\n

can see the square cross sections, the x axis\n

y axis is in that direction. This picture\n

where they're only about eight or 10 slices,\n

on the front and the back. better picture\n

are infinitely thin, but they're still square\n

a way that they're perpendicular to the x\n

the integral from a to b of area as a function\n

is negative two, and the maximum x value is\n

of integration. Also, I know that the area\n

So I can write my cross sectional area as\n

of the square as a function of x. Notice that\n

different. But my side length is always twice\n

axis to the y value on the ellipse. So I'll\n

of x squared. I can simplify this a little\n

dx. Now all I need to do is find a formula\n

this equation up here, relating y and x, all\n

In fact, I can get by solving for y squared,\n

solving for y squared, I have y squared over\n

four, which means that y squared is equal\n

Now I plug this into my volume equation. And\n

times nine, one minus x squared over four\n

36. and integrate. Plugging in values and\n

Now, how would this problem be different if\n

to the y axis instead of the x axis? Well,\n

bit different. Since our squares would now\n

squares are now fin in the y direction, instead\n

the width of a slab B delta y, and to compute\n

y. Our bounds of integration now and they\n

the minimum y value of negative three to the\n

area should also be written in terms of y.\n

our side length is actually twice our x value,\n

our x value squared in terms of our y value\n

Therefore, our area, which is our side length\n

4x squared, is going to be equal to 16 times\n

to calculate our volume by taking the integral\n

minus y squared over nine d y. If we integrate\n

answer from the answer to our first problem,\n

answer, because we now have a different three\n

and a different volume. In this video, we\n

into slices, then the volume of the three\n

cross sectional area, dx. In this video, I'll\n

a curve given as a function y equals f of\n

of a bunch of straight line segments, it's\n

the distance formula to find the length of\n

the distance between two points x one, y one,\n

of x two minus x one squared plus y two minus\n

first line segment, connecting the points\n

of the square root of two minus one squared\n

square root of two, the length of the next\n

and the next piece has length two, we don't\n

And the last line segment has a length of\n

of these four line segments, we get a total\n

root of five, plus the square root of two\nplus two

Now I plug this into my volume equation. And\n

times nine, one minus x squared over four\n

36. and integrate. Plugging in values and\n

Now, how would this problem be different if\n

to the y axis instead of the x axis? Well,\n

bit different. Since our squares would now\n

squares are now fin in the y direction, instead\n

the width of a slab B delta y, and to compute\n

y. Our bounds of integration now and they\n

the minimum y value of negative three to the\n

area should also be written in terms of y.\n

our side length is actually twice our x value,\n

our x value squared in terms of our y value\n

Therefore, our area, which is our side length\n

4x squared, is going to be equal to 16 times\n

to calculate our volume by taking the integral\n

minus y squared over nine d y. If we integrate\n

answer from the answer to our first problem,\n

answer, because we now have a different three\n

and a different volume. In this video, we\n

into slices, then the volume of the three\n

cross sectional area, dx. In this video, I'll\n

a curve given as a function y equals f of\n

of a bunch of straight line segments, it's\n

the distance formula to find the length of\n

the distance between two points x one, y one,\n

of x two minus x one squared plus y two minus\n

first line segment, connecting the points\n

of the square root of two minus one squared\n

square root of two, the length of the next\n

and the next piece has length two, we don't\n

And the last line segment has a length of\n

of these four line segments, we get a total\n

root of five, plus the square root of two\nplus two

we can use the same process to approximate\n

into n small pieces, and approximate the length\n

line segment. And using the distance formula\n

in this picture, the curve is divided up into\n

points on the x axis, x zero for a x 1x, two,\n

B. And I can label the points on my curve.\n

xx, and then y coordinate will be f of x x,\n

equals f of x. More generally, I have N sub\n

piece of I. The point before it is then a\n

ice segment is given by the distance between\n

square root of x sub i minus x sub i minus\n

sub i minus one squared by the distance formula.\n

by adding the lengths of all these line segments\n

N for the N line segments of these lengths.\n

a Riemann sum because of the Sigma sign, but\n

out here. So I'm going to use a trick, I'm\n

by x sub i minus x sub i minus one divided\n

change the value of my expression, but it\n

my equation, because delta x represents the\n

to x by minus x, y minus one, I'm going to\n

the square root sign, notice I have to square\n

sign. Now I'm going to rewrite my fraction\n

fraction is just one, and the second fraction\n

And the second expression should look familiar\n

of a secant line. At effects, I n x i minus\n

of that secant line is very close to the slope\n

In fact, you might recall that the mean value\n

line is actually exactly equal to the slope\n

it x i star in that interval. So I'll rewrite\n

the curve. And since x i minus x i minus one\n

sum here. So if I want to find the exact arc\n

Riemann sum. This is the limit as the number\n

the limit of a Riemann sum is given by an\n

integral of the square root of one plus f\n

x value of A to the last x value of b. And\n

Sometimes, this formula is also written with\n

of f prime of x. Let's use the arc length\n

equals x to the three halves between x equals\n

we can use the same process to approximate\n

into n small pieces, and approximate the length\n

line segment. And using the distance formula\n

in this picture, the curve is divided up into\n

points on the x axis, x zero for a x 1x, two,\n

B. And I can label the points on my curve.\n

xx, and then y coordinate will be f of x x,\n

equals f of x. More generally, I have N sub\n

piece of I. The point before it is then a\n

ice segment is given by the distance between\n

square root of x sub i minus x sub i minus\n

sub i minus one squared by the distance formula.\n

by adding the lengths of all these line segments\n

N for the N line segments of these lengths.\n

a Riemann sum because of the Sigma sign, but\n

out here. So I'm going to use a trick, I'm\n

by x sub i minus x sub i minus one divided\n

change the value of my expression, but it\n

my equation, because delta x represents the\n

to x by minus x, y minus one, I'm going to\n

the square root sign, notice I have to square\n

sign. Now I'm going to rewrite my fraction\n

fraction is just one, and the second fraction\n

And the second expression should look familiar\n

of a secant line. At effects, I n x i minus\n

of that secant line is very close to the slope\n

In fact, you might recall that the mean value\n

line is actually exactly equal to the slope\n

it x i star in that interval. So I'll rewrite\n

the curve. And since x i minus x i minus one\n

sum here. So if I want to find the exact arc\n

Riemann sum. This is the limit as the number\n

the limit of a Riemann sum is given by an\n

integral of the square root of one plus f\n

x value of A to the last x value of b. And\n

Sometimes, this formula is also written with\n

of f prime of x. Let's use the arc length\n

equals x to the three halves between x equals\n

Here's the general formula for arc length.\n

x to the one half, we get that arc length\n

root of one plus three halves x to the one\n

bit, we can use use substitution to rewrite\n

to 10 of the square root of u times four ninths\n

divided by three halves, times four ninths,\n

after some computation works out to 1/27 times\n

of 13 or approximately 7.6 units, which seems\n

account the fact that this scale here is by\n

the formula for the arc length of a curve.\n

x equals b, then the arc length is given by\n

of one plus f prime of x squared dx. This\n

and the key role of integration in doing work\n

to move an object a distance d, then the work\n

equals force times distance, or in symbols,\n

be given in metric units, or in bold fashion,\n

the units of force are going to be units of\n

acceleration, meters per second squared. This\n

In English units, force is given typically,\n

since work is force times distance, and distance\n

for work of kilograms meters squared per second\n

meters. And these collection of units is also\n

we're using English units for work, work again\n

pounds times feet are usually this is written\n

my weight in pounds, I weigh about 140 pounds.\n

So that 140 pounds, my weight is also telling\n

tell you instead, that my mass is 63.5 kilograms,\n

force. So if I want to know the force, due\n

that 63.5 by the acceleration due to gravity,\n

product works out to be 622.3 kilogram meters\n

have the right units for force, we could also\n

Here's the general formula for arc length.\n

x to the one half, we get that arc length\n

root of one plus three halves x to the one\n

bit, we can use use substitution to rewrite\n

to 10 of the square root of u times four ninths\n

divided by three halves, times four ninths,\n

after some computation works out to 1/27 times\n

of 13 or approximately 7.6 units, which seems\n

account the fact that this scale here is by\n

the formula for the arc length of a curve.\n

x equals b, then the arc length is given by\n

of one plus f prime of x squared dx. This\n

and the key role of integration in doing work\n

to move an object a distance d, then the work\n

equals force times distance, or in symbols,\n

be given in metric units, or in bold fashion,\n

the units of force are going to be units of\n

acceleration, meters per second squared. This\n

In English units, force is given typically,\n

since work is force times distance, and distance\n

for work of kilograms meters squared per second\n

meters. And these collection of units is also\n

we're using English units for work, work again\n

pounds times feet are usually this is written\n

my weight in pounds, I weigh about 140 pounds.\n

So that 140 pounds, my weight is also telling\n

tell you instead, that my mass is 63.5 kilograms,\n

force. So if I want to know the force, due\n

that 63.5 by the acceleration due to gravity,\n

product works out to be 622.3 kilogram meters\n

have the right units for force, we could also\n

Now that we familiarize ourselves with units\n

example, how much work is done to lift a two\n

five feet high? Well, we know that work is\n

a unit of force. And distance is five feet.\n

do the same problem in metric units. The two\n

kilogram book, and we're lifting it off the\n

high. Well, work is still force times distance.\n

times the acceleration due to gravity 9.8\n

of 1.5 meters. That gives us a product of\n

squared, or in other words, 13.23 jewels.\n

so we could just multiply force by distance\n

force is not constant. Let's say a particle\n

a to a point x equals b. According to a force,\n

with x. How much work is done in moving a\n

on the whole interval from a to b, if we divide\n

intervals, each of width delta x, then on\n

to be approximately constant, it's not going\n

interval. As usual, let's pick a sample point\n

little sub interval. exabytes star could be\n

right endpoint or any point in the middle.\n

approximately constant is approximately equal\n

that I sub interval is approximately equal\n

distance that the particle is going on that\n

length of the sub interval delta x. Instead\n

going all the way from A to B, I'm thinking\n

with the approximately constant force at a\n

go the second sub interval, again, the forces\n

of delta x. And then we'll do some more work\n

proximately, constant force times delta x.\n

I'll get another little chunk of work. And\n

adding all those little chunks of work up.\n

sum from i equals one to n, where n is the\n

each sub interval, which is f of x sub i star\n

is approximately the total work, in order\n

take a limit as we use more and more skinnier\n

n goes to infinity of this Riemann sum, the\n

we've got the integral of the force f of x,\n

x value of a and the maximum x value of b.\n

Now that we familiarize ourselves with units\n

example, how much work is done to lift a two\n

five feet high? Well, we know that work is\n

a unit of force. And distance is five feet.\n

do the same problem in metric units. The two\n

kilogram book, and we're lifting it off the\n

high. Well, work is still force times distance.\n

times the acceleration due to gravity 9.8\n

of 1.5 meters. That gives us a product of\n

squared, or in other words, 13.23 jewels.\n

so we could just multiply force by distance\n

force is not constant. Let's say a particle\n

a to a point x equals b. According to a force,\n

with x. How much work is done in moving a\n

on the whole interval from a to b, if we divide\n

intervals, each of width delta x, then on\n

to be approximately constant, it's not going\n

interval. As usual, let's pick a sample point\n

little sub interval. exabytes star could be\n

right endpoint or any point in the middle.\n

approximately constant is approximately equal\n

that I sub interval is approximately equal\n

distance that the particle is going on that\n

length of the sub interval delta x. Instead\n

going all the way from A to B, I'm thinking\n

with the approximately constant force at a\n

go the second sub interval, again, the forces\n

of delta x. And then we'll do some more work\n

proximately, constant force times delta x.\n

I'll get another little chunk of work. And\n

adding all those little chunks of work up.\n

sum from i equals one to n, where n is the\n

each sub interval, which is f of x sub i star\n

is approximately the total work, in order\n

take a limit as we use more and more skinnier\n

n goes to infinity of this Riemann sum, the\n

we've got the integral of the force f of x,\n

x value of a and the maximum x value of b.\n

Let's look at a physical example. How much\n

from the Earth's surface to an altitude of\n

surface. We're given that the gravitational\n

lowercase m divided by r squared, where m\n

mass of the satellite, r is the distance between\n

and g is the gravitational constant. We're\n

the mass of the Earth, and the gravitational\n

problem of lifting the calculus book. When\n

force of gravity was essentially constant\n

the equation work equals force times distance.\n

satellite a larger distance, the force of\n

to use work as the integral of this force\n

in this problem is R. So I'll rewrite this\n

Let's look at a physical example. How much\n

from the Earth's surface to an altitude of\n

surface. We're given that the gravitational\n

lowercase m divided by r squared, where m\n

mass of the satellite, r is the distance between\n

and g is the gravitational constant. We're\n

the mass of the Earth, and the gravitational\n

problem of lifting the calculus book. When\n

force of gravity was essentially constant\n

the equation work equals force times distance.\n

satellite a larger distance, the force of\n

to use work as the integral of this force\n

in this problem is R. So I'll rewrite this\n

I'm starting at the Earth's surface. So that's\n

from the center of the earth, since that's\n

of two times 10 to the six meters above the\n

6.4 plus two, or 8.4 times 10 to the six meters\n

this integral is R. So let me pull out the\n

squared as r to the minus two. Now I can integrate\n

one over minus one Rewrite one more time and\n

of negative GE, capital M lowercase n times\n

Now I still need to plug in for capital G,\n

here. And I have G is 6.67 times 10 to the\n

is six times 10 to the 24, and lowercase and\n

Multiplying all these numbers together, gives\n

10 to the 10th. jewels. To put this number\n

of work done by a car in a year, or by the\n

this video, we saw that for a constant force\n

But for a variable force work is equal to\n

This video introduces the idea of an average\n

a finite list of numbers, we just add the\n

numbers. In summation notation, we write the\n

by n. But defining the average value of a\n

Because a function can take on infinitely\n

could estimate the average value of the function\n

spaced x values. I'll call them x one through\n

a distance of delta x apart, then the average\n

the sum of the values of f divided by n, the\n

the sum from i equals one to n of f of x i\n

value of f, since we're just using n sample\n

as the number of sample points n gets bigger\n

as the limit as n goes to infinity of the\n

more like a Riemann sum. So I need to get\n

the top and the bottom by delta x. And notice\n

the interval b minus a. Now as the number\n

the distance between them goes to zero. So\n

x goes to zero of the sum of FX II times delta\n

Riemann sum in the numerator is just the integral\n

value of the function is given by the integral\n

I'm starting at the Earth's surface. So that's\n

from the center of the earth, since that's\n

of two times 10 to the six meters above the\n

6.4 plus two, or 8.4 times 10 to the six meters\n

this integral is R. So let me pull out the\n

squared as r to the minus two. Now I can integrate\n

one over minus one Rewrite one more time and\n

of negative GE, capital M lowercase n times\n

Now I still need to plug in for capital G,\n

here. And I have G is 6.67 times 10 to the\n

is six times 10 to the 24, and lowercase and\n

Multiplying all these numbers together, gives\n

10 to the 10th. jewels. To put this number\n

of work done by a car in a year, or by the\n

this video, we saw that for a constant force\n

But for a variable force work is equal to\n

This video introduces the idea of an average\n

a finite list of numbers, we just add the\n

numbers. In summation notation, we write the\n

by n. But defining the average value of a\n

Because a function can take on infinitely\n

could estimate the average value of the function\n

spaced x values. I'll call them x one through\n

a distance of delta x apart, then the average\n

the sum of the values of f divided by n, the\n

the sum from i equals one to n of f of x i\n

value of f, since we're just using n sample\n

as the number of sample points n gets bigger\n

as the limit as n goes to infinity of the\n

more like a Riemann sum. So I need to get\n

the top and the bottom by delta x. And notice\n

the interval b minus a. Now as the number\n

the distance between them goes to zero. So\n

x goes to zero of the sum of FX II times delta\n

Riemann sum in the numerator is just the integral\n

value of the function is given by the integral\n

Notice the similarity between the formula\n

formula for the average value of a list of\n

to the summation sign for the list of numbers.\n

the function corresponds to n, the number\n

work an example. For the function g of x equals\n

two to five. We know that the average value\n

five of one over one minus 5x dx divided by\n

use use of the tuition to integrate. So I'm\n

u is negative five dx. In other words, dx\n

of integration, when x is equal to two, u\n

is negative nine. And when x is equal to five,\n

my integral, I get the integral from negative\n

1/5. Do and that's divided by three. Now dividing\n

as I integrate, I'm going to pull the negative\n

over u, that's ln of the absolute value of\n

nine. The absolute value signs are important\n

to take the natural log of negative numbers.\n

of 24 minus ln of nine, I can use my log rules\n

over nine, that's negative 1/15 ln of eight\n

negative 0.0654. So I found the average value\n

achieve that average value, in other words,\n

to five for which GFC equals its average value?\n

equal to GS average value. In other words,\n

1/15 ln of eight thirds, and try to solve\n

equation. But I'm going to take the reciprocal\n

and divide by negative five. This simplifies\n

What which is approximately 3.25. And that\n

two to five. So we've demonstrated that g\n

But in fact, we could have predicted this\n

between GS minimum value and maximum value\n

on the interval from two to five, it has to\n

minimum and maximum, including its average\n

continuous function, the function must achieve\n

is known as the mean value theorem for integrals.\n

on an interval from a to b, there has to be\n

that f of c equals its average value, or,\n

a to b of f of x dx divided by b minus a.\n

value of a function, and stated the mean value\n

for average value a little, then we can see\n

The area of the box with height the average\nvalue

Notice the similarity between the formula\n

formula for the average value of a list of\n

to the summation sign for the list of numbers.\n

the function corresponds to n, the number\n

work an example. For the function g of x equals\n

two to five. We know that the average value\n

five of one over one minus 5x dx divided by\n

use use of the tuition to integrate. So I'm\n

u is negative five dx. In other words, dx\n

of integration, when x is equal to two, u\n

is negative nine. And when x is equal to five,\n

my integral, I get the integral from negative\n

1/5. Do and that's divided by three. Now dividing\n

as I integrate, I'm going to pull the negative\n

over u, that's ln of the absolute value of\n

nine. The absolute value signs are important\n

to take the natural log of negative numbers.\n

of 24 minus ln of nine, I can use my log rules\n

over nine, that's negative 1/15 ln of eight\n

negative 0.0654. So I found the average value\n

achieve that average value, in other words,\n

to five for which GFC equals its average value?\n

equal to GS average value. In other words,\n

1/15 ln of eight thirds, and try to solve\n

equation. But I'm going to take the reciprocal\n

and divide by negative five. This simplifies\n

What which is approximately 3.25. And that\n

two to five. So we've demonstrated that g\n

But in fact, we could have predicted this\n

between GS minimum value and maximum value\n

on the interval from two to five, it has to\n

minimum and maximum, including its average\n

continuous function, the function must achieve\n

is known as the mean value theorem for integrals.\n

on an interval from a to b, there has to be\n

that f of c equals its average value, or,\n

a to b of f of x dx divided by b minus a.\n

value of a function, and stated the mean value\n

for average value a little, then we can see\n

The area of the box with height the average\nvalue

is the same as the area under the curve. This\n

for integrals. the mean value theorem for\n

f of x, defined on an interval from a to b,\n

that f of c is equal to the average value\n

us is the intermediate value theorem. Recall\n

if we have a continuous function f defined\n

If we have some number l in between f of x\n

the value l somewhere between x one and x\n

theorem, let's turn our attention back to\n

it's possible that our function f of x might\n

if that's true, then our mean value theorem\n

just equal to that constant, which is equal\n

assume that f is not constant. Well, like\n

to have a minimum value and a maximum value,\n

we know that F's average value on the interval\n

minimum value. If you don't believe this,\n

the interval have to lie between big M and\n

we get little m times b minus a is less than\n

or equal to big M times b minus a. Notice\n

just integrating a constant. Now if I divide\n

little m is less than or equal to the average\n

as I wanted. Now, I just need to apply the\n

as my number L and little m and big M as my\n

value theorem says that F average is achieved\n

x two. And therefore, for some C in my interval\n

for integrals. Now I'm going to give a second\n

And this time, it's going to be as a corollary\n

Recall that the mean value theorem for functions,\n

interval, and differentiable on the interior\n

c in the interval, such that the derivative\n

change of G, across the whole interval from\n

for functions in mind, and turn our attention\n

I'm going to define a function g of x to be\n

F is the function given to us in the statement\n

that g of A is just the integral from a to\n

from a to b of our function. Now, by the fundamental\n

continuous and differentiable on the interval\n

is the same as the area under the curve. This\n

for integrals. the mean value theorem for\n

f of x, defined on an interval from a to b,\n

that f of c is equal to the average value\n

us is the intermediate value theorem. Recall\n

if we have a continuous function f defined\n

If we have some number l in between f of x\n

the value l somewhere between x one and x\n

theorem, let's turn our attention back to\n

it's possible that our function f of x might\n

if that's true, then our mean value theorem\n

just equal to that constant, which is equal\n

assume that f is not constant. Well, like\n

to have a minimum value and a maximum value,\n

we know that F's average value on the interval\n

minimum value. If you don't believe this,\n

the interval have to lie between big M and\n

we get little m times b minus a is less than\n

or equal to big M times b minus a. Notice\n

just integrating a constant. Now if I divide\n

little m is less than or equal to the average\n

as I wanted. Now, I just need to apply the\n

as my number L and little m and big M as my\n

value theorem says that F average is achieved\n

x two. And therefore, for some C in my interval\n

for integrals. Now I'm going to give a second\n

And this time, it's going to be as a corollary\n

Recall that the mean value theorem for functions,\n

interval, and differentiable on the interior\n

c in the interval, such that the derivative\n

change of G, across the whole interval from\n

for functions in mind, and turn our attention\n

I'm going to define a function g of x to be\n

F is the function given to us in the statement\n

that g of A is just the integral from a to\n

from a to b of our function. Now, by the fundamental\n

continuous and differentiable on the interval\n

And by the mean value theorem for functions,\n

b minus g of a over b minus a, for some number\n

the three facts above, into our equation below,\n

a to b of f of t dt minus zero over b minus\n

wanted to reach. This shows that the mean\n

mean value theorem for functions where our\n

the second proof of the mean value theorem\n

value theorem for integrals in two different\n

of calculus along the way. In this video,\n

inner By parts. integration by parts is based\n

Recall that the product rule says that when\n

functions, that's equal to the derivative\n

plus the first function times the derivative\n

formula, by solving for this last term, we\n

f of x g of x prime minus f prime of x g of\n

equation with respect to x. The integral of\n

So I can break up this right hand side into\n

of f times g is just equal to f times g, by\n

carry the rest of the formula down. And now\n

f times g prime to the integral of f prime\n

something that might be tricky to integrate,\n

to integrate. Although I've written this formula,\n

of integration, it would be just as easy to\n

theorem tells us that the integral of the\n

evaluated from A to B. There's another version\n

to remember. If we let u equal f of x, and\n

of x dx using differential notation, and dv\n

we can rewrite the formula as the integral\n

the integral of v, d U. since V is our G of\n

we can include bounds of integration, if we're\n

be our key formula for this section. and integrating\n

parts. Let's use integration by parts to find\n

for integration by parts, says the integral\n

of the do. So we need to split up our inner\n

going to call you, and a part that we're going\n

is to let u equal x and dv equal e to the\n

u equal to e to the x and dv equal to x dx\n

product, x e to the x and leave dv as just\ndx.

And by the mean value theorem for functions,\n

b minus g of a over b minus a, for some number\n

the three facts above, into our equation below,\n

a to b of f of t dt minus zero over b minus\n

wanted to reach. This shows that the mean\n

mean value theorem for functions where our\n

the second proof of the mean value theorem\n

value theorem for integrals in two different\n

of calculus along the way. In this video,\n

inner By parts. integration by parts is based\n

Recall that the product rule says that when\n

functions, that's equal to the derivative\n

plus the first function times the derivative\n

formula, by solving for this last term, we\n

f of x g of x prime minus f prime of x g of\n

equation with respect to x. The integral of\n

So I can break up this right hand side into\n

of f times g is just equal to f times g, by\n

carry the rest of the formula down. And now\n

f times g prime to the integral of f prime\n

something that might be tricky to integrate,\n

to integrate. Although I've written this formula,\n

of integration, it would be just as easy to\n

theorem tells us that the integral of the\n

evaluated from A to B. There's another version\n

to remember. If we let u equal f of x, and\n

of x dx using differential notation, and dv\n

we can rewrite the formula as the integral\n

the integral of v, d U. since V is our G of\n

we can include bounds of integration, if we're\n

be our key formula for this section. and integrating\n

parts. Let's use integration by parts to find\n

for integration by parts, says the integral\n

of the do. So we need to split up our inner\n

going to call you, and a part that we're going\n

is to let u equal x and dv equal e to the\n

u equal to e to the x and dv equal to x dx\n

product, x e to the x and leave dv as just\ndx.

Whatever choice we make, we need a u times\n

need dx to be part of dv in order to use proper\n

first choice first. If u is equal to x, then\n

x dx, then we can find V by integrating this\n

the x. plugging into our integration by parts\n

x, e to the x dx is equal to u times v x into\n

the x dx. Well, this is looking very promising,\n

dx. It's just e to the x plus a constant of\n

allowed us to compute our integral, let's\n

of what we got the derivative of x, e to the\n

derivative of x, that's one, times e to the\n

x, which is dx, minus the derivative of iliacs,\n

of C, which is just zero. And since this term\n

x e to the x, which is exactly what we started\n

that our work is correct. Notice that in order\n

up having to use the product rule. And that's\n

by parts is really just the product rule used\n

this integral using our first choice of u\n

let's see what would have happened if we used\n

equal to either the X and dv equal to x dx,\n

do to be either the x dx, and we would have\n

x squared over two. plugging this into our\n

get the integral of udv, that's e to the x\n

to the x times x squared over two, minus the\n

times e to the x, dx. Now in order to go any\n

integral of x squared over two times e to\n

that's just a constant, but the integral of\n

than the integral of e to the x times x that\n

direction here. And this turns out to be a\n

that the third choice of u and dv that I suggested\n

instead of simpler. So the first choice turned\n

integration by parts works by replacing an\n

expression, u times v minus the integral of\n

are a lot of trig identities that come in\n

good news is, if you're willing to do a little\n

you really have to memorize. In this video,\n

and show you how to derive a bunch more with\n

going to tell you the three trig identities\n

all three of them are very easy to remember.\n

That's the identity that says sine squared\n

to one. That one's easy to remember, because\n

triangles in the context of the unit circle.

Whatever choice we make, we need a u times\n

need dx to be part of dv in order to use proper\n

first choice first. If u is equal to x, then\n

x dx, then we can find V by integrating this\n

the x. plugging into our integration by parts\n

x, e to the x dx is equal to u times v x into\n

the x dx. Well, this is looking very promising,\n

dx. It's just e to the x plus a constant of\n

allowed us to compute our integral, let's\n

of what we got the derivative of x, e to the\n

derivative of x, that's one, times e to the\n

x, which is dx, minus the derivative of iliacs,\n

of C, which is just zero. And since this term\n

x e to the x, which is exactly what we started\n

that our work is correct. Notice that in order\n

up having to use the product rule. And that's\n

by parts is really just the product rule used\n

this integral using our first choice of u\n

let's see what would have happened if we used\n

equal to either the X and dv equal to x dx,\n

do to be either the x dx, and we would have\n

x squared over two. plugging this into our\n

get the integral of udv, that's e to the x\n

to the x times x squared over two, minus the\n

times e to the x, dx. Now in order to go any\n

integral of x squared over two times e to\n

that's just a constant, but the integral of\n

than the integral of e to the x times x that\n

direction here. And this turns out to be a\n

that the third choice of u and dv that I suggested\n

instead of simpler. So the first choice turned\n

integration by parts works by replacing an\n

expression, u times v minus the integral of\n

are a lot of trig identities that come in\n

good news is, if you're willing to do a little\n

you really have to memorize. In this video,\n

and show you how to derive a bunch more with\n

going to tell you the three trig identities\n

all three of them are very easy to remember.\n

That's the identity that says sine squared\n

to one. That one's easy to remember, because\n

triangles in the context of the unit circle.

If I draw a right triangle with angle theta\n

one, because a unit circle means a circle\n

length cosine of theta, because by definition,\n

point on the unit circle at angle theta. Similarly,\n

because sine of theta means the y coordinate\n

theorem for right triangles, we know that\n

equals one, which is exactly the bit beggary\n

should know are the even and odd identities.\n

identities but they go together, so I'm counting\n

of negative theta is equal to cosine of Beta.\n

And the identity says that sine of negative\n

In other words, sine is an odd function. One\n

at the graphs of sine and cosine. Y equals\n

an even function. And if I look at the value\n

theta, my graph has the same height. So cosine\n

The graph of y equals sine x does not have\n

degree rotational symmetry, which is characteristic\n

values, that theta and negative theta, I see\n

the opposite as the y value at theta, and\n

to negative sine of theta. Another way to\n

by going back to the unit circle. If I look\n

of negative theta, the x coordinate, which\n

angles. So that means cosine of negative theta\n

instead is the opposite. For negative theta,\n

one of them is negative, but they have the\n

theta is the negative of sine of theta. My\n

sum formula. Again, there are really two,\n

go together, so I'll consider them a single\n

because there's a song that goes with them.\n

cosine, cosine, minus sign sign, you may recognize\n

So remember it 123 go sine cosine cosine sine,\n

you have to remember when you sing the song\n

and then the cosine of A plus B. So these\n

everybody should memorize. Next, I'll show\n

Pretty simply, from these three. I've written\n

top for easy reference. First of all, we can\n

the beggary and identity. Let's start by dividing\n

theta. We can break up the fraction on the\n

theta plus cosine squared theta over cosine\n

theta. But sine squared theta over cosine\n

cosine squared theta over cosine squared theta\n

is seacon squared theta, since secant theta\n

If I draw a right triangle with angle theta\n

one, because a unit circle means a circle\n

length cosine of theta, because by definition,\n

point on the unit circle at angle theta. Similarly,\n

because sine of theta means the y coordinate\n

theorem for right triangles, we know that\n

equals one, which is exactly the bit beggary\n

should know are the even and odd identities.\n

identities but they go together, so I'm counting\n

of negative theta is equal to cosine of Beta.\n

And the identity says that sine of negative\n

In other words, sine is an odd function. One\n

at the graphs of sine and cosine. Y equals\n

an even function. And if I look at the value\n

theta, my graph has the same height. So cosine\n

The graph of y equals sine x does not have\n

degree rotational symmetry, which is characteristic\n

values, that theta and negative theta, I see\n

the opposite as the y value at theta, and\n

to negative sine of theta. Another way to\n

by going back to the unit circle. If I look\n

of negative theta, the x coordinate, which\n

angles. So that means cosine of negative theta\n

instead is the opposite. For negative theta,\n

one of them is negative, but they have the\n

theta is the negative of sine of theta. My\n

sum formula. Again, there are really two,\n

go together, so I'll consider them a single\n

because there's a song that goes with them.\n

cosine, cosine, minus sign sign, you may recognize\n

So remember it 123 go sine cosine cosine sine,\n

you have to remember when you sing the song\n

and then the cosine of A plus B. So these\n

everybody should memorize. Next, I'll show\n

Pretty simply, from these three. I've written\n

top for easy reference. First of all, we can\n

the beggary and identity. Let's start by dividing\n

theta. We can break up the fraction on the\n

theta plus cosine squared theta over cosine\n

theta. But sine squared theta over cosine\n

cosine squared theta over cosine squared theta\n

is seacon squared theta, since secant theta\n

So we found a new identity, a variation on\n

and secant. Now, what if we wanted an identity\n

Well, as you might be thinking, we could just\n

time, divide both sides by sine squared of\n

we get one plus cotangent squared theta equals\n

cosine over sine, and cosecant is one of our\n

great identities from the angle some formulas.\n

formula. If we want a formula for sine of\n

B for B In our angle some formula. So we get\n

of A sine of negative B. But we know that\n

cosine is 11. And sine of negative b is negative\n

to a negative sign. And we have an angle difference\n

angle sum formula for cosine, plugging in\n

to get an angle difference formula for cosine.\n

double angle formula. If we want a formula\n

as being the sine of theta plus theta. So\n

B in our angle sum formula. That gives us\n

to two sine theta cosine theta. That's the\n

out so well, let's try the same thing for\n

plus theta. And using the angle sum formula,\n

can rewrite this as cosine squared theta minus\n

of the double angle formula for cosine. But\n

to replace cosine squared theta by one minus\n

replace sine squared theta by one minus cosine\n

we get cosine of two theta is one minus sine\n

other words, cosine of two theta is one minus\n

to the original, and replace sine squared\n

get cosine of two theta is cosine squared\n

theta, which simplifies to cosine of two theta\n

we found one version of a double angle formula\n

using the angle sum formula and the Pythagorean\n

useful for integration when they're rewritten\n

sine squared of theta, I get two sine squared\n

theta. So sine squared of theta is one half\n

the analogous thing with this formula and\n

So we found a new identity, a variation on\n

and secant. Now, what if we wanted an identity\n

Well, as you might be thinking, we could just\n

time, divide both sides by sine squared of\n

we get one plus cotangent squared theta equals\n

cosine over sine, and cosecant is one of our\n

great identities from the angle some formulas.\n

formula. If we want a formula for sine of\n

B for B In our angle some formula. So we get\n

of A sine of negative B. But we know that\n

cosine is 11. And sine of negative b is negative\n

to a negative sign. And we have an angle difference\n

angle sum formula for cosine, plugging in\n

to get an angle difference formula for cosine.\n

double angle formula. If we want a formula\n

as being the sine of theta plus theta. So\n

B in our angle sum formula. That gives us\n

to two sine theta cosine theta. That's the\n

out so well, let's try the same thing for\n

plus theta. And using the angle sum formula,\n

can rewrite this as cosine squared theta minus\n

of the double angle formula for cosine. But\n

to replace cosine squared theta by one minus\n

replace sine squared theta by one minus cosine\n

we get cosine of two theta is one minus sine\n

other words, cosine of two theta is one minus\n

to the original, and replace sine squared\n

get cosine of two theta is cosine squared\n

theta, which simplifies to cosine of two theta\n

we found one version of a double angle formula\n

using the angle sum formula and the Pythagorean\n

useful for integration when they're rewritten\n

sine squared of theta, I get two sine squared\n

theta. So sine squared of theta is one half\n

the analogous thing with this formula and\n

to cosine squared theta is equal to one plus\n

is equal to one half, plus one half cosine\n

we're gonna need. It may seem like a lot,\n

from my three favorites at the top. Well,\n

that are of particular importance for techniques\n

and identity and its various forms. And the\n

favorite trig identities, the Pythagorean\n

angle some formulas. From these I derived\n

that will be particularly useful as we do\n

some formulas, those are the formulas for\n

A plus B. I like to sing them, sine, cosine,\n

This video gives a geometric proof of those\n

angle some formulas, but I'd like to share\n

are interested. Alright, the angle some formulas\n

prove. To prove these formulas, let me start\n

that. Next, I'm going to draw a line perpendicular\n

the top line until it meets that perpendicular,\n

a rectangle around that right triangle that\n

now divided up into four right triangles.\n

so that the high partners of my middle triangle\n

to think about the angles of these other triangles.\n

are parallel lines, and this high partners\n

have the same measure as a plus b down here.\n

same measure as a down here. Because this\n

this angle here. And this angle A is also\n

a triangle minus 90 degrees minus that same\nangle.

to cosine squared theta is equal to one plus\n

is equal to one half, plus one half cosine\n

we're gonna need. It may seem like a lot,\n

from my three favorites at the top. Well,\n

that are of particular importance for techniques\n

and identity and its various forms. And the\n

favorite trig identities, the Pythagorean\n

angle some formulas. From these I derived\n

that will be particularly useful as we do\n

some formulas, those are the formulas for\n

A plus B. I like to sing them, sine, cosine,\n

This video gives a geometric proof of those\n

angle some formulas, but I'd like to share\n

are interested. Alright, the angle some formulas\n

prove. To prove these formulas, let me start\n

that. Next, I'm going to draw a line perpendicular\n

the top line until it meets that perpendicular,\n

a rectangle around that right triangle that\n

now divided up into four right triangles.\n

so that the high partners of my middle triangle\n

to think about the angles of these other triangles.\n

are parallel lines, and this high partners\n

have the same measure as a plus b down here.\n

same measure as a down here. Because this\n

this angle here. And this angle A is also\n

a triangle minus 90 degrees minus that same\nangle.

So I'll label this skinny angle with a. Next,\n

So I'll label this skinny angle with a. Next,\n

Based on the middle right triangle with high\n

down here must be cosine of B, since adjacent\n

side length here must be sine of B. Since\n

we see that sine of B is the hypotenuse of\n

little side here has measure sign a time sign\n

of this angle has to equal sign a. A similar\n

measure cosine A time sign B. Please pause\n

side length of this right triangle. And this\n

A cosine B cosine A cosine B sine of A plus\n

have a rectangle here. So the opposite sides\n

of A plus B has to equal sine of A cosine\n

exactly the first angle sum formula. Also,\n

is exactly the difference of this side length\n

of A sine B. And that's the second angle sum\n

geometric proof of the angle some formulas.\n

that involve at least one odd power of sine\n

of sine to the fourth x times cosine x dx.\n

If we let u equal sine of x Then d u is equal\n

integral as the integral of u to the fourth\n

the fifth plus C. And I can convert things\n

to get 1/5 sine to the fifth of x plus C.\n

the fourth x times cosine cubed x dx can also\n

with a bit more work. First, I'm going to\n

to the fourth x times cosine squared x times\n

of x as cosine squared of x times cosine x\n

as my d u, like I did in the previous problem.\n

going to need you to be sine of x. But unlike\n

sine of the fourth x with you to the fourth\n

I've got this cosine squared x hanging around\n

have that cosine squared x, I wish everything\n

sine. Now, as you might realize, it's not\n

Pythagorean identity, that cosine squared\n

So cosine squared of x can be written as one\n

I can now rewrite my integral entirely in\n

out, this becomes easy to integrate, I get\n

plus C. And I can rewrite this in terms of\nx.

Based on the middle right triangle with high\n

down here must be cosine of B, since adjacent\n

side length here must be sine of B. Since\n

we see that sine of B is the hypotenuse of\n

little side here has measure sign a time sign\n

of this angle has to equal sign a. A similar\n

measure cosine A time sign B. Please pause\n

side length of this right triangle. And this\n

A cosine B cosine A cosine B sine of A plus\n

have a rectangle here. So the opposite sides\n

of A plus B has to equal sine of A cosine\n

exactly the first angle sum formula. Also,\n

is exactly the difference of this side length\n

of A sine B. And that's the second angle sum\n

geometric proof of the angle some formulas.\n

that involve at least one odd power of sine\n

of sine to the fourth x times cosine x dx.\n

If we let u equal sine of x Then d u is equal\n

integral as the integral of u to the fourth\n

the fifth plus C. And I can convert things\n

to get 1/5 sine to the fifth of x plus C.\n

the fourth x times cosine cubed x dx can also\n

with a bit more work. First, I'm going to\n

to the fourth x times cosine squared x times\n

of x as cosine squared of x times cosine x\n

as my d u, like I did in the previous problem.\n

going to need you to be sine of x. But unlike\n

sine of the fourth x with you to the fourth\n

I've got this cosine squared x hanging around\n

have that cosine squared x, I wish everything\n

sine. Now, as you might realize, it's not\n

Pythagorean identity, that cosine squared\n

So cosine squared of x can be written as one\n

I can now rewrite my integral entirely in\n

out, this becomes easy to integrate, I get\n

plus C. And I can rewrite this in terms of\nx.

To recap, we separated out one copy of cosine\n

the rest of the cosines in the signs isn't\n

do use of the tuition with u equal to sine\n

with some modifications works on a lot of\n

tried separating out one copy of cosine of\n

because we just have one copy of cosine x\n

can to convert a single cosine into signs.\n

like this. But then we'd have to introduce\n

into our integrand, which would make things\n

if we can just use the cosine squared x is\n

to replace our cosines with signs. But this\n

of cosines leftover that we want to convert.\n

cosine x, let's say about a copy of sine x\n

the integral of sine to the fourth of x times\n

x and the dx. Now we want the sine x to be\n

of x. That way d u is equal to negative sine\n

d U. since u is equal to cosine x, this time,\n

with cosines. We know that sine squared of\n

x by the Pythagorean identity. So let's rewrite\n

x squared. That allows us to substitute in\n

of x And now we can replace everything with\n

and multiply things out. One minus u squared\n

to the fourth. And this multiplies out to\n

u to the sixth. Now I can integrate, distribute\n

of x for you. That completes this problem.\n

and the Pythagorean identity. To evaluate\n

or cosine. The idea is to separate off one\n

part of our D U. and the remaining even power\n

In this case, we convert sine squared into\n

do the use substitution and evaluate the integral.\n

trig functions involving only even powers\n

that will come in handy here, the first ones\n

The second one's an identity that allows us\n

cosine of 2x. And the third one is in nd that\n

of cosine of 2x. The only difference between\n

cosine squared and one sine squared. But the\n

To recap, we separated out one copy of cosine\n

the rest of the cosines in the signs isn't\n

do use of the tuition with u equal to sine\n

with some modifications works on a lot of\n

tried separating out one copy of cosine of\n

because we just have one copy of cosine x\n

can to convert a single cosine into signs.\n

like this. But then we'd have to introduce\n

into our integrand, which would make things\n

if we can just use the cosine squared x is\n

to replace our cosines with signs. But this\n

of cosines leftover that we want to convert.\n

cosine x, let's say about a copy of sine x\n

the integral of sine to the fourth of x times\n

x and the dx. Now we want the sine x to be\n

of x. That way d u is equal to negative sine\n

d U. since u is equal to cosine x, this time,\n

with cosines. We know that sine squared of\n

x by the Pythagorean identity. So let's rewrite\n

x squared. That allows us to substitute in\n

of x And now we can replace everything with\n

and multiply things out. One minus u squared\n

to the fourth. And this multiplies out to\n

u to the sixth. Now I can integrate, distribute\n

of x for you. That completes this problem.\n

and the Pythagorean identity. To evaluate\n

or cosine. The idea is to separate off one\n

part of our D U. and the remaining even power\n

In this case, we convert sine squared into\n

do the use substitution and evaluate the integral.\n

trig functions involving only even powers\n

that will come in handy here, the first ones\n

The second one's an identity that allows us\n

cosine of 2x. And the third one is in nd that\n

of cosine of 2x. The only difference between\n

cosine squared and one sine squared. But the\n

Let's start with the simplest possible even\n

According to my calculator, this integral\n

x over two. And if I look it up in the back\n

x plus one for a sine of 2x. So what's going\n

same? Well, yes, because sine of 2x is equal\n

that, here, I get this second answer is equivalent\n

x, which is equivalent to the first answer.\n

the back of the book omit the plus c the cost\n

them for indefinite integrals. Now let's see\n

the integral of cosine squared x dx by hand\n

computer integral table, the hint is that\n

2x over two, or sometimes I like to write\n

So I'll use that identity to rewrite my integral.\n

about integration. The integral of one half\n

of 2x is one half times sine of 2x. It's possible\n

u equal to x. Or as a shortcut, we can observe\n

of 2x times two, the times two comes from\n

inside. So to get rid of that, or counteract\n

half. In any case, our final answer is then\n

is the same answer that we can find using\n

Please pause the video for a minute and try\n

dx. Using a similar technique. You'll want\n

to one half minus one Half cosine of 2x. If\n

like before, and get an answer of one half\n

can be used to evaluate a lot of other integrals\n

often with a lot more effort. Let's try evaluating\n

know that sine squared of x is equal to one\n

six is an even power, we know we can rewrite\n

raised to a power, in this case, the power\n

I think I'm going to factor out the one half.\n

eight out of the integral side and multiply\n

pieces. And I'll try to handle each piece\n

easy, that's x, and the integral of cosine\n

of 2x, like before. Now to integrate cosine\n

used to integrate cosine squared of x, cosine\n

one half cosine of two times 2x. So that's\n4x.

Let's start with the simplest possible even\n

According to my calculator, this integral\n

x \nover two. And if I look it up in the back

of my book, I get the integral to be one half\n

on here? Are these two answers really the\n

to two times sine x cosine x. So if I replace\n

to one half x plus 1/4 times two sine x cosine\n

Notice that the calculator on the table in\n

of integration, but you should always include\n

where these answers come from. So let's compute\n

that is, without the aid of a calculator or\n

cosine squared x is equal to one plus cosine\n

this as one half plus one half cosine of 2x.\n

Now I can split up the integral using rules\n

is just one half x, and the integral of cosine\n

to get that answer by use substitution with\n

that the derivative of sine of 2x is cosine\n

the chain rule taking the derivative of the\n

that, times two, we have to multiply by one\n

one half x plus 1/4 sine of 2x plus C. This\n

the integral table in the back of the book.\n

to compute the integral of sine squared x\n

to use the identity sine squared of x is equal\n

you rewrite the integral, you can integrate\n

x minus 1/4 sine of 2x plus C, the same technique\n

involving even powers of sine and cosine,\n

the integral of sine to the sixth x dx. We\n

half minus one half cosine of 2x. And since\n

sine to the sixth as sine squared to the x\n

of three. Now let's substitute in our identity.\n

One half cubed is 1/8. So I'll pull the one\n

out. Now let me divide up my integral into\n

separately. The integral of one dx, that's\n

of 2x, that's going to be one half times sine\n

squared of 2x, we can use the same trick we\n

squared of 2x is going to be one half, plus\n

And finally, to integrate cosine cubed of\n

we can use our odd power trick and save aside\n

cosine squared into one minus sine squared.\n

minus sine squared of 2x. Because we're replacing\n

part down. Now integrate one half to get one\n

of forex. To get one half times 1/4 sine of\n

four that we get by taking the derivative\n

can finish off the integration by using a\n

And so d u is two cosine of 2x. So we can\n

integral of one minus us squared times, one\n

and integrate that last part, I'll pull out\n

u cubed, I need a plus C. Now. I'll plug back\n

I get a final answer of five sixteenths x\n

sine of 4x plus 148, sine cubed of 2x. Oh,\n

answer. And it was a complicated computation.\n

as before, using identities like this one,\n

and using the Pythagorean identity to handle\n

two identities to rewrite even powers of cosine\n

This trick can be used to compute the integrals\n

provided that you have a lot of time and patience\n

trig integrals, namely, the integral of tangent\n

x. these integrals are special only in the\n

to integrate them. When I come across the\n

wishing that I could integrate secant squared\n

x is easy. It's just tangent x plus c since\n

But happily, I know how to rewrite tangent\n

squared is secant squared minus one. So I'll\n

of secant squared minus one. And that integrates\n

same sort of trick works to integrate cotangent\n

relating cotangent and cosecant every now\n

x. There's several possible tricks that can\n

secant x by secant x plus tangent x in the\n

we get seacon squared x plus seek index tangent\n

x in the denominator. Since secant squared\n

is the root of a secant. We can say u equal\n

right where we want it in the integrand.

And finally, to integrate cosine cubed of\n

we can use our odd power trick and save aside\n

cosine squared into one minus sine squared.\n

minus sine squared of 2x. Because we're replacing\n

part down. Now integrate one half to get one\n

of forex. To get one half times 1/4 sine of\n

four that we get by taking the derivative\n

can finish off the integration by using a\n

And so d u is two cosine of 2x. So we can\n

integral of one minus us squared times, one\n

and integrate that last part, I'll pull out\n

u cubed, I need a plus C. Now. I'll plug back\n

I get a final answer of five sixteenths x\n

sine of 4x plus 148, sine cubed of 2x. Oh,\n

answer. And it was a complicated computation.\n

as before, using identities like this one,\n

and using the Pythagorean identity to handle\n

two identities to rewrite even powers of cosine\n

This trick can be used to compute the integrals\n

provided that you have a lot of time and patience\n

trig integrals, namely, the integral of tangent\n

x. these integrals are special only in the\n

to integrate them. When I come across the\n

wishing that I could integrate secant squared\n

x is easy. It's just tangent x plus c since\n

But happily, I know how to rewrite tangent\n

squared is secant squared minus one. So I'll\n

of secant squared minus one. And that integrates\n

same sort of trick works to integrate cotangent\n

relating cotangent and cosecant every now\n

x. There's several possible tricks that can\n

secant x by secant x plus tangent x in the\n

we get seacon squared x plus seek index tangent\n

x in the denominator. Since secant squared\n

is the root of a secant. We can say u equal\n

right where we want it in the integrand.

So now we just have to integrate one over\n

plugging back in for you, we get that our\n

to the natural log of secant x plus tangent\n

used to evaluate the integral of cosecant\n

squared, and the integral of secant. This\n

to evaluate integrals involving square root\n

identities that are especially useful for\n

identity. And the second is the related identity\n

also a third related identity that involves\n

used in the method of tree substitution, although\n

As our first example, let's look at the integral\n

x squared. According to Wolfram Alpha, this\n

a square root expression kind of an as expected\n

seems to come out of the blue here. Let's\n

trig substitution. The inverse sine function\n

want to substitute in something related to\n

seven sine theta. If x is seven sine theta,\n

d theta. Now I'll substitute in for x and\n

sine theta squared over the square root of\n

cosine theta d theta. let me simplify a little.\n

sine squared theta cosine theta. And the nominator\n

factor out the 49 here. And since the square\n

of the square root sign. Now here is where\n

one minus sine squared theta is equal to cosine\n

and the square root of cosine squared theta\n

it's equal to the absolute value of cosine\n

if cosine theta is positive. Or in other words,\n

pi over two, for example, I would really like\n

squared theta by just cosine theta, not the\n

going to assume that theta is between negative\n

substitution. This might seem like cheating,\n

about the unit circle, as theta ranges from\n

cosine is always positive like we want it\n

negative one to one. And so we can actually\n

seven and seven this way, which are the only\n

Now if we go back to our integral, we can\n

sine squared theta with a simple cosine theta,\n

tricky square root sign, which is the whole\n

simplify, we just have to integrate seven\n

is a familiar problem that we can solve using\n

So now we just have to integrate one over\n

plugging back in for you, we get that our\n

to the natural log of secant x plus tangent\n

used to evaluate the integral of cosecant\n

squared, and the integral of secant. This\n

to evaluate integrals involving square root\n

identities that are especially useful for\n

identity. And the second is the related identity\n

also a third related identity that involves\n

used in the method of tree substitution, although\n

As our first example, let's look at the integral\n

x squared. According to Wolfram Alpha, this\n

a square root expression kind of an as expected\n

seems to come out of the blue here. Let's\n

trig substitution. The inverse sine function\n

want to substitute in something related to\n

seven sine theta. If x is seven sine theta,\n

d theta. Now I'll substitute in for x and\n

sine theta squared over the square root of\n

cosine theta d theta. let me simplify a little.\n

sine squared theta cosine theta. And the nominator\n

factor out the 49 here. And since the square\n

of the square root sign. Now here is where\n

one minus sine squared theta is equal to cosine\n

and the square root of cosine squared theta\n

it's equal to the absolute value of cosine\n

if cosine theta is positive. Or in other words,\n

pi over two, for example, I would really like\n

squared theta by just cosine theta, not the\n

going to assume that theta is between negative\n

substitution. This might seem like cheating,\n

about the unit circle, as theta ranges from\n

cosine is always positive like we want it\n

negative one to one. And so we can actually\n

seven and seven this way, which are the only\n

Now if we go back to our integral, we can\n

sine squared theta with a simple cosine theta,\n

tricky square root sign, which is the whole\n

simplify, we just have to integrate seven\n

is a familiar problem that we can solve using\n

Now we've got an integral we can compute.\n

and the integral of cosine of two theta is\n

my constant of integration. And I'm almost\n

answer is in terms of theta, and my original\n

x and theta are related, according to this\n

get sine theta is equal to x over seven. So\n

seven. So I could just plug in for theta here\n

half sine inverse x over seven minus one half\n

inverse x over seven plus C. This is a correct\n

of twice sine inverse x over seven, there's\n

we can't just pull the two out and cancel\n

does work that way. But instead, we can use\n

sine of two theta as twice sine theta cosine\n

And I'll rewrite the above line as 49 times\n

Now we've got an integral we can compute.\n

and the integral of cosine of two theta is\n

my constant of integration. And I'm almost\n

answer is in terms of theta, and my original\n

x and theta are related, according to this\n

get sine theta is equal to x over seven. So\n

seven. So I could just plug in for theta here\n

half sine inverse x over seven minus one half\n

inverse x over seven plus C. This is a correct\n

of twice sine inverse x over seven, there's\n

we can't just pull the two out and cancel\n

does work that way. But instead, we can use\n

sine of two theta as twice sine theta cosine\n

And I'll rewrite the above line as 49 times\n

Now there's one more trick we need to use.\n

I'm going to label the sides of that triangle\n

made, or this equivalent forms a little easier\n

if I label one of my angles as theta, then\n

of x, and I have partners should have a measure\n

Theorem, that means that this bottom side\n

of 49 minus x squared. Notice that that's\n

original integrand. Once we have the triangle\n

for sine of theta and cosine of theta in terms\n

hypothesis. So that's x over seven. Well,\n

is adjacent over hypotenuse. So that's the\n

We can use these two equations to substitute\n

get rid of the, this naked theta here, we'll\n

sine of x over seven. Those substitutions\n

simplification leads us to the same answer\n

problem. But the key step was to use this\n

us to use a trig identity in order to rewrite\n

root. After that, it was a fairly routine\n

where we had to substitute back in for theta,\n

us figure out what to do. In this video, we\n

with a square root and that This video introduces\n

integrate many rational functions. Recall\n

the form a polynomial divided by another polynomial.\n

divided by x squared plus 2x minus three.\n

to this expression involving the log of x\n

what do x minus one and x plus three have\n

I challenge you to pause the video for a moment\n

noticed that the denominator x squared plus\n

x plus three. Let's look at what happens if\n

x minus one and one with denominator x plus\n

of these two fractions is x minus one times\n

together using standard algebra, I should\n

minus one times x plus three. Well, my original\n

So it seems plausible, that I might be able\n

partial fractions add up to this original\n

and b might make this work. To solve for A\n

by multiplying both sides of my equation by\n

plus three. When I distribute on the left\n

one times B. And on the right side, my x minus\n

3x plus two. I'll distribute the left side\n

terms that involve x and the terms that don't\n

true for all values of x, then I need the\n

Now there's one more trick we need to use.\n

I'm going to label the sides of that triangle\n

made, or this equivalent forms a little easier\n

if I label one of my angles as theta, then\n

of x, and I have partners should have a measure\n

Theorem, that means that this bottom side\n

of 49 minus x squared. Notice that that's\n

original integrand. Once we have the triangle\n

for sine of theta and cosine of theta in terms\n

hypothesis. So that's x over seven. Well,\n

is adjacent over hypotenuse. So that's the\n

We can use these two equations to substitute\n

get rid of the, this naked theta here, we'll\n

sine of x over seven. Those substitutions\n

simplification leads us to the same answer\n

problem. But the key step was to use this\n

us to use a trig identity in order to rewrite\n

root. After that, it was a fairly routine\n

where we had to substitute back in for theta,\n

us figure out what to do. In this video, we\n

with a square root and that This video introduces\n

integrate many rational functions. Recall\n

the form a polynomial divided by another polynomial.\n

divided by x squared plus 2x minus three.\n

to this expression involving the log of x\n

what do x minus one and x plus three have\n

I challenge you to pause the video for a moment\n

noticed that the denominator x squared plus\n

x plus three. Let's look at what happens if\n

x minus one and one with denominator x plus\n

of these two fractions is x minus one times\n

together using standard algebra, I should\n

minus one times x plus three. Well, my original\n

So it seems plausible, that I might be able\n

partial fractions add up to this original\n

and b might make this work. To solve for A\n

by multiplying both sides of my equation by\n

plus three. When I distribute on the left\n

one times B. And on the right side, my x minus\n

3x plus two. I'll distribute the left side\n

terms that involve x and the terms that don't\n

true for all values of x, then I need the\n

So I need a plus b to equal three. Similarly,\n

the left and the right. So I need three A\n

equations in the two unknowns, a and b. So\n

equations for A and B. For example, if I add\n

b cancel out to give me four A equals five,\n

substitute in this five fourths into either\n

was able to find an a value of a and b that\n

partial fractions. Let me fill those values\n

of my original expression, I can calculate\n

I split up my integrals here. Now the integral\n

absolute value of x minus one, you can check\n

do a simple use of the tuition where you as\n

the integral of one over x plus three is natural\n

completes the computation of the integral\n

that after factoring our denominator, the\n

a and b that made this equation hold. And\n

time. But can we always find numbers a and\n

had been something different? What if our\n

say 7x minus 15? Could we have still found\n

we still would have gotten two linear equations\n

been able to solve them, we'd get different\n

been able to find a solution. Even if our\n

we still couldn't use the same method, we\n

And we could have used the same equations\n

zero, and three, A minus B would have to equal\n

unknowns to solve for. In fact, if you think\n

yourself that this method will always work\n

factors into distinct linear factors. By a\n

can be written like a number times x plus\n

in it. And by distinct, I just mean that these\n

The second condition is that the degree of\n

the denominator. The second condition guarantee\n

here, A and B, as we do coefficients here,\n

that we'll be able to solve for our unknowns.\n

the linear factors guarantees that we won't\n

two conditions, then we can proceed to integrate,\n

method even works if we have more than two\n

have three or four, any number of distinct\n

more complicated, because we'll have more\n

those constants. There are also several related\n

functions, even if the denominator factors\n

distinct. Or even if we can't get all the\n

factors have squares in them, or even if the\n

won't get into the details of those related\n

about them in the book or wait for class or\n

In this video, we integrated a rational function\n

So I need a plus b to equal three. Similarly,\n

the left and the right. So I need three A\n

equations in the two unknowns, a and b. So\n

equations for A and B. For example, if I add\n

b cancel out to give me four A equals five,\n

substitute in this five fourths into either\n

was able to find an a value of a and b that\n

partial fractions. Let me fill those values\n

of my original expression, I can calculate\n

I split up my integrals here. Now the integral\n

absolute value of x minus one, you can check\n

do a simple use of the tuition where you as\n

the integral of one over x plus three is natural\n

completes the computation of the integral\n

that after factoring our denominator, the\n

a and b that made this equation hold. And\n

time. But can we always find numbers a and\n

had been something different? What if our\n

say 7x minus 15? Could we have still found\n

we still would have gotten two linear equations\n

been able to solve them, we'd get different\n

been able to find a solution. Even if our\n

we still couldn't use the same method, we\n

And we could have used the same equations\n

zero, and three, A minus B would have to equal\n

unknowns to solve for. In fact, if you think\n

yourself that this method will always work\n

factors into distinct linear factors. By a\n

can be written like a number times x plus\n

in it. And by distinct, I just mean that these\n

The second condition is that the degree of\n

the denominator. The second condition guarantee\n

here, A and B, as we do coefficients here,\n

that we'll be able to solve for our unknowns.\n

the linear factors guarantees that we won't\n

two conditions, then we can proceed to integrate,\n

method even works if we have more than two\n

have three or four, any number of distinct\n

more complicated, because we'll have more\n

those constants. There are also several related\n

functions, even if the denominator factors\n

distinct. Or even if we can't get all the\n

factors have squares in them, or even if the\n

won't get into the details of those related\n

about them in the book or wait for class or\n

In this video, we integrated a rational function\n

This video is about improper integrals, especially\n

Two examples of improper integrals are the\n

x squared dx, and the integral from zero to\n

integrals improper? Well, in the first example,\n

And then the second example, is the fact that\n

on the interval from zero to pi over two where\n

improper. If either of these two situations\n

if we're integrating over an infinite interval.\n

infinity somewhere in the bounds of integration.\n

integral. A type two improper integral occurs\n

has an infinite discontinuity on the interval.\n

is going to infinity or negative infinity,\n

This vertical asymptote could occur in the\n

over. Or, as in this example, it could occur\n

Now, it's possible that both of these situations\n

also an improper integral. This video will\n

one improper integral, asks us to integrate\n

take the integral over larger and larger finite\n

to find the integral from one to infinity\n

integral from one to some finite number T,\n

In symbols, we can write the limit as t goes\n

of one over x squared dx. Since one over x\n

two, we can integrate it to get negative x\n

T, and then take that limit or rewrite this\n

bounds of integration. as t goes to infinity,\n

comes from this expression, which evaluates\n

representing area, this is a little surprising.\n

long region, the area still evaluates to a\n

say that the improper integral converges.\n

some finite number a \nto infinity of f of x dx is defined as the

limit as t goes to infinity of the integral\n

integral converges if this limit exists as\n

diverges if the limit is infinity, or negative\n

we evaluate the integral from negative infinity\n

bigger intervals that extend off to negative\n

as the limit as the left endpoint t goes to\n

b of f of x dx. We say that this integral\n

number, and diverges otherwise. So to evaluate\n

one of one over x dx, we take the limit as\n

This video is about improper integrals, especially\n

Two examples of improper integrals are the\n

x squared dx, and the integral from zero to\n

integrals improper? Well, in the first example,\n

And then the second example, is the fact that\n

on the interval from zero to pi over two where\n

improper. If either of these two situations\n

if we're integrating over an infinite interval.\n

infinity somewhere in the bounds of integration.\n

integral. A type two improper integral occurs\n

has an infinite discontinuity on the interval.\n

is going to infinity or negative infinity,\n

This vertical asymptote could occur in the\n

over. Or, as in this example, it could occur\n

Now, it's possible that both of these situations\n

also an improper integral. This video will\n

one improper integral, asks us to integrate\n

take the integral over larger and larger finite\n

to find the integral from one to infinity\n

integral from one to some finite number T,\n

In symbols, we can write the limit as t goes\n

of one over x squared dx. Since one over x\n

two, we can integrate it to get negative x\n

T, and then take that limit or rewrite this\n

bounds of integration. as t goes to infinity,\n

comes from this expression, which evaluates\n

representing area, this is a little surprising.\n

long region, the area still evaluates to a\n

say that the improper integral converges.\n

some finite number a to infinity of f of x\n

of the integral from a to T of f of x dx.\n

limit exists as a finite number. And we say\n

infinity, or negative infinity, or if it doesn't\n

from negative infinity to some finite number\n

extend off to negative infinity. That is,\n

left endpoint t goes to negative infinity\n

We say that this integral converges if the\n

otherwise. So to evaluate the integral from\n

x dx, we take the limit as t goes to negative\ninfinity

of the integral from t to negative one of\n

x is ln of the absolute value of x, which\n

one and take a limit. We evaluate here we\n

one, that's ln of one which is zero. A graph\n

this expression. as t goes to negative infinity,\n

And so ln is also going to infinity. Therefore,\n

so the integral diverges. In this video, we\n

that we're integrating over is infinite by\n

finite intervals and taking a limit. This\n

These are integrals for which the interval\n

the function that we're integrating goes to\n

type to improper integral like this one, or\n

larger and larger sub intervals on which the\n

So in this first picture, where the function\n

the left, we can call that moving endpoint\n

as that right endpoint t approaches B from\n

of x dx. The same definition works if f of\n

infinity. In the second picture here, we again\n

intervals on which the function is finite.\n

infinity or negative infinity, as x goes to\n

from a to b of f of x dx is going to be the\n

integral from t to b of f of x dx. As an example,\n

x over the square root of x squared minus\n

x equals two. That area can be described as\n

But this is an improper integral, because\n

to one from the right. So we'll evaluate it\n

the right of the integral from t to two of\n

u substitution, where u is x squared minus\n

in my integrand, here, I'll solve for x dx,\n

I'm also going to change my bounds of integration,\n

to t squared minus one. And when x is equal\n

three. So I'll rewrite my integral x dx is\n

root of x squared minus one becomes a square\n

by pulling the one half out of the integral\n

denominator as u to the negative one half.\n

the one half evaluated between three and t\n

and I'll plug in my bounds of integration\n

t squared is also going to one, so t squared\n

of the integral from t to negative one of\n

x is ln of the absolute value of x, which\n

one and take a limit. We evaluate here we\n

one, that's ln of one which is zero. A graph\n

this expression. as t goes to negative infinity,\n

And so ln is also going to infinity. Therefore,\n

so the integral diverges. In this video, we\n

that we're integrating over is infinite by\n

finite intervals and taking a limit. This\n

These are integrals for which the interval\n

the function that we're integrating goes to\n

type to improper integral like this one, or\n

larger and larger sub intervals on which the\n

So in this first picture, where the function\n

the left, we can call that moving endpoint\n

as that right endpoint t approaches B from\n

of x dx. The same definition works if f of\n

infinity. In the second picture here, we again\n

intervals on which the function is finite.\n

infinity or negative infinity, as x goes to\n

from a to b of f of x dx is going to be the\n

integral from t to b of f of x dx. As an example,\n

x over the square root of x squared minus\n

x equals two. That area can be described as\n

But this is an improper integral, because\n

to one from the right. So we'll evaluate it\n

the right of the integral from t to two of\n

u substitution, where u is x squared minus\n

in my integrand, here, I'll solve for x dx,\n

I'm also going to change my bounds of integration,\n

to t squared minus one. And when x is equal\n

three. So I'll rewrite my integral x dx is\n

root of x squared minus one becomes a square\n

by pulling the one half out of the integral\n

denominator as u to the negative one half.\n

the one half evaluated between three and t\n

and I'll plug in my bounds of integration\n

t squared is also going to one, so t squared\n

Therefore, my limit is just three to the one\n

the area underneath my curve. This video was\n

to compute them as the limit of integrals\n

the function is finite. Sometimes we don't\n

integral is, we just want to know whether\n

or diverges. In this situation, the comparison\n

theorem can allow us to determine if an integral\n

to evaluate the integral instead by comparing\n

converges or diverges. So suppose that g of\n

They're both greater than zero for all x's\n

that g of x is less than f of x on that interval\n

infinity. So in the picture, we'll call this\n

f of x. And let's consider the interval from\n

f of x, and both of them are bigger than zero.\n

of f of x on this interval converges to a\n

than f of x also has to converge to a finite\n

then g of x converges. If we turn this around,\n

doesn't converge to a finite number, then\n

where the integral of the bigger function\n

function diverges, then we can make some conclusions\n

you have to be a little careful about this.\n

function diverges, then we really can't make\n

of the smaller function, it could also diverged\n

similarly, if the integral of the smaller\n

know anything about the integral of the bigger\n

Let's look at an example. Suppose we want\n

of two plus sine x over square root of x dx\n

evaluate it, which could get tricky, because\n

here, I'm going to just try to compare it\n

The first thing that I notice is that sine\n

negative one. And that means that the numerator\n

three and one. Therefore, the function two\n

to be between three over square root of x\n

general idea of the picture. Now the comparison\n

than a function whose interval converges,\n

will converge. And if our function is greater\n

the integral of our function will diverged.\n

we want to use depends on what happens to\n

Now we know that the integral of one over\n

has to die verge. That's because this is a\n

which is less than one, the integral from\n

of x dx also diverges since it's just three\n

compare our function to a function whose integral\n

divergent integral in order to get any useful\n

whose integral diverges doesn't tell us anything.\n

Therefore, my limit is just three to the one\n

the area underneath my curve. This video was\n

to compute them as the limit of integrals\n

the function is finite. Sometimes we don't\n

integral is, we just want to know whether\n

or diverges. In this situation, the comparison\n

theorem can allow us to determine if an integral\n

to evaluate the integral instead by comparing\n

converges or diverges. So suppose that g of\n

They're both greater than zero for all x's\n

that g of x is less than f of x on that interval\n

infinity. So in the picture, we'll call this\n

f of x. And let's consider the interval from\n

f of x, and both of them are bigger than zero.\n

of f of x on this interval converges to a\n

than f of x also has to converge to a finite\n

then g of x converges. If we turn this around,\n

doesn't converge to a finite number, then\n

where the integral of the bigger function\n

function diverges, then we can make some conclusions\n

you have to be a little careful about this.\n

function diverges, then we really can't make\n

of the smaller function, it could also diverged\n

similarly, if the integral of the smaller\n

know anything about the integral of the bigger\n

Let's look at an example. Suppose we want\n

of two plus sine x over square root of x dx\n

evaluate it, which could get tricky, because\n

here, I'm going to just try to compare it\n

The first thing that I notice is that sine\n

negative one. And that means that the numerator\n

three and one. Therefore, the function two\n

to be between three over square root of x\n

general idea of the picture. Now the comparison\n

than a function whose interval converges,\n

will converge. And if our function is greater\n

the integral of our function will diverged.\n

we want to use depends on what happens to\n

Now we know that the integral of one over\n

has to die verge. That's because this is a\n

which is less than one, the integral from\n

of x dx also diverges since it's just three\n

compare our function to a function whose integral\n

divergent integral in order to get any useful\n

whose integral diverges doesn't tell us anything.\n

And now we can say that, since the integral\n

root of x dx diverges by the P test, the integral\n

test. In this video, we saw that if we have\n

always less than or equal to the other function\n

integral diverges, the bigger functions integral\n

integral converges, the smaller functions\n

comparison there. In this video, I'll give\n

A sequence is a list of numbers in a particular\n

sequence that gives the digits of pi. A sequence\n

in place of numbers, a sub one, a sub two,\n

by writing a sub n with these curly brackets.\n

one all the way up towards infinity. Sometimes\n

curly brackets. Here, it's implied that n\n

these sequences, given by formulas, let's\n

n equals one, and we get a sub one is three\n

That is, for over three factorial. Recall\n

three and then multiply with consecutive numbers\n

four, six, or two thirds. To find the next\n

seven over four factorial, which is 720 fourths.\n

which is 10 over 120, or 112. So the first\n

720, fourths, and 112. For the second example,\n

I'll call the first term, a sub two, and just\n

negative one squared is positive one. Plugging\n

to negative two nights. For as of four, we\n

one to the fourth is positive. And in fact,\n

to alternate between negative numbers and\n

one to the K in the definition. Sometimes,\n

in terms of previous terms. This is called\n

few terms of this recursive sequence, we're\n

sub two, we just use the recursive formula\n

one is two, that's four minus a half, or seven\n

recursive formula again, four minus one over\n

possible, describe a sequence with either\n

recursive formula. For example, if I look\n

recursively by saying a sub one is two, and\n

plus two. Or I can describe as a closed form\n

form two times n, where n starts at one.

And now we can say that, since the integral\n

root of x dx diverges by the P test, the integral\n

test. In this video, we saw that if we have\n

always less than or equal to the other function\n

integral diverges, the bigger functions integral\n

integral converges, the smaller functions\n

comparison there. In this video, I'll give\n

A sequence is a list of numbers in a particular\n

sequence that gives the digits of pi. A sequence\n

in place of numbers, a sub one, a sub two,\n

by writing a sub n with these curly brackets.\n

one all the way up towards infinity. Sometimes\n

curly brackets. Here, it's implied that n\n

these sequences, given by formulas, let's\n

n equals one, and we get a sub one is three\n

That is, for over three factorial. Recall\n

three and then multiply with consecutive numbers\n

four, six, or two thirds. To find the next\n

seven over four factorial, which is 720 fourths.\n

which is 10 over 120, or 112. So the first\n

720, fourths, and 112. For the second example,\n

I'll call the first term, a sub two, and just\n

negative one squared is positive one. Plugging\n

to negative two nights. For as of four, we\n

one to the fourth is positive. And in fact,\n

to alternate between negative numbers and\n

one to the K in the definition. Sometimes,\n

in terms of previous terms. This is called\n

few terms of this recursive sequence, we're\n

sub two, we just use the recursive formula\n

one is two, that's four minus a half, or seven\n

recursive formula again, four minus one over\n

possible, describe a sequence with either\n

recursive formula. For example, if I look\n

recursively by saying a sub one is two, and\n

plus two. Or I can describe as a closed form\n

form two times n, where n starts at one.

Now let's practice writing out a formula for\n

this first sequence, notice that each term\n

is like a linear function with slope three,\n

go up by three. And so I can write a sub n\n

like my y intercept and a linear equation,\n

one that corresponds to an N value of one,\n

general formula is three times n plus four,\n

plugging in a few values of n, like we did\n

it works. Notice that it would also be possible\n

If we're willing to start with n equals zero\n

then our first term functions like our y intercept.\n

a sequence for which consecutive terms have\n

if A is the first term, and d is the common\n

the form d times k plus a, if our index is\n

Or if we'd rather start with an index of one,\n

plus a. Notice these two expressions are exactly\n

one, in particular, the starting value of\n

I get the equivalent starting value of K to\n

sequences. In this first sequence, notice\n

ratio of 1/10. In other words, each time n\n

by 1/10. This is the same property that exponential\n

a sub n in the form of an exponential function\n

the right initial value, so that when n is\n

correct initial value is 30. As usual, we\n

values of n, n equals 123. And making sure\n

we prefer to start with our index at zero,\n

the nth power. Since a value of zero for n\n

a value of one for n in this formula gives\n

have a common ratio, if I divide the second\n

halves. And that's the same ratio as I get\n

term. So if I use an index starting with zero,\n

is the first term times five halves to the\n

at one, one way to do this is to let K equal\n

When n is 0k, will be one. And since K is\n

n with k minus one. This gives the following\n

example has a common ratio of negative two\n

term of three times that ratio, negative two\n

at zero. Or, as above, we can write it as\n

Now let's practice writing out a formula for\n

this first sequence, notice that each term\n

is like a linear function with slope three,\n

go up by three. And so I can write a sub n\n

like my y intercept and a linear equation,\n

one that corresponds to an N value of one,\n

general formula is three times n plus four,\n

plugging in a few values of n, like we did\n

it works. Notice that it would also be possible\n

If we're willing to start with n equals zero\n

then our first term functions like our y intercept.\n

a sequence for which consecutive terms have\n

if A is the first term, and d is the common\n

the form d times k plus a, if our index is\n

Or if we'd rather start with an index of one,\n

plus a. Notice these two expressions are exactly\n

one, in particular, the starting value of\n

I get the equivalent starting value of K to\n

sequences. In this first sequence, notice\n

ratio of 1/10. In other words, each time n\n

by 1/10. This is the same property that exponential\n

a sub n in the form of an exponential function\n

the right initial value, so that when n is\n

correct initial value is 30. As usual, we\n

values of n, n equals 123. And making sure\n

we prefer to start with our index at zero,\n

the nth power. Since a value of zero for n\n

a value of one for n in this formula gives\n

have a common ratio, if I divide the second\n

halves. And that's the same ratio as I get\n

term. So if I use an index starting with zero,\n

is the first term times five halves to the\n

at one, one way to do this is to let K equal\n

When n is 0k, will be one. And since K is\n

n with k minus one. This gives the following\n

example has a common ratio of negative two\n

term of three times that ratio, negative two\n

at zero. Or, as above, we can write it as\n

Where k starts at one. Sometimes people like\n

negative one to the power makes a series alternate\n

three sequences are all examples of geometric\n

terms have the same common ratio. And in general,\n

ratio than a geometric sequence can be written\n

starts at zero, or as a times r to the n minus\n

sequences are neither arithmetic sequences\n

neither have common differences nor common\n

for the nth term just by looking for the pattern.\n

if I start at n equals one, I know that 90\n

start with a negative and then alternate,\n

like it's just twice n. And the denominator\n

with the square of three, so I'll write that\n

have a simple closed form formula, but I can\n

is negative six, a two is five. And in general,\n

terms, a sub n minus one, plus a sub n minus\n

standard Fibonacci sequence, which has the\n

values. This video gave an introduction to\n

geometric sequences, and recursively defined\n

of a series and how to find it sum. For any\n

the sum of its terms, a sub one plus a sub\n

a series. Often this series is written in\n

to infinity of a sub n. Let's look at the\n

one to infinity. If we add together all the\n

one to infinity of one over two to the N,\n

that's one half, plus one over two squared,\n

1/16, plus 1/32, and so on. But what does\n

numbers? How can we figure out what this infinite\n

add up finitely, many at a time, may write\n

is the ace of ends. And I'll keep adding more\n

But just add the first term, well, that's\n

gives me a song of three fourths. If I add\n

And then I'll add the next one. I get 15 sixteenths,\n

This process of repeated addition gives me\n

called the sequence of partial sums. And it's\n

in the sequence of partial sums as S sub one,\n

the second partial sum S sub two, adding together\n

the first three terms, and so on. Let me contrast\n

of terms that we started out with. Those are\n

first term, a sub two, the second term, and\nso on.

Where k starts at one. Sometimes people like\n

negative one to the power makes a series alternate\n

three sequences are all examples of geometric\n

terms have the same common ratio. And in general,\n

ratio than a geometric sequence can be written\n

starts at zero, or as a times r to the n minus\n

sequences are neither arithmetic sequences\n

neither have common differences nor common\n

for the nth term just by looking for the pattern.\n

if I start at n equals one, I know that 90\n

start with a negative and then alternate,\n

like it's just twice n. And the denominator\n

with the square of three, so I'll write that\n

have a simple closed form formula, but I can\n

is negative six, a two is five. And in general,\n

terms, a sub n minus one, plus a sub n minus\n

standard Fibonacci sequence, which has the\n

values. This video gave an introduction to\n

geometric sequences, and recursively defined\n

of a series and how to find it sum. For any\n

the sum of its terms, a sub one plus a sub\n

a series. Often this series is written in\n

to infinity of a sub n. Let's look at the\n

one to infinity. If we add together all the\n

one to infinity of one over two to the N,\n

that's one half, plus one over two squared,\n

1/16, plus 1/32, and \nso on. But what does it really mean to add

have infinitely many numbers? How can we figure\n

the start out, we could add up finitely, many\n

or so terms, that is the ace of ends. And\n

one more at a time. But just add the first\n

the next term on, that gives me a song of\n

goes up to seven eights. And then I'll add\n

on just adding one more term each time. This\n

sequence down at the bottom. That's called\n

denoted by s sub n. So the first term in the\n

adding together the first term, here's the\n

the first two terms. So three means add together,\n

the sequence of partial sums with the sequence\n

denoted a sub n. So here's a sub one, the\n

Although I can't physically add up infinitely\n

up more and more numbers, my partial sums\n

limit, as the number of terms I add up and\n

equal to one. So it makes sense that if I\n

I should get an exact sum of one, the sum\n

this particular series, there's a nice way\n

If I draw a square with side length one and\n

an area of one half. Now if I draw a line\n

1/4. Here's an area of 1/8 1/16 and I keep\n

exactly match the terms of this series. In\n

R square, which has an area of one. In this\n

by evaluating the limit of the partial sums.\n

of any series. For any series, the partial\n

s sub n, where S sub one is equal to just\n

to the sum of the first two terms is a one\n

first three terms. And in general, s sub n\n

write this in sigma notation, as the sum of\n

a different letter K here as the index, just\n

number of terms that I'm adding up. That sum\n

of partial sums converges as a sequence. That\n

the sub ends, exists as a finite number. Otherwise,\n

is infinity or negative infinity, then the\n

that we're talking about the limit of the\n

terms as a bad, it's important to keep in\n

for any series, they're actually two sequences\n

the ace of ends. And then there's the sequence\n

of partial sums is telling us what the sum\n

sums converges to a number L, then we say\n

words, the sum of the series is out. Let's\n

plus n, please pause the video and take a\n

the first four partial sums. The first four\n

and a sub four. So plugging in one for n,\n

half, a sub two is one over two squared plus\n

being 1/12. And a sub four is 1/20. Now S\n

that's the same as a sub one is just one half

Although I can't physically add up infinitely\n

up more and more numbers, my partial sums\n

limit, as the number of terms I add up and\n

equal to one. So it makes sense that if I\n

I should get an exact sum of one, the sum\n

this particular series, there's a nice way\n

If I draw a square with side length one and\n

an area of one half. Now if I draw a line\n

1/4. Here's an area of 1/8 1/16 and I keep\n

exactly match the terms of this series. In\n

R square, which has an area of one. In this\n

by evaluating the limit of the partial sums.\n

of any series. For any series, the partial\n

s sub n, where S sub one is equal to just\n

to the sum of the first two terms is a one\n

first three terms. And in general, s sub n\n

write this in sigma notation, as the sum of\n

a different letter K here as the index, just\n

number of terms that I'm adding up. That sum\n

of partial sums converges as a sequence. That\n

the sub ends, exists as a finite number. Otherwise,\n

is infinity or negative infinity, then the\n

that we're talking about the limit of the\n

terms as a bad, it's important to keep in\n

for any series, they're actually two sequences\n

the ace of ends. And then there's the sequence\n

of partial sums is telling us what the sum\n

sums converges to a number L, then we say\n

words, the sum of the series is out. Let's\n

plus n, please pause the video and take a\n

the first four partial sums. The first four\n

and a sub four. So plugging in one for n,\n

half, a sub two is one over two squared plus\n

being 1/12. And a sub four is 1/20. Now S\n

that's the same as a sub one is just one half

S sub two, is what I get when I add one half\n

To get a sub three, I need to add on the next\n

fourths. And finally, as sub four, I need\n

fifths. There's a nice pattern going on with\n

the numerator is just n, and the denominator\n

to infinity of the partial sums, is the limit\n

which is just one. That means that the sum\n

coincidence the same sound as in the previous\n

the sum of an infinite series, we have to\n

more and more finitely many terms, our partial\n

in fact, the sum of our infinite sum is defined\n

partial sum. This video gives some more definitions\n

of bounded and the definition of monotonic\n

are less than or equal to some number in other\n

the term a sub n is less than or equal to\n

A sequence is bounded below if all of its\n

In other words, there's a number lowercase\n

to lowercase m, for all and we say that a\n

and below. In other words, all of its terms\n

the video and decide which of these three\n

is bounded above by three. Since all of its\n

course, we could have also used four as an\n

three is the tightest upper bound that we\n

since all of the terms are positive. So we\n

sequence is bounded below by one, but it's\n

grow past any potential bound. The third sequence\n

graph n on the x axis and a sub n on the y\n

positive and negative values. But since we're\n

to get from one term to the next, the oscillations\n

the terms can never get above three or below\n

if each term is less than the next term, that\n

for all n. sequences called non decreasing\n

next term. So a sub n is less than or equal\n

is like increasing, it's just we allow equality\n

S sub two, is what I get when I add one half\n

To get a sub three, I need to add on the next\n

fourths. And finally, as sub four, I need\n

fifths. There's a nice pattern going on with\n

the numerator is just n, and the denominator\n

to infinity of the partial sums, is the limit\n

which is just one. That means that the sum\n

coincidence the same sound as in the previous\n

the sum of an infinite series, we have to\n

more and more finitely many terms, our partial\n

in fact, the sum of our infinite sum is defined\n

partial sum. This video gives some more definitions\n

of bounded and the definition of monotonic\n

are less than or equal to some number in other\n

the term a sub n is less than or equal to\n

A sequence is bounded below if all of its\n

In other words, there's a number lowercase\n

to lowercase m, for all and we say that a\n

and below. In other words, all of its terms\n

the video and decide which of these three\n

is bounded above by three. Since all of its\n

course, we could have also used four as an\n

three is the tightest upper bound that we\n

since all of the terms are positive. So we\n

sequence is bounded below by one, but it's\n

grow past any potential bound. The third sequence\n

graph n on the x axis and a sub n on the y\n

positive and negative values. But since we're\n

to get from one term to the next, the oscillations\n

the terms can never get above three or below\n

if each term is less than the next term, that\n

for all n. sequences called non decreasing\n

next term. So a sub n is less than or equal\n

is like increasing, it's just we allow equality\n

a sequence is decreasing if each term is greater\n

a sub n plus one for all n. And a sequence\n

or equal to a sub n plus one for all n. Again,\n

allow for equality between consecutive terms.\n

a sub n on the y axis, then increasing looks\n

whereas decreasing would go down non decreasing\n

and mnandi increasing would go down possibly\n

is a stronger condition than non decreasing\n

stronger than just being less than or equal\n

it is also non decreasing. And similarly,\n

increasing. A sequence is called monotonic,\n

Please pause the video and try to decide which\n

The first two sequences are monotonic. The\n

since we never increase when we go from one\n

that it's monotonically decreasing. Since\n

term to the next, we never have equality between\n

non decreasing. Since we never go down, as\n

go up or stay at the same level. In this case,\n

is monotonically increasing because of the\n

terms. The third sequence is not monotonic,\n

and negative numbers, and therefore sometimes\n

increasing. And the fourth sequence is not\n

However, from the fifth term on the terms\n

could also say monotonically increasing. In\n

and monotonic sequences. We'll see later that\n

when a sequence converges. In class, we talked\n

n squared, where n goes from one to infinity.\n

And if it's bounded. If we compute the first\n

increasing. But in this case, appearances\n

whether it's monotonic is to use calculus\n

x equals x minus five over x squared, it is\n

let me take the derivative. Using the quotient\n

is equal to minus x squared plus 10x over\n

is F prime of X greater than zero for x bigger\n

my sequence will, will be increasing. To check\nif f prime of x

is greater than zero, I'll first set f prime\n

equal to zero, which means my numerator needs\n

I get that x equals zero, or x equals 10.

a sequence is decreasing if each term is greater\n

a sub n plus one for all n. And a sequence\n

or equal to a sub n plus one for all n. Again,\n

allow for equality between consecutive terms.\n

a sub n on the y axis, then increasing looks\n

whereas decreasing would go down non decreasing\n

and mnandi increasing would go down possibly\n

is a stronger condition than non decreasing\n

stronger than just being less than or equal\n

it is also non decreasing. And similarly,\n

increasing. A sequence is called monotonic,\n

Please pause the video and try to decide which\n

The first two sequences are monotonic. The\n

since we never increase when we go from one\n

that it's monotonically decreasing. Since\n

term to the next, we never have equality between\n

non decreasing. Since we never go down, as\n

go up or stay at the same level. In this case,\n

is monotonically increasing because of the\n

terms. The third sequence is not monotonic,\n

and negative numbers, and therefore sometimes\n

increasing. And the fourth sequence is not\n

However, from the fifth term on the terms\n

could also say monotonically increasing. In\n

and monotonic sequences. We'll see later that\n

when a sequence converges. In class, we talked\n

n squared, where n goes from one to infinity.\n

And if it's bounded. If we compute the first\n

increasing. But in this case, appearances\n

whether it's monotonic is to use calculus\n

x equals x minus five over x squared, it is\n

let me take the derivative. Using the quotient\n

is equal to minus x squared plus 10x over\n

is F prime of X greater than zero for x bigger\n

my sequence will, will be increasing. To check\n

first set f prime of x equal to zero. So I'll\n

my numerator needs to be equal to zero. And\n

Now if I draw my number line, since f prime\n

positive and negative in between these values.\n

one, one and 11, I can see that f prime is\n

x between zero and 10, and negative for x\n

when x increases from one to 10. And then\n

to our sequence. Therefore, the sequence is\n

check if the sequence is bounded. Our first\n

of x has a maximum at x equals 10. At least\n

one to infinity. And that's all that's relevant\n

is bounded above by its value at 10, which\n

Now notice that our sequence and minus five\n

for n bigger than five since the numerator\n

situation. And since there are only finitely,\n

just use the minimum of these terms and zero\n

four terms is negative four, which is less\n

bound. So the sequence is in fact bounded.\n

of derivative and maximum and minimum. In\n

limit as x goes to two of x minus two over\n

limit just by plugging In two for x, because\n

four goes to zero as x goes to two. This is\n

It's called indeterminate because we can't\n

fact that the numerator goes to zero and the\n

fast the numerator and the denominator are\n

the final limit of the quotient could be any\n

it could not even exist. In the past, we've\n

over zero indeterminate form by using algebraic\n

we'll introduce lopatok rule, which is a very\n

indeterminate forms. A limit of the form the\n

is called a zero over zero indeterminate form.\n

to zero, and the limit as x goes to a of g\n

is called an infinity over infinity indeterminate\n

x is equal to infinity or minus infinity.\n

equal to infinity or minus infinity. We saw\n

form in the introductory slide. One example\n

form is the limit as x goes to infinity of\n

negative 2x squared plus 16. Notice that as\n

infinity while the denominator goes to negative\n

form, it's possible for a to be a negative\n

but it doesn't have to be loopy. talls rule\n

functions. And the derivative of g is nonzero\n

Now if I draw my number line, since f prime\n

positive and negative in between these values.\n

one, one and 11, I can see that f prime is\n

x between zero and 10, and negative for x\n

when x increases from one to 10. And then\n

to our sequence. Therefore, the sequence is\n

check if the sequence is bounded. Our first\n

of x has a maximum at x equals 10. At least\n

one to infinity. And that's all that's relevant\n

is bounded above by its value at 10, which\n

Now notice that our sequence and minus five\n

for n bigger than five since the numerator\n

situation. And since there are only finitely,\n

just use the minimum of these terms and zero\n

four terms is negative four, which is less\n

bound. So the sequence is in fact bounded.\n

of derivative and maximum and minimum. In\n

limit as x goes to two of x minus two over\n

limit just by plugging In two for x, because\n

four goes to zero as x goes to two. This is\n

It's called indeterminate because we can't\n

fact that the numerator goes to zero and the\n

fast the numerator and the denominator are\n

the final limit of the quotient could be any\n

it could not even exist. In the past, we've\n

over zero indeterminate form by using algebraic\n

we'll introduce lopatok rule, which is a very\n

indeterminate forms. A limit of the form the\n

is called a zero over zero indeterminate form.\n

to zero, and the limit as x goes to a of g\n

is called an infinity over infinity indeterminate\n

x is equal to infinity or minus infinity.\n

equal to infinity or minus infinity. We saw\n

form in the introductory slide. One example\n

form is the limit as x goes to infinity of\n

negative 2x squared plus 16. Notice that as\n

infinity while the denominator goes to negative\n

form, it's possible for a to be a negative\n

but it doesn't have to be loopy. talls rule\n

functions. And the derivative of g is nonzero\n

under these conditions, if the limit as x\n

zero or infinity over infinity indeterminate\n

x over g of x is the same thing as the limit\n

of x, provided that the second limit exists,\n

loopy towers rule in action. In this example,\n

to infinity and the denominator three to the\n

over infinity indeterminate form. So let's\n

should equal the limit as x goes to infinity\n

is one divided by the derivative of the denominator,\n

provided that the second limit exists or as\n

limit, the numerators just fixed at one. And\n

infinity. Therefore, the second limit is just\n

to zero as well. In this example, we have\n

as x goes to zero, sine of x and x, both go\n

goes to zero in the denominator. So using\n

the limit I get by taking the derivative of\n

The derivative of sine x minus x is cosine\n

x cubed is three times sine x squared times\n

try to evaluate the limit again. As x goes\n

here goes to zero. As x goes to zero, sine\n

one, so the denominator also goes to zero.\n

form. And I might as well try applying leptitox\n

out that cosine of x is going to one. So the\n

limit. And in fact, I could rewrite my limit\n

the second limit is just one and can be ignored\n

this first limit, which is a little bit easier\n

of the top is minus sine x. And the derivative\n

x. Now let's try to evaluate again, as x goes\n

our denominator is also going to zero. But\n

rule again, because we can actually just simplify\n

with the sine x on the bottom. And we can\n

one over six cosine of x, which evaluates\n

I want to emphasize that it's a good idea\n

talls rule. If you don't simplify, like we\n

lopatok rule and additional time when you\n

more complicated. Instead of simpler to solve\n

over zero and infinity over infinity indeterminate\n

g of x with the limit of f prime of x over\n

This trick is known as lopi tels rule. We've\n

limits of the form zero over zero, or infinity\n

to use loopy towels rule to evaluate additional\n

infinity to the 00 to the zero, and one to\nthe infinity.

under these conditions, if the limit as x\n

zero or infinity over infinity indeterminate\n

x over g of x is the same thing as the limit\n

of x, provided that the second limit exists,\n

loopy towers rule in action. In this example,\n

to infinity and the denominator three to the\n

over infinity indeterminate form. So let's\n

should equal the limit as x goes to infinity\n

is one divided by the derivative of the denominator,\n

provided that the second limit exists or as\n

limit, the numerators just fixed at one. And\n

infinity. Therefore, the second limit is just\n

to zero as well. In this example, we have\n

as x goes to zero, sine of x and x, both go\n

goes to zero in the denominator. So using\n

the limit I get by taking the derivative of\n

The derivative of sine x minus x is cosine\n

x cubed is three times sine x squared times\n

try to evaluate the limit again. As x goes\n

here goes to zero. As x goes to zero, sine\n

one, so the denominator also goes to zero.\n

form. And I might as well try applying leptitox\n

out that cosine of x is going to one. So the\n

limit. And in fact, I could rewrite my limit\n

the second limit is just one and can be ignored\n

this first limit, which is a little bit easier\n

of the top is minus sine x. And the derivative\n

x. Now let's try to evaluate again, as x goes\n

our denominator is also going to zero. But\n

rule again, because we can actually just simplify\n

with the sine x on the bottom. And we can\n

one over six cosine of x, which evaluates\n

I want to emphasize that it's a good idea\n

talls rule. If you don't simplify, like we\n

lopatok rule and additional time when you\n

more complicated. Instead of simpler to solve\n

over zero and infinity over infinity indeterminate\n

g of x with the limit of f prime of x over\n

This trick is known as lopi tels rule. We've\n

limits of the form zero over zero, or infinity\n

to use loopy towels rule to evaluate additional\n

infinity to the 00 to the zero, and one to\nthe infinity.

In this example, we want to evaluate the limit\n

from the positive side, sine x goes to zero,\n

the graph of y equals ln x. So this is actually\n

Even though the second factor is going to\n

times infinity and indeterminate form, you\n

for either positive or negative infinity.\n

the sine x factor is pulling the product towards\n

product towards large negative numbers. And\n

product will actually be. But the great thing\n

look like an infinity over infinity and determinant\n

form. Instead of sine x times ln x, I can\n

sine x. Now as x goes to zero, my numerator\n

x is going to zero through positive numbers,\n

positive infinity. So I have an infinity over\n

instead choose to leave the sine x in the\n

x in the denominator. If I do this, then as\n

x goes to zero. And since ln x goes to negative\n

so I have a zero over zero indeterminate form.\n

of these two ways to rewrite a product as\n

version that makes it easier to take the derivative\n

trick is just to try one of the ways and if\n

I'm going to use the first method of rewriting\n

x can be written as cosecant of x and I know\n

lopi towels rule On this infinity over infinity\n

as the limit of what I get when I take the\n

x divided by the derivative of the denominator,\n

always, I want to simplify my expression before\n

in the denominator in terms of sine and cosine.\n

x is cosine of x over sine of x. Now flipping\n

to zero plus of one over x times the sine\n

other words, the limit of negative sine squared\n

x goes to one as x goes to zero. So I can\n

squared x over x times the limit of something\n

over zero and determinant form. And I can\n

of the top, I get negative two sine x, cosine\n

one. Now I'm in a good position just to evaluate\n

numerator, I get negative two times zero times\n

limit is zero. In this limit, we have a battle\n

over x is going to zero. So one plus one over\n

to infinity, it's hard to tell what's going\n

number, that would be one. But anything slightly\n

and bigger powers, we would expect to get\ninfinity.

In this example, we want to evaluate the limit\n

from the positive side, sine x goes to zero,\n

the graph of y equals ln x. So this is actually\n

Even though the second factor is going to\n

times infinity and indeterminate form, you\n

for either positive or negative infinity.\n

the sine x factor is pulling the product towards\n

product towards large negative numbers. And\n

product will actually be. But the great thing\n

look like an infinity over infinity and determinant\n

form. Instead of sine x times ln x, I can\n

sine x. Now as x goes to zero, my numerator\n

x is going to zero through positive numbers,\n

positive infinity. So I have an infinity over\n

instead choose to leave the sine x in the\n

x in the denominator. If I do this, then as\n

x goes to zero. And since ln x goes to negative\n

so I have a zero over zero indeterminate form.\n

of these two ways to rewrite a product as\n

version that makes it easier to take the derivative\n

trick is just to try one of the ways and if\n

I'm going to use the first method of rewriting\n

x can be written as cosecant of x and I know\n

lopi towels rule On this infinity over infinity\n

as the limit of what I get when I take the\n

x divided by the derivative of the denominator,\n

always, I want to simplify my expression before\n

in the denominator in terms of sine and cosine.\n

x is cosine of x over sine of x. Now flipping\n

to zero plus of one over x times the sine\n

other words, the limit of negative sine squared\n

x goes to one as x goes to zero. So I can\n

squared x over x times the limit of something\n

over zero and determinant form. And I can\n

of the top, I get negative two sine x, cosine\n

one. Now I'm in a good position just to evaluate\n

numerator, I get negative two times zero times\n

limit is zero. In this limit, we have a battle\n

over x is going to zero. So one plus one over\n

to infinity, it's hard to tell what's going\n

number, that would be one. But anything slightly\n

and bigger powers, we would expect to get\ninfinity.

So our limit has an independent permanent\n

is going to be one infinity, or maybe something\n

a variable in the base, and a variable in\n

If we set y equal to one plus one over x to\n

sides, I can use my log roles to rewrite that\n

I wanted to take the limit as x goes to infinity\n

x times ln one plus one over x. As x goes\n

One plus one over x goes to just one and ln\n

times zero indeterminate form, which we can\n

or a zero over zero indeterminate form. Let's\n

over x divided by one of our x. This is indeed\n

can use lobi tiles rule and take the derivative\n

of the top is one over one plus one over x\n

would be negative one over x squared. And\n

one over x is negative one over x squared,\n

and rewrite our limit as the limit as x goes\n

x, which is just equal to one since one over\n

of ln y is equal to one, but we're really\n

we can think of as e to the ln y. Since ln\n

to e to the one. In other words, E. So we\n

E. And in fact, you may recognize that this\n

So our limit has an independent permanent\n

is going to be one infinity, or maybe something\n

a variable in the base, and a variable in\n

If we set y equal to one plus one over x to\n

sides, I can use my log roles to rewrite that\n

I wanted to take the limit as x goes to infinity\n

x times ln one plus one over x. As x goes\n

One plus one over x goes to just one and ln\n

times zero indeterminate form, which we can\n

or a zero over zero indeterminate form. Let's\n

over x divided by one of our x. This is indeed\n

can use lobi tiles rule and take the derivative\n

of the top is one over one plus one over x\n

would be negative one over x squared. And\n

one over x is negative one over x squared,\n

and rewrite our limit as the limit as x goes\n

x, which is just equal to one since one over\n

of ln y is equal to one, but we're really\n

we can think of as e to the ln y. Since ln\n

to e to the one. In other words, E. So we\n

E. And in fact, you may recognize that this\n

In the previous example, we had a one to the\n

logs and used log rules to write that as an\n

the same thing can be done if we have an infinity\n

to the zero and determinant form. So one to\n

to the zero are all indeterminate forms that\n

video, we saw that a zero times infinity indeterminate\n

or infinity over infinity indeterminate form\n

divided by one over g of x, or as g of x divided\n

lopi talls rule on these three sorts of indeterminate\n

f of x to the g of x that we want to take\n

for deciding whether a sequence converges.\n

if the limit as n goes to infinity of the\n

Otherwise, we say the sequence diverges. In\n

is infinity, or negative infinity, or does\n

as n goes to infinity of a sub n equals L,\n

number capital N, such that when the index\n

N, a sub n is within distance epsilon of owl.\n

of L is the same thing as saying that the\n

than epsilon. Let me draw this as a picture.\n

the y axis, we can plot our terms a sub n\n

are settling at a particular value, I'll draw\n

say that the limit of a sub n is equal to\n

I've tried to draw a distance epsilon above\n

or a sub ends within epsilon of L by requiring\n

I've chosen here, are a sub ends are trapped\n

than equal to three. So there is that big\n

such that when little n is bigger than that\n

of L. And if I chosen a tinier epsilon, I\n

out to make sure that my sequence was trapped\n

definition of a limit. Formally, we say the\n

infinity. If for any big number, omega, there's\n

than capital N, for a little n bigger than\n

matter how big an Omega I originally pick,\n

omega. Let me draw a picture for this one\n

And now if I pick a certain height omega,\n

And if I pick a different, bigger value of\n

than omega, I might just need to go further\nout in my sequence.

In the previous example, we had a one to the\n

logs and used log rules to write that as an\n

the same thing can be done if we have an infinity\n

to the zero and determinant form. So one to\n

to the zero are all indeterminate forms that\n

video, we saw that a zero times infinity indeterminate\n

or infinity over infinity indeterminate form\n

divided by one over g of x, or as g of x divided\n

lopi talls rule on these three sorts of indeterminate\n

f of x to the g of x that we want to take\n

for deciding whether a sequence converges.\n

if the limit as n goes to infinity of the\n

Otherwise, we say the sequence diverges. In\n

is infinity, or negative infinity, or does\n

as n goes to infinity of a sub n equals L,\n

number capital N, such that when the index\n

N, a sub n is within distance epsilon of owl.\n

of L is the same thing as saying that the\n

than epsilon. Let me draw this as a picture.\n

the y axis, we can plot our terms a sub n\n

are settling at a particular value, I'll draw\n

say that the limit of a sub n is equal to\n

I've tried to draw a distance epsilon above\n

or a sub ends within epsilon of L by requiring\n

I've chosen here, are a sub ends are trapped\n

than equal to three. So there is that big\n

such that when little n is bigger than that\n

of L. And if I chosen a tinier epsilon, I\n

out to make sure that my sequence was trapped\n

definition of a limit. Formally, we say the\n

infinity. If for any big number, omega, there's\n

than capital N, for a little n bigger than\n

matter how big an Omega I originally pick,\n

omega. Let me draw a picture for this one\n

And now if I pick a certain height omega,\n

And if I pick a different, bigger value of\n

than omega, I might just need to go further\nout in my sequence.

For this first value of omega, I would just\n

Once my little ends are all bigger than six,\n

and for this bigger value of omega, I need\n

once my little ends are bigger than about\n

this omega. There's a similar definition for\n

equaling negative infinity. Now we just need\n

there's a number. And such that a sub n is\n

or equal to capital N. Please take a moment\n

that converges, a sequence that diverges to\n

that's bounded but still diverges. One example\n

n, this sequence converges to zero, since\n

N is zero. A divergent sequence is two to\n

of two to the n is infinity. One example of\n

is negative one to the n. The sequence alternates\n

whether n is odd, or even. So it's bounded\n

diverges because the limit does not exist.\n

value. The rest of this video will give some\n

The first technique is to use standard calculus\n

though a sequence is only defined on positive\n

a function defined on all positive real numbers\n

In other words, the terms as of n are equal\n

then if the limit as x goes to infinity of\n

to infinity of a sub n also equals out, the\n

as the blue function. So a lot of times, we\n

replacing the terms a sub n with F of X for\n

lobaton ruler, other tricks from calculus\n

Let's try that for the following example.\n

we'll assume that n starts at one and goes\n

converges, let's instead look at the function\n

x, where x is a real number. Now let's look\n

x. That's the limit as x goes to infinity\n

as x goes to infinity in the x goes to infinity,\n

which means ln of that goes to infinity. So\n

to infinity. And so as the denominator, we\n

form. So we can apply lobby towels rule and\n

the derivative of the denominator, the derivative\n

e to the x times two times e to the x using\n

denominator is just one. We can take derivatives\n

number, not just an integer. Simplifying,\n

form. So let's take the derivatives again

For this first value of omega, I would just\n

Once my little ends are all bigger than six,\n

and for this bigger value of omega, I need\n

once my little ends are bigger than about\n

this omega. There's a similar definition for\n

equaling negative infinity. Now we just need\n

there's a number. And such that a sub n is\n

or equal to capital N. Please take a moment\n

that converges, a sequence that diverges to\n

that's bounded but still diverges. One example\n

n, this sequence converges to zero, since\n

N is zero. A divergent sequence is two to\n

of two to the n is infinity. One example of\n

is negative one to the n. The sequence alternates\n

whether n is odd, or even. So it's bounded\n

diverges because the limit does not exist.\n

value. The rest of this video will give some\n

The first technique is to use standard calculus\n

though a sequence is only defined on positive\n

a function defined on all positive real numbers\n

In other words, the terms as of n are equal\n

then if the limit as x goes to infinity of\n

to infinity of a sub n also equals out, the\n

as the blue function. So a lot of times, we\n

replacing the terms a sub n with F of X for\n

lobaton ruler, other tricks from calculus\n

Let's try that for the following example.\n

we'll assume that n starts at one and goes\n

converges, let's instead look at the function\n

x, where x is a real number. Now let's look\n

x. That's the limit as x goes to infinity\n

as x goes to infinity in the x goes to infinity,\n

which means ln of that goes to infinity. So\n

to infinity. And so as the denominator, we\n

form. So we can apply lobby towels rule and\n

the derivative of the denominator, the derivative\n

e to the x times two times e to the x using\n

denominator is just one. We can take derivatives\n

number, not just an integer. Simplifying,\n

form. So let's take the derivatives again

the derivative of the numerator is now two\n

denominator is also two times e to the x.\n

converges to one, our sequence also converges\n

the derivative of the numerator is now two\n

denominator is also two times e to the x.\n

converges to one, our sequence also converges\n

Another technique for proving that sequences\n

trap the sequence between two simpler sequences\n

since sine and cosine are both bounded in\n

cosine of n plus sine of n can't be any bigger\n

negative two. If I divide all sides of this\n

see that the original sequence is bounded\n

thirds and two over n to the two thirds. Notice\n

since as usual, we're assuming that n starts\n

by n to the two thirds does not switch the\n

the limit as n goes to infinity of negative\n

as n goes to infinity, and to the two thirds\n

as n goes to infinity of two over n to the\n

are the same, we know by the squeeze theorem,\n

has to exist at equals zero. Also, it's a\n

if a sub n is bounded, and monotonic, then\n

for this fact, by looking at a graph. If the\n

for example, but are bounded, then there's\n

can't oscillate up and down. Because they're\n

sense that they have to settle on some limit.\n

starts point 1.1 2.12 3.1234, and so on. Where\n

numbers as our decimal, the sequence is certainly\n

Since every term of the sequence is greater\n

So we have a bounded monotonic sequence. And\n

it actually converges to is a little mysterious,\n

already familiar with like, like, point six,\n

it does converge to some real number. And\n

constant. And it has some interesting properties.\n

that's easy to to come up with, since you\n

numbers. If you can recognize a sequence to\n

easy to decide whether it converges or diverges.\n

of the form a times r to the n minus one where\n

it's written as a times r to the n, where\n

figure out for what values of our the sequence\n

here whether and starts at zero or starts\n

we're talking about the behavior of the terms\n

really don't matter. if r is greater than\n

sequence. In fact, for the sequence r to the\n

function f of x equals r to the x, that's\n

r is greater than one, the base for our exponential\n

that the limit as x goes to infinity of our\n

sequence r to the n also has to diverge to\ninfinity.

Another technique for proving that sequences\n

trap the sequence between two simpler sequences\n

since sine and cosine are both bounded in\n

cosine of n plus sine of n can't be any bigger\n

negative two. If I divide all sides of this\n

see that the original sequence is bounded\n

thirds and two over n to the two thirds. Notice\n

since as usual, we're assuming that n starts\n

by n to the two thirds does not switch the\n

the limit as n goes to infinity of negative\n

as n goes to infinity, and to the two thirds\n

as n goes to infinity of two over n to the\n

are the same, we know by the squeeze theorem,\n

has to exist at equals zero. Also, it's a\n

if a sub n is bounded, and monotonic, then\n

for this fact, by looking at a graph. If the\n

for example, but are bounded, then there's\n

can't oscillate up and down. Because they're\n

sense that they have to settle on some limit.\n

starts point 1.1 2.12 3.1234, and so on. Where\n

numbers as our decimal, the sequence is certainly\n

Since every term of the sequence is greater\n

So we have a bounded monotonic sequence. And\n

it actually converges to is a little mysterious,\n

already familiar with like, like, point six,\n

it does converge to some real number. And\n

constant. And it has some interesting properties.\n

that's easy to to come up with, since you\n

numbers. If you can recognize a sequence to\n

easy to decide whether it converges or diverges.\n

of the form a times r to the n minus one where\n

it's written as a times r to the n, where\n

figure out for what values of our the sequence\n

here whether and starts at zero or starts\n

we're talking about the behavior of the terms\n

really don't matter. if r is greater than\n

sequence. In fact, for the sequence r to the\n

function f of x equals r to the x, that's\n

r is greater than one, the base for our exponential\n

that the limit as x goes to infinity of our\n

sequence r to the n also has to diverge to\ninfinity.

If instead, r is equal to one, then r to the\n

of ones, so that converges to one. if r is\n

decreasing. This time, it's like the exponential\n

base between zero and one. And so the limit\n

infinity is going to be equals zero. Therefore,\n

if r equals exactly zero, and \n

so it also converges to zero. Next, let's\n

one and zero. Now, the sequence of rd ends\n

negative numbers that get smaller and smaller\n

because we get from one number to the next\n

of magnitude less than one. So our limit as\n

going to be zero again. Another way of thinking\n

squeeze theorem, since r to the n is always\n

r to the n, which is o n is always greater\n

of r to the n. Technically, r to the n is\n

are the n when n is even, so that art of the\n

to the negative of the absolute value of r,\n

is negative. But in any case, the inequality\n

since by the squeeze theorem, the limit of\n

right terms is zero. As we noted before, the\n

So our sequence converges to zero. Now, if\n

r to the n is just negative one to the n,\n

one. And so that sequence diverges. Finally,\n

from one term to the next, we're multiplying\n

than one. And so our terms are going to oscillate\n

they're going to be going up in magnitude.\n

a sub n, does not exist. And we see that our\n

these cases, we see that the sequence r to\n

negative one and one. it converges to one\n

when r is bigger than one or less than negative\n

almost the same thing as true, when we look\n

A is any real number, the sequence a times\n

negative one and one. it converges to a, when\n

is less than negative one, or greater than\n

the terms in the sequence by a just multiplies\n

one times i is a. So anytime you encounter\n

can be written in the form of a times r to\nthe n

If instead, r is equal to one, then r to the\n

of ones, so that converges to one. if r is\n

decreasing. This time, it's like the exponential\n

base between zero and one. And so the limit\n

infinity is going to be equals zero. Therefore,\n

if r equals exactly zero, and the sequence\n

to zero. Next, let's look at the case when\n

sequence of rd ends are going to oscillate\n

get smaller and smaller and magnitude as n\n

one number to the next by multiplying by r,\n

than one. So our limit as n goes to infinity\n

again. Another way of thinking about this\n

since r to the n is always less than or equal\n

is o n is always greater than or equal to\n

Technically, r to the n is always exactly\n

n is even, so that art of the N is positive.\n

of the absolute value of r, the N, when n\n

But in any case, the inequality still does\n

by the squeeze theorem, the limit of the left\n

terms is zero. As we noted before, the limit\n

our sequence converges to zero. Now, if r\n

r to the n is just negative one to the n,\n

one. And so that sequence diverges. Finally,\n

from one term to the next, we're multiplying\n

than one. And so our terms are going to oscillate\n

they're going to be going up in magnitude.\n

a sub n, does not exist. And we see that our\n

these cases, we see that the sequence r to\n

negative one and one. it converges to one\n

when r is bigger than one or less than negative\n

almost the same thing as true, when we look\n

A is any real number, the sequence a times\n

negative one and one. it converges to a, when\n

is less than negative one, or greater than\n

the terms in the sequence by a just multiplies\n

one times i is a. So anytime you encounter\n

can be written in the form of a times r to\nthe n

you can know that it converges if r is bigger\n

one. This sequence here, although it looks\n

sequence in disguise. One way to see this\n

this is negative one to the t, e to the t\n

T times three squared. This is the same thing\n

over three squared times E. Now this is looking\n

where A is one over three squared times E,\n

say the tail end, because we have T starting\n

less than three, the magnitude of our is got\n

as a negative number, that's between negative\n

of \nthis geometric sequence converges. It's kind

of interesting to note that we could also\n

to, using an index and going from zero to\n

do that the are the common ratio stays the\n

version starts at t equals three, the first\n

times one over three squared E. And that becomes\n

here, I get this value. And when t is three,\n

So these sequences are equivalent. But in\n

ratio, R is negative B over three, and a sequence\n

I want to mention for deciding whether sequences\n

limit laws about addition, subtraction, and\n

So for example, if the limit as n goes to\n

of b sub n is am than the limit of the sum,\n

to L plus m. And the limit of a sub n times\n

a sub n where C is some constant is going\n

subtraction and division. I want to emphasize\n

that the limits of the component sequences\n

laws to decide if this sequence converges,\n

difference of the limits, provided those limits\n

the degree of the numerator is less than the\n

limit is also zero. Since this is a geometric\n

fifths is between zero and one. Therefore,\n

zero. In this video, we saw several ways to\n

we could use calculus techniques like lopi\n

its associated function defined on real numbers,\n

theorem. we noted that all sequences that\n

we saw that geometric sequences always converge\n

than or equal to one. Finally, we saw that\n

and other conglomerations of sequences. This\n

they converge, and when they diverged.

you can know that it converges if r is bigger\n

one. This sequence here, although it looks\n

sequence in disguise. One way to see this\n

this is negative one to the t, e to the t\n

T times three squared. This is the same thing\n

over three squared times E. Now this is looking\n

where A is one over three squared times E,\n

say the tail end, because we have T starting\n

less than three, the magnitude of our is got\n

as a negative number, that's between negative\n

of this geometric sequence converges. It's\n

also rewrite this geometric sequence if we\n

to infinity. And one way to figure out how\n

the same as negative e over three. But since\n

first term here is really minus e over three\n

that becomes our value of A. Notice that when\n

t is three, here, I get the same value for\n

But in any case, for either a sequence, the\n

and a sequence converges. Therefore, the final\n

whether sequences converge or diverge is limit\n

subtraction, and so on hold for sequences\n

the limit as n goes to infinity of a sub n\n

the limit of the sum, a sub n plus b sub n\n

limit of a sub n times b sub n is L times\n

C is some constant is going to be c times\n

and division. I want to emphasize that these\n

the limits of the component sequences exist\n

to decide if this sequence converges, since\n

of the limits, provided those limits exist.\n

of the numerator is less than the degree of\n

is also zero. Since this is a geometric sequence,\n

is between zero and one. Therefore, the limit\n

this video, we saw several ways to prove that\n

use calculus techniques like lopi taas rule.\n

function defined on real numbers, we also\n

we noted that all sequences that are bounded\n

geometric sequences always converge if r is\n

equal to one. Finally, we saw that we can\n

and other conglomerations of sequences. This\n

they converge, and when they diverged.

A geometric sequence is a sequence of the\n

r cubed, and so on. For some numbers a and\n

a times r to the k, where k goes from zero\n

r is one half, the sequence would be three,\n

and so on, which could be written as three\n

zero to infinity. A geometric series is the\n

up. So that would be a plus a times r plus\n

written in summation notation is the sum from\n

the K. For our example, our series would be\n

sum from K equals zero to infinity of three\n

get a geometric series that's in disguise,\n

terms, notice that we're told to start with\n

here, I get negative one, squared over three\n

over three. If I plug in i equals three, I\n

to the two times three minus three, or negative\n

go four, I get one over 243. If I look at\n

divided by 1/3, is negative one night. Similarly,\n

suggesting that we might have a geometric\n

term coming from here, of 1/3. But there's\n

intensive, and makes use of exponent rules.\n

it using exponent rules, this is three to\n

I can rewrite the three to the two is three\n

three and the denominator becomes a three\n

all the pieces that are raised to the nth\n

i times three cubed, or 27 times negative\n

see that every time I increases by one, I\n

in my expression. And therefore, the this\n

one nine, as I saw before, now, you might\n

remember that it doesn't start at zero, it\n

to be what I get I get when I plug in the\n

one nine squared, which works out to 1/3.\n

with first term of 1/3, a common ratio of\n

in a more standard form as the sum of 1/3\n

from zero to infinity. Since k equals zero,\n

two in this expression, k is equal to, can\n

this is really a reindexing tugann simplification\n

A geometric sequence is a sequence of the\n

r cubed, and so on. For some numbers a and\n

a times r to the k, where k goes from zero\n

r is one half, the sequence would be three,\n

and so on, which could be written as three\n

zero to infinity. A geometric series is the\n

up. So that would be a plus a times r plus\n

written in summation notation is the sum from\n

the K. For our example, our series would be\n

sum from K equals zero to infinity of three\n

get a geometric series that's in disguise,\n

terms, notice that we're told to start with\n

here, I get negative one, squared over three\n

over three. If I plug in i equals three, I\n

to the two times three minus three, or negative\n

go four, I get one over 243. If I look at\n

divided by 1/3, is negative one night. Similarly,\n

suggesting that we might have a geometric\n

term coming from here, of 1/3. But there's\n

intensive, and makes use of exponent rules.\n

it using exponent rules, this is three to\n

I can rewrite the three to the two is three\n

three and the denominator becomes a three\n

all the pieces that are raised to the nth\n

i times three cubed, or 27 times negative\n

see that every time I increases by one, I\n

in my expression. And therefore, the this\n

one nine, as I saw before, now, you might\n

remember that it doesn't start at zero, it\n

to be what I get I get when I plug in the\n

one nine squared, which works out to 1/3.\n

with first term of 1/3, a common ratio of\n

in a more standard form as the sum of 1/3\n

from zero to infinity. Since k equals zero,\n

two in this expression, k is equal to, can\n

this is really a reindexing tugann simplification\n

As you may know, a geometric sequence converges\n

negative one and one, it converges to a, when\n

when r is less than or equal to negative one,\n

here, that A is not equal zero, since otherwise,\n

sequence of all zeros. I want to restate this\n

we'll use it later. But it's saying is that\n

limit, as K goes to infinity of A times r\n

converge to zero. when r is equal to one,\n

than or equal to negative one or greater than\n

number. Now I'd like to find similar rules\n

converges or diverges. Again, we'll assume\n

I just have a sum of a bunch of zeros, a boring\n

is going to be to find a formula for the nth\n

limit of partial sums. Since by definition,\n

or diverges. Before we carry out that strategy,\n

r is equal to one, then the series is just\n

if A is positive, or to negative infinity,\n

that A is not zero. so far is one our series\n

R is not equal to one. Let's look at a few\n

one, just the first term A, S sub two is a\n

nth partial sum s sub n is a plus a times\n

let's see, the last term will be a times r\n

be a times r to the n minus two. Notice that\n

r to the n minus one. Since we're starting\n

write the partial sum in a nicer form, so\n

to do that, I'm going to use a trick, I'm\n

by R. So on the left, I get r times s sub\n

each term by R. So the first term becomes\n

a times r cubed, and so on, the second to\n

one, and the last term becomes a times r to\n

equation under it have a lot of terms in common,\n

subtract the second equation from the first,\n

minus r times s sub n. But on the right side,\n

cancel the next one, this one cancels with\n

just a minus a times r to the n, I get a minus\n

equation. Now I can solve for a sub n, I can\n

just to keep things tidy. And then I get that\n

one minus R. I don't have to worry about dividing\n

remember, I'm assuming that R is not one.\n

sub n, I can proceed to take the limit as\n

of partial sums converges or diverges sets\n

only part of this formula that depends on\n

all constants as far as n is concerned. So\n

of the limit, I can rewrite the limit as a\n

r to the n or even better as a over one minus\n

Now this limit I've seen before, right on\n

considering when I was looking at convergence\n

my A is equal to one. So we know that this\n

one and one, and does not exist as a finite\n

one, or r is greater than one. Therefore,\n

equal a over one minus r times one minus zero,\n

between negative one and one, and that limit,\n

is less than or equal to negative one, or\n

doesn't exist as a finite number, when r is\n

now I've got all this cases for our covered\n

series converges to a over one minus R.

As you may know, a geometric sequence converges\n

negative one and one, it converges to a, when\n

when r is less than or equal to negative one,\n

here, that A is not equal zero, since otherwise,\n

sequence of all zeros. I want to restate this\n

we'll use it later. But it's saying is that\n

limit, as K goes to infinity of A times r\n

converge to zero. when r is equal to one,\n

than or equal to negative one or greater than\n

number. Now I'd like to find similar rules\n

converges or diverges. Again, we'll assume\n

I just have a sum of a bunch of zeros, a boring\n

is going to be to find a formula for the nth\n

limit of partial sums. Since by definition,\n

or diverges. Before we carry out that strategy,\n

r is equal to one, then the series is just\n

if A is positive, or to negative infinity,\n

that A is not zero. so far is one our series\n

R is not equal to one. Let's look at a few\n

one, just the first term A, S sub two is a\n

nth partial sum s sub n is a plus a times\n

let's see, the last term will be a times r\n

be a times r to the n minus two. Notice that\n

r to the n minus one. Since we're starting\n

write the partial sum in a nicer form, so\n

to do that, I'm going to use a trick, I'm\n

by R. So on the left, I get r times s sub\n

each term by R. So the first term becomes\n

a times r cubed, and so on, the second to\n

one, and the last term becomes a times r to\n

equation under it have a lot of terms in common,\n

subtract the second equation from the first,\n

minus r times s sub n. But on the right side,\n

cancel the next one, this one cancels with\n

just a minus a times r to the n, I get a minus\n

equation. Now I can solve for a sub n, I can\n

just to keep things tidy. And then I get that\n

one minus R. I don't have to worry about dividing\n

remember, I'm assuming that R is not one.\n

sub n, I can proceed to take the limit as\n

of partial sums converges or diverges sets\n

only part of this formula that depends on\n

all constants as far as n is concerned. So\n

of the limit, I can rewrite the limit as a\n

r to the n or even better as a over one minus\n

Now this limit I've seen before, right on\n

considering when I was looking at convergence\n

my A is equal to one. So we know that this\n

one and one, and does not exist as a finite\n

one, or r is greater than one. Therefore,\n

equal a over one minus r times one minus zero,\n

between negative one and one, and that limit,\n

is less than or equal to negative one, or\n

doesn't exist as a finite number, when r is\n

now I've got all this cases for our covered\n

series converges to a over one minus R.

r between negative one and one, I can also\n

one, and it diverges for the absolute value\n

this fact in this example. Remember that we\n

a common ratio r equal to negative one night,\n

in i equals two, and I got that first term\n

value of negative 1/9 is 1/9, which is less\n

and it converges to a over one minus r, so\n

which is 1/3 over one plus one night, that's\n

tenths. In this video, we looked at geometric\n

we saw that a geometric series will converge\n

And they'll diverged if the absolute value\n

case that it converges, it converges to a\n

and r is the common ratio. This video explains\n

or diverges using an integral. Let's start\n

Please pause the video for a moment and think\n

The sum from n equals one to infinity of one\n

from one to infinity of one over x squared\n

I've graphed the function y equals one over\n

bunch of rectangles, I've divided the x axis\n

rectangle has a base of length one, and a\n

right endpoint of the sub interval. So the\n

of one, so it has an area of one, the second\n

one over two squared, so that's 1/4. So the\n

of one again, and a height of 1/9, and so\n

same as its height, and its height is just\n

value of n In other words, if I write out\n

the same as the areas of my rectangles. And\n

total green area. Now my integral can also\n

represents the area from one to infinity.\n

is finite, because we know that this integral\n

if I just look at the rectangles, starting\n

rectangles will lie below the blue curve,\n

so they will have a smaller area, and therefore\n

one on has to converge to a finite number.\n

of one over n squared converges. We're interested\n

just the area of this single rectangle plus\n

one more than this sum here. So it also converges\n

logic goes like this. First, we solve the\n

of the P test. From that, we can conclude\n

of one over n squared represents a smaller\n

figure out that the sum from n equals one\n

So our series converges to a finite number.\n

n equals one to infinity of one over the square\nroot of n.

r between negative one and one, I can also\n

one, and it diverges for the absolute value\n

this fact in this example. Remember that we\n

a common ratio r equal to negative one night,\n

in i equals two, and I got that first term\n

value of negative 1/9 is 1/9, which is less\n

and it converges to a over one minus r, so\n

which is 1/3 over one plus one night, that's\n

tenths. In this video, we looked at geometric\n

we saw that a geometric series will converge\n

And they'll diverged if the absolute value\n

case that it converges, it converges to a\n

and r is the common ratio. This video explains\n

or diverges using an integral. Let's start\n

Please pause the video for a moment and think\n

The sum from n equals one to infinity of one\n

from one to infinity of one over x squared\n

I've graphed the function y equals one over\n

bunch of rectangles, I've divided the x axis\n

rectangle has a base of length one, and a\n

right endpoint of the sub interval. So the\n

of one, so it has an area of one, the second\n

one over two squared, so that's 1/4. So the\n

of one again, and a height of 1/9, and so\n

same as its height, and its height is just\n

value of n In other words, if I write out\n

the same as the areas of my rectangles. And\n

total green area. Now my integral can also\n

represents the area from one to infinity.\n

is finite, because we know that this integral\n

if I just look at the rectangles, starting\n

rectangles will lie below the blue curve,\n

so they will have a smaller area, and therefore\n

one on has to converge to a finite number.\n

of one over n squared converges. We're interested\n

just the area of this single rectangle plus\n

one more than this sum here. So it also converges\n

logic goes like this. First, we solve the\n

of the P test. From that, we can conclude\n

of one over n squared represents a smaller\n

figure out that the sum from n equals one\n

So our series converges to a finite number.\n

n equals one to infinity of one over the square\nroot of n.

Please pause the video for a moment and think\n

if this series converges or diverges. A natural\n

one to infinity of one over the square root\n

where p is equal to one half, which is less\n

areas. The blue curve here is the graph of\n

root of x. And here, once again, I've drawn\n

to get the heights of the rectangles. So the\n

terms in my series. As in the previous problem,\n

the rest of the rectangles have an area that's\n

to infinity. But there's a serious problem\n

one over the square root of x dx diverges\n

the second one on might be less than my area\n

something that diverges to infinity, it tells\n

it could diverged. But don't give up hope.\n

we'll be able to get something that we can\n

left endpoints instead of right endpoints\n

my function y equals one over the square root\n

endpoints, makes my rectangles have a larger\n

section of the curve. let me label the rectangles\n

correspond to the terms in my series. But\n

that total area, that total area is bigger\n

Please pause the video for a moment and think\n

if this series converges or diverges. A natural\n

one to infinity of one over the square root\n

where p is equal to one half, which is less\n

areas. The blue curve here is the graph of\n

root of x. And here, once again, I've drawn\n

to get the heights of the rectangles. So the\n

terms in my series. As in the previous problem,\n

the rest of the rectangles have an area that's\n

to infinity. But there's a serious problem\n

one over the square root of x dx diverges\n

the second one on might be less than my area\n

something that diverges to infinity, it tells\n

it could diverged. But don't give up hope.\n

we'll be able to get something that we can\n

left endpoints instead of right endpoints\n

my function y equals one over the square root\n

endpoints, makes my rectangles have a larger\n

section of the curve. let me label the rectangles\n

correspond to the terms in my series. But\n

that total area, that total area is bigger\n

Since this integral diverges, and this series\n

of comparing a series to an integral is a\n

It's known as the interval test. The integral\n

and decreasing function on the interval from\n

equal to f evaluated at n, then if the integral\n

the series from one to infinity of a sub n\n

infinity of f of x dx diverges, then the series\n

of this theorem, the logic behind is the same\n

If the interval converges, we use the picture\n

The area of each rectangle is the same as\n

and the height of each rectangle is just f\n

is just f sub one, which is a sub one, the\n

is a sub two, and so on. If we focus on the\n

area of those rectangles is less than the\n

say the sum from n equals two to infinity\n

So the integral converges by assumption, this\n

our original series from n equals one to infinity\n

diverges, then we use the other picture, and\n

the areas of the rectangles are still given\n

time, the combined area of the green rectangles,\n

the integral of our function. Since this integral\n

well. That's the idea behind the integral\n

able to integrate the function that corresponds\n

is continuous, positive and decreasing. Actually,\n

eventually continuous, positive and decreasing.\n

on some interval from R to infinity for some\n

I can always draw the same pictures, just\n

and get the integral converges if and only\n

the same arguments we use before, but the\n

if the series starting at one converges, since\n

series doesn't affect whether it converges\n

converges if and only if the integral from\n

adding on a finite little piece of area from\n

of the integral. So by this chain of logic,\n

for a while, as long as it's eventually positive,\n

of the integral test and action. We want to\n

of ln n over n, converges or diverges. So\n

Since this integral diverges, and this series\n

of comparing a series to an integral is a\n

It's known as the interval test. The integral\n

and decreasing function on the interval from\n

equal to f evaluated at n, then if the integral\n

the series from one to infinity of a sub n\n

infinity of f of x dx diverges, then the series\n

of this theorem, the logic behind is the same\n

If the interval converges, we use the picture\n

The area of each rectangle is the same as\n

and the height of each rectangle is just f\n

is just f sub one, which is a sub one, the\n

is a sub two, and so on. If we focus on the\n

area of those rectangles is less than the\n

say the sum from n equals two to infinity\n

So the integral converges by assumption, this\n

our original series from n equals one to infinity\n

diverges, then we use the other picture, and\n

the areas of the rectangles are still given\n

time, the combined area of the green rectangles,\n

the integral of our function. Since this integral\n

well. That's the idea behind the integral\n

able to integrate the function that corresponds\n

is continuous, positive and decreasing. Actually,\n

eventually continuous, positive and decreasing.\n

on some interval from R to infinity for some\n

I can always draw the same pictures, just\n

and get the integral converges if and only\n

the same arguments we use before, but the\n

if the series starting at one converges, since\n

series doesn't affect whether it converges\n

converges if and only if the integral from\n

adding on a finite little piece of area from\n

of the integral. So by this chain of logic,\n

for a while, as long as it's eventually positive,\n

of the integral test and action. We want to\n

of ln n over n, converges or diverges. So\n

This is a continuous function, because it's\n

and we're starting at an x value of one, so\n

being zero. It's also a positive function.\n

zero for x bigger than one, and therefore\n

This is a continuous function, because it's\n

and we're starting at an x value of one, so\n

being zero. It's also a positive function.\n

zero for x bigger than one, and therefore\n

Finally, let's check if our function is decreasing.\n

If f of x is ln x over x, then f prime of\n

x minus ln x times one over x squared. This\n

which is negative when one minus ln x is less\n

x, that is E is less than x. So the function\n

whenever x is bigger than E, so it's eventually\n

Finally, let's check if our function is decreasing.\n

If f of x is ln x over x, then f prime of\n

x minus ln x times one over x squared. This\n

which is negative when one minus ln x is less\n

x, that is E is less than x. So the function\n

whenever x is bigger than E, so it's eventually\n

They're met, so we can apply the integral\n

This is an improper integral. So by definition,\n

integral from one to T of ln x over x. We\n

where u is equal to ln x, d u is equal to\n

u is equal to ln of one, that's zero, when\n

in, we get the integral from zero to ln T\n

two evaluated between ln T and zero.

Substituting in our bounds of integration,\n

ln t squared over two minus zero. Now as t\n

So ln t squared over two goes to infinity,\n

the integral test, the series also diverges.\n

if and only if the corresponding integral\n

function is eventually continuous, positive\n

whether series converge or diverge by comparing\n

the sum of a sub n and the sum of b sub n\n

the series are always greater than or equal\n

equal to b sub n for all n. In the pictures\n

to represent the ACE events, and the heights\n

the base events. If we put the pictures together,\n

less than the heights of the green bars. So\n

or equal to a sub n is less than or equal\n

length one, the height of each bar is the\n

the sum from n equals one to infinity, of\n

blue rectangles added up the total blue area.\n

infinity of b sub n, this represents the area\n

area. Because the blue bars have a smaller\n

conclusions. First of all, if the total green\n

area. In other words, if the sum of the b\n

A sub N. Furthermore, if the total blue area\n

So we can also say, if the sum of the ace\n

the base events. These facts are known as\n

useful in establishing convergence. But we\n

too far. In particular, if the smaller series\n

say anything about the larger series of these\n

red could diverge. Also, if the larger series\n

anything about the smaller series of Ace events,\n

could diverged. When using the comparison\n

it's handy to compare your unfamiliar series\n

converges or diverges. The following series\n

First, the geometric series Sum of A times\n

value of r is less than one. And second, the\n

when p is greater than one. Let's use a comparison\n

to the n over five to the n plus n squared\n

far as whether a series converges or diverges\n

close to infinity, the behavior of the terms\n

terms doesn't make any difference as far as\n

So I'm going to focus on what happens to these\n

to the n goes to infinity, and five to the\n

But between five to the n and n squared, five\n

So I'm going to say that this five to the\n

important. And for that reason, the behavior\n

to the behavior of the series, three to the\n

out the n squared term, which is insignificant\n

So I'm going to compare our given series to\n

the second series we know converges, because\n

the absolute value of three fifths is less\n

I'm going to need to compare the terms of\n

I want to show that these terms are less than\n

than a convergent series will guarantee convergence\n

don't have to worry about that. And it's also\n

bigger than or equal to five to the n. When\n

ratio. So three to the n over five to the\n

equal to three to the n over five to the n,\n

And so by the comparison theorem, since the\n

so does our original series. We've established\n

This video was about the comparison test,\n

equal to A ad less than or equal to bn, and\n

of the smaller series A ends converges. And\n

sum of the larger series diverges. The limit\n

regular ordinary comparison test for series.\n

the sum from n equals one to infinity of three\n

And we use the ordinary comparison test and\n

to the n over five to the n. This worked out\n

less than the terms here. And this series\n

ensures convergence. But if we change the\n

the sum of three to the n over five to the\n

go wrong. If we now try to compare to the\n

to the n minus n squared is less than or equal\n

the n over five to the n minus n squared is\n

five to the n. Since dividing by a smaller\n

unfortunately, is not useful to us, being\n

guarantee convergence or divergence. So we\n

the limit comparison test gives us one way\n

the following. Suppose that sum of a n and\n

If the limit as n goes to infinity of the\n

L is a finite number that's bigger than zero,\n

or both diverge, so they have the same convergence\n

on the problem we were just working on. We\n

three to the n over five to the n.

But this time, we're going to try a limit\n

as n goes to infinity of the ratio of terms,\n

by three to the n over five to the n minus\n

term goes on the top and which goes on the\n

other way, whatever limit we get, when we\n

of the limit, we get When do the ratio this\n

bigger than zero, this ratio is reciprocal\n

than zero. So I'll just stick with the first\n

multiplying, I can cancel my three to the\n

as the limit of one minus n squared over five\n

the same as one minus the limit of n squared\n

is an infinity over infinity form. So using\n

and get the limit of what I get when I take\n

of the denominator, I've still got an infinity\n

use Libby toss rule again, the derivative\n

denominator is ln five times ln five times\n

at two, while the denominator goes to infinity\n

goes to zero. And my final limit is one. Since\n

it's less than infinity. The limit comparison\n

my comparison series, either both converge,\n

is a geometric series with ratio three fifths,\n

the limit comparison test, our given series\n

theorem in action. The limit comparison test\n

terms, if the limit of the ratio of the terms\n

and less than infinity, then the two theories\n

they either both converge or both diverged.\n

when the ordinary comparison test doesn't\n

compare to, but we can't get the inequality\n

I'll prove that the limit comparison test\n

if the sum of the ACE events, and the sum\n

terms. And if the limit as n goes to infinity\n

to L, where L is a finite number that's bigger\n

or both series diverged. To prove this theorem,\n

are true. That is, we'll assume all the stuff\n

to infinity of a sub n over B sub n equals\n

so on, on the x axis, so those are the values\n

B sub n, on the y axis, those ratios are going\n

to infinity. Using more technical mathematical\n

number, epsilon that's bigger than zero, we\n

long as we go out far enough for our values\n

N, such that n over bn is between L plus epsilon.\n

or equal to capital N. In the picture here,\n

for all little ends bigger than or equal to\n

minus epsilon, which is right here, and l\n

Let's pick a small enough epsilon. So this\n

to zero through zero on the y axis, it just\n

owl. Recall that that l itself is a positive\n

interval. That's all positive numbers. For\n

that way, the interval that I'm drawing here\n

is equal to L over two, and our extend up\n

two. So we have that l over two is less than\n

four little n bigger than or equal to our\n

Now, I'm going to multiply all three sides\n

b sub n is a positive number, all the series\n

around the inequalities at all. So we get\n

n is less than three L over two times b sub\n

than zero, since all the ACE events and B\n

this inequality tells us. First of all, if\n

so does the sum of three L over two times\n

series by a constant still convergent. But\n

of a convergent series. So by the ordinary\n

ACE events converges also. Furthermore, if\n

we can focus on this part of the inequality,\n

the ace of ends are bigger than the terms\n

bands must diverged, just using the ordinary\n

about the fact that this inequality only holds\n

capital N. So we could rewrite this argument\n

one to infinity of the base events converges,\n

n to infinity of b sub n, because adding or\n

never changes the convergence status of a\n

so does the sum from n equals capital n to\n

And so then by the ordinary comparison test,\n

n bigger than or equal to capital N, we can\n

capital n to infinity of a sub n converges.\n

of a sub n converges Also, since again, adding\n

beginning of a series doesn't change anything\n

the second part of the argument using precise\n

of the B ends and the sum of the A ends either\n

completes the proof of the theorem. In this\n

using the ordinary comparison test. This video\n

related to convergence for a series. A series\n

of absolute values of the terms converges.\n

to decide which of the following series are\n

convergent. The first series is convergent,\n

r equal to negative 0.8. It's also absolutely\n

absolute values, that's the same thing as\n

of 0.8, which is also convergent. The second\n

of k is not convergent, we can see this by\n

which is less than one. It's also not absolutely\n

values of the terms is just the same as the\n

said diverges. The third series is the sum\n

This is a convergent series by the alternating\n

harmonic series. What about absolute convergence?\n

of the terms, that's just the same thing as\n

So this series is not absolutely convergent.\n

to have a series that's convergent, but not\n

for a moment and try to answer this question.\n

of such a series, the alternating harmonic\n

theories. And in fact, there's a special name\n

A series is called conditionally convergent\n

In symbols, that is, the sum of the a ns converges,\n

a ns diverges. Next question for you. Is it\n

convergent, but not convergent? This is a\n

for a moment and think about your answer.

The answer to this one is no. It's a fact\n

convergent. Let me prove to you why that's\n

that's absolutely convergent. That is the\n

We know that the A ends might be positive\n

the absolute value of a n and the negative\n

that a sub n is either equal to its absolute\n

I'm ready This way with inequalities to help\n

Now, I can't quite use a comparison test here\n

because even though the ace of ends are less\n

series, they're not necessarily positive or\n

test only applies to series whose terms are\n

But there's a nice trick to get around that\n

value of a sub n to all the sides of the inequality.\n

a sub n plus the absolute value of a sub n,\n

value of a sub n. Now, since this sum of the\n

the sum of twice the absolute value of a sub\n

on this inequality, which now does involve\n

I can conclude that the sum of a sub n plus\n

on the ordinary comparison test. But now,\n

of ends can be written as a difference. And\n

the terms of two convergent series itself\n

the difference of the sobs. We've proved that\n

it must be convergent. The fact that absolute\n

handy when you're trying to prove that a series\n

this example, we want to prove that this series\n

and sine make things a little bit tricky.\n

converge since cosine and sine are bounded.\n

like the sum of one over n cubed, which converges\n

But we can't just compare our series to the\n

comparison test like we've done in similar\n

is that our terms are not always going to\n

positive and negative. And so can there's\n

think about absolute convergence instead.\n

which is the same as the sum of the absolute\n

Since cosine of n is between one and negative\n

and negative one, we know that the som cosine\n

to two and bigger than or equal to negative\n

to two or all the way down to negative two,\n

but this is good enough for our purposes,\n

value of cosine n plus sine of n is less than\n

here by n cubed, I have an inequality involving\n

one over n cubed converges by the P test,\n

also converges. It's just a constant multiple.\n

value of cosine n plus sine n over n cubed\n

we know that our series is absolutely convergent.\n

an absolutely convergent series is always\n

series converges. In this video, we saw that\n

it has to be convergent But not vice versa.\n

that can be used to prove that a series converges\n

at the ratio of consecutive terms. To figure\n

about geometric series first, recall it for\n

terms is given by this number R. And if R\n

converges. Why while if R has absolute value\n

series, the ratio of consecutive terms is\n

at the limit as n goes to infinity, of the\n

terms, and if we get a limit of L, which is\n

just like a geometric series. In fact, the\n

sum of the absolute values of the a ns converges,\n

also, if instead, the limit as n goes to infinity\n

terms, is a number l that's greater than one.\n

a geometric series, the series diverges. Finally,\n

ratio of consecutive terms is exactly equal\n

the ratio test is inconclusive. That is, the\n

may diverge. And to figure out, which will\n

argument, let's apply the ratio test to this\n

n goes to infinity of the absolute value of\n

that's the limit as n goes to infinity, have\n

times negative 10 to the n plus one divided\n

times negative 10 to the n over n factorial.\n

formula to get the a sub n plus one term.\n

And I'm going to rearrange my factors. This\n

I've just arranged factors so that similar\n

make it easier to cancel things. Now, n factorial\n

and so on. And n plus one factorial means\n

so on. So if I divide n factorial by n plus\n

will cancel with factors from the denominator.\n

one. Also, negative 10 to the n plus one over\n

10 to the One Power, so I can rewrite my limit\n

negative 10 times one over n plus one, I'm\n

a product of limits. Now as n goes to infinity,\n

which is equivalent to the square of n plus\n

the absolute value of negative 10 is just\n

one over n plus one is zero. Therefore our\n

zero. And since zero is less than one by the\n

absolutely. This video was about the ratio\n

to infinity of the absolute value of the ratio\n

this limit is less than one greater than one,\n

the series converges or diverges. Or if we\n

prove the ratio test for convergence and divergence\nof series.

The ratio test says that for a series, if\n

of consecutive terms is equal to a number\n

absolutely convergent, and therefore convergent.\n

of the ratio of consecutive terms is a number\n

infinity, then the series is divergent. Although\n

to one exactly, or if the limit doesn't exist,\n

be used to establish convergence or divergence.\n

first assume that the limit is less than one.\n

and my absolute value of the ratio of consecutive\n

that settle at a value of L. and this number\n

epsilon, so that when I add epsilon to L,\n

pick epsilon greater than zero, such that\n

of limit, if I go far enough to the right\n

to be trapped in between this epsilon interval\n

there exists a number capital N, such that\n

a limiting to L is between L plus epsilon.\n

I'm going to multiply all three sides of this\n

little m, I'm going to focus on the right\n

I want to prove that my series is absolutely\n

to show that my terms are smaller than things\n

things. This inequality is true for all values\n

to capital N. So in particular, it's true\n

also true when little n is equal to capital\n

N plus one for a little n here. And when I\n

one plus one, which is capital N plus two.\n

essentially, I'm substituting in this inequality,\n

which simplifies to give me the absolute value\n

plus epsilon squared times the absolute value\n

time. Going back to my original inequality\n

two for lowercase n. And stringing these two\n

less than l plus epsilon cube times the absolute\n

the same reasoning shows that the absolute\n

l plus epsilon to the i times the asset value\n

inequalities, I'm gradually building up to\n

sum of the absolute value of a sub capital\n

to infinity. And the second series is the\n

value of a so capital that again, let's start\n

series is a geometric series, where r is equal\n

because remember, we chose epsilon to make\nsure that was

Therefore, this series converges. But that's\n

we can show that series one converges, just\n

that these terms are bigger than or equal\n

test. But series one is just the tail end\n

infinity absolute value of a sub little n.\n

finally many terms added on to a convergent\n

series converges absolutely. This proves the\n

the second part. And let's start by assuming\n

since l is bigger than one, we can pick a\n

from L by epsilon, we don't go as far as the\n

greater than zero, such that l minus epsilon\n

use the definition of limit and a little algebra\n

This time, I'm going to focus on the left\n

epsilon is greater than one. This tells me\n

a sub n plus one is always bigger than the\n

bigger than or equal to capital N. This means\n

n is less than the absolute value of a sub\n

absolute value of capital A sub capital N\n

terms is actually ultimately an increasing\n

of terms cannot converge to zero, limit as\n

a sub little n cannot be zero. And so the\n

little n cannot be zero either. But that means\n

has to diverged by the divergence test, we\n

that the limit of the absolute value of ratios\n

we just used, works almost exactly the same.\n

And here, and just note that there exists\n

bigger than say to for all little n bigger\n

are heading towards infinity, they're certainly\n

gives us the same inequality that we need,\n

we can conclude that a sub n plus one is absolute\n

absolute value. In fact, it's greater than\n

to capital N. And we can make the same conclusion\n

That so that the terms can't go to zero, and\n

test. This concludes the proof of the ratio\n

part of the ratio test by comparing our series\n

the divergent part of the ratio test using\n

of all the convergence tests we've talked\n

in the order that I would try to apply them.\n

Usually it's pretty easy to check if the limit\n

to zero. And if not, you're done because the\n

test can only be used to check for divergence,\n

if the limits of the terms is equal to zero,\n

diverge. The next thing I do is to check if\n

series. Remember, a p series is a series of\n

variable P is some number like two or 5.8.\n

it converges. If p is greater than one and\n

this is the kind of the form a times r to\n

start at zero here. So it's the really the\n

this one's easy to check to because it converges\n

and diverges otherwise. If the series happens\n

is a good one to apply next. Be careful, this\n

If the series is actually alternating. And\n

absolute value of the terms is going to zero\n

the series converges. But if some of those\n

assume that the series diverges. Well, not\n

size doesn't go to zero, then we should have\n

the divergence test were really, really applying\n

size is not decreasing, not even ultimately\n

inconclusive. It doesn't apply, we don't know\n

we have to look for another test. Now if the\n

or alternating series, my go to test is going\n

good for series with n factorials in them

two to the ends and things with sort of geometric\n

good candidates for the ratio test. But be\n

for like all p like series, so series that\n

maybe the square roots of ends and things\n

are not good candidates for the ratio test.\n

save some time by not trying the ratio test\n

candidate or ends up being inconclusive, what\n

So that would be like what I call the ordinary\n

want to compare two series that we know a\n

of so we would generally want to compare to\n

tests are especially good for p like series\n

to figure out what to compare to it's a good\n

highest power terms. One thing you need to\n

is that it only applies to series with positive\n

matter for series convergence. So it's okay\n

terms. But if the terms never become always\n

so the alternating test doesn't apply. We\n

fact that absolute convergence implies convergence.\n

seventh test to try. So you can just take\n

maybe use the comparison test. And if that\n

series will converge also. Another method\n

laws to split up the series. So if you have\n

a geometric series, then a natural thing to\n

and use a different method for each piece.\n

situation, then the psalm also converges.\n

the other diverges, then the sun will diverged.\n

both pieces diverged, then the psalm may still\n

might be cancellation, one piece might be\n

diverging to negative infinity. And that's\n

this stuff has worked so far, I might look\n

to an integral. This is especially handy in\n

So something and also it has to be a series\n

know, something like ln n over n, if you instead\n

pretty easy to compute using use substitution.\n

the integral test. Be aware that the integral\n

can be thought of as the functions values\n

continuous, and decreasing. Last on my list\n

it last only because it's kind of a hassle\n

have some good points. First of all, using\n

actually compute the sum rather than just\n

other tool on this list that will actually\n

geometric series test where we have a formula\n

reason to use the telescoping series is if\n

difference of related expressions. So something\n

minus e to the one over N might be a good\n

like the sum of one over n squared minus one\n

you can rewrite it using the method of partial\n

the telescoping series stuff will help you\n

to know convergence, it'll be a lot easier\n

n squared for this one. So that's pretty much\n

for series. If you want to keep watching,\n

page. Before you keep watching, please take\n

decide which convergence or divergence test\nyou might try.

aware that for many of these series, there\n

because you pick a different one than I do\n

this first example can be conquered using\n

that limit and use lopi talls rule, we get\n

though would be to use the ratio test. Because\nthis is

term that is has a geometric piece as well\n

an alternating series. So my first try is\n

my recollection is it does work to prove convergence\n

the kind that I call a p like series, because\n

the highest power terms, I could compare to\n

cubed rooted or in other words, one over n\n

I'm gonna expect my original one also diverged,\n

test because I don't think the inequalities\n

comparison test. This next one's a perfect\n

This one, the second piece, I could use the\n

And this first one says it has an n factorial\n

This next one's kind of tricky for me. At\n

a candidate for the integral test. If this\n

I might be able to integrate using use substitution.\n

test would be more tricky to do or I might\n

So I'm gonna stay away from that. And I'm\n

this does have a geometric kind of like piece\n

use the integral test here because I know\n

the use substitution u equals ln x, apply\n

that takes practice, the more you do it, the\n

might apply. But a lot of times, there's no\n

it doesn't work, it's where it's inconclusive.\n

you in class. This video introduces some of\n

One of the main ideas behind Taylor series,\n

So suppose we have some function f of x, we\n

and we'd like the approximation to be good\n

that f prime of zero exists, and f double\n

exists at zero, all of its derivatives, we're\n

if we just want to get F value, right at zero

we can approximate F with the constant function,\n

function as being a degrees zero polynomial\n

we can do much better than that, in fact,\n

a degree one polynomial, that's a linear function.\n

zero is a linear function that provides a\n

when x is near zero least that's the best\n

function. The equation for the tangent line\n

of zero times x. This comes straight from\n

the slope of the tangent line, that's the\n

one is just the point zero, f of zero. So\n

times x minus zero, which simplifies to that\n

Notice that the tangent line has the same\n

the same slope as f of x at x equals zero.\n

I'd like to approximate my function f of x\n

the same slope or first derivative, and the\n

zero. It turns out, I can do this with a degree\n

general, a degree two polynomial, also called\n

of x equals c sub zero, plus c sub one times\n

have to figure out what values of the constants\n

my polynomial, agree with my function in its\n

Well, if I want P of zero to equal f of zero,\n

one times zero plus c sub two times zero squared\n

c sub zero had better be equal to f of zero.\n

f prime of zero. P prime of x is equal to\n

using my derivative rules on the equation\n

Therefore, to evaluate p prime of zero, I\n

but this needs to equal f prime of zero, and\n

zero. Finally, I want p double prime of zero\n

find p double prime of x by taking the derivative\n

C two, but that needs to equal f double prime\n

f double prime of zero divided by two. So\n

P has the same value as zero as f forces c\n

that the polynomial and F had the same first\n

one, two equal f prime of zero, and requiring\n

the same second derivative at zero forces\n

zero divided by two. So we've shown that there\n

value first derivative, the second derivative\n

such polynomial, and it's given by p of x\nequals f of zero

plus f prime of zero times x, plus f double\n

Visually, that second degree polynomials going\n

like this. Let's play this game again. But\n

p of x, such that P of zero is the same as\n

as f prime of zero, p double prime of zero\n

third derivative at zero is the same as F,\n

Graphically, that's going to be a cubic polynomial\n

to be a pretty close approximation. We know\n

the form c sub zero plus c sub 1x, plus c\n

need to find the values of all the constant\n

can figure out the values of those constants,\n

of the value of f and its derivatives at zero.\n

get the following expressions. And if we evaluate\n

terms with x's in them vanish. So we get these\n

value, and its derivatives to match the value\n

equations from which we can solve for our\n

one is f prime of zero, c sub two is f double\n

the third derivative of f at zero divided\n

multiplying three times two, and that three\n

of exponents, when I took derivatives, we\n

c sub three is the third derivative of f of\n

So the third degree polynomial that approximates\n

plus f prime of zero times x plus f double\n

f third derivative at zero over three factorial\n

to remember it's the third degree polynomial.\n

degree polynomial whose value add zero is\n

derivatives at zero are the same as f first\n

video and either work out expressions for\n

make an educated guess what those coefficients\n

should get the fourth degree polynomial has\n

and has a final term of f, fourth derivative\n

fourth. If we continue this process forever\n

that match more and more derivatives of f.\n

terms that look like this. This is an infinite\n

notation as the sum from n equals zero to\n

divided by n factorial times x to the nth\n

that the zeroeth derivative means just the\n

and that x to the zero is just equal to one,\n

called the Maclaurin series for f of x. And\n

x centered at x equals zero. Now, so far,\n

its derivatives at x equals zero. What if\n

a, please pause the video and write down what\n

equals A should look like. This is a series\n

and we want all of its derivatives to match\n

series should have a formula similar to the\n

derivatives at a instead of derivatives at\n

powers of x minus a, instead of powers of\n

And to verify that this series really does\n

In this video, we tried to approximate a function\n

derivatives at x equals zero. And we ended\n

equals zero, which we generalized to a Taylor\n

defines power series. informally, a power\n

often the letter X, and it looks like a polynomial\n

we look at the series, the sum from n equals\n

to the n over three to the n minus one, that's\n

that out by plugging in values of n, we get\n

to the minus 1x to the zero is one and three\n

same as three on the numerator. So we can\n

term, when n equals one, is three times x\n

rewrite this as 3x, since three to the zero\n

and we can continue like this. I want to point\n

to the zero is always taken to be one, even\n

end up being zero, and zero to the zero is\n

working with power series, x to the zero for\n

out to one plus five times x minus six, and\n

centered at six because of all the factors\n

centered at a is a series of the form the\n

n times x minus a to the n, where x is the\n

they're constants, called the coefficients,\n

called the center. If I expand out the power\n

looks like this, where c sub zero is the constant\n

taken to be one, even when x equals a. If\n

we just set a equal to zero, we can write\n

form. Sometimes you might see a power series\n

That's perfectly legit. It just means there's\n

think of the constant term as being zero.\n

But it's not considered a power series if\n

in x's in the denominator. That's all for\n

a power series is a series with a variable\n

A are supposed to be real numbers that are\n

That's the only place where I can plug in\n

video explores the question of for what values\n

and for what values of x does it diverged.\n

values of x does this power series converge.\n

it converges for, please pause the video for\n

I'm thinking of the series definitely converges\n

few terms of the series, I get this expression.\n

vanish to zero, except for my constant term\n

to its constant turn. And in fact, this is\n

converge at their center. But let's see what\n

we have many tests for convergence in our\n

test to use to determine where of power series\n

take the limit as n goes to infinity of the\n

terms. For our example, this is n plus one\n

one divided by n factorial times x minus three\n

We get n plus one, times x minus three. Now\n

it's a nonzero number, since I already dealt\n

a nonzero number that stays fixed as n goes\n

infinity. So the absolute value of the product\n

we have, other than the x value of three.\n

infinity, the series diverges. Therefore,\n

x except for three. The only place where it\n

next example, the center of this series is\n

when x equals negative four. Let's use the\n

of x make it converge. So we'll take the limit\n

of a sub n plus one over a sub n. This works\n

two to the n plus one times x plus four to\n

all over negative two to the n x plus four\n

by flipping and multiplying and rearranging\n

of the opposite value of negative two times\n

of this expression doesn't depend on an n,\n

the denominator goes to infinity. Therefore,\n

and larger numbers, and so this limit is equal\n

on x as value It's always zero no matter what\n

one, the series converges for all values of\nx.

Here's our third and last example, in this\n

out what the center is, one thing we can do\n

form by factoring out the negative five, then\n

all raised to the nth power over N, I can\n

n times x minus two fifths to the n over n.\n

to recognize that the center is two fifths.\n

figure out the value of x, that makes terms\n

5x plus two, two equals zero, we need X to\n

must be the center, like we found before.\n

two fifths, for sure, but it might converge\n

test. To find other values of x that make\n

way as usual, by taking a limit of a ratio\n

simplify by flipping and multiplying. And\n

5x plus two times n over n plus one as n goes\n

And negative 5x plus two doesn't depend on\n

value of negative 5x plus two. So by the ratio\n

is less than one, and it diverges when the\n

is inconclusive when the absolute value of\n

so we'll worry about that case later. Let's\n

When the absolute value of something is less\n

the absolute value sign has to be between\n

absolute value inequality as negative one\n

less than one, we can solve this for x by\n

dividing by a negative number reverses the\n

that our series converges for these values\n

value and equality. The one that tells us\n

diverges when the absolute value of negative\n

absolute value of something is greater than\n

absolute value sign has to either be less\n

we can replace our absolute value in equality\n

is less than negative one, or negative 5x\n

these inequalities by subtracting two and\n

the other side. So our series diverges when\n

1/5. That makes sense, it's kind of the opposite\n

an absolute value inequality with the opposite\n

together, we see that our series converges\n

divert On either side of this interval

we still don't know what happens when x is\n

three fifths, since those values correspond\n

So let's turn our attention to the x values\n

down our original power series, it was the\n

If we want to know if this power series converges\n

1/5. This simplifies to the sum of one to\n

series, which diverges. If we plug in x equals\n

to the alternating harmonic series. So it\n

converges when x equals three fifths and diverges\n

that the power series converges on the interval\n

open bracket, to denote that we exclude the\n

there, and the closed bracket, square bracket\n

fifths where the series does converge. I want\n

this example. Notice that the midpoint of\n

at the beginning of the problem, we calculated\n

also two fifths. We'll see in a moment that\n

of convergence is always centered at the center\n

describe the interior of this interval of\n

that center is less than 1/5. Bet is all x\n

interval center. We've seen three examples\n

a very different way. In general, it turns\n

types of convergence that we've already seen.\n

that a series might converge only at its center.\n

converge for all values of x. This is what\n

of these two cases hold, then the only other\n

R, such that R is series converges anytime,\n

the power series diverges for any x values\n

array. In symbols, I can right there exists\n

when the absolute value of x minus a is less\n

of x minus a is greater than R, since the\n

distance between x and a. This was a situation\n

case, we say that the radius of convergence\n

example, we say the radius of convergence\n

say the radius of convergence is our since\n

of the interval, sort of like the radius of\n

the center. Now the interval of convergence\n

the power series converges. So in our first\n

just the number a it's not really an interval,\n

of convergence anyway, in the second situation,\n

from now negative infinity to infinity. And\n

convergence includes this entire interval\n

So our interval could be the open interval,\n

one or more endpoints, so it could be the\n

the left endpoint, or just the right endpoint.\n

using the ratio test to figure out what x\n

stated the fact that there are only three\n

convergence at the center only convergence\n

finite interval centered at the center of\n

of computing the interval of convergence and\n

To compute the radius of convergence and interval\n

by using the ratio test. So we need to find\n

value of a sub n plus one over a sub n, where\n

series, we can compute a sub n plus one, by\n

see an N in this expression. So that's negative\n

to the two times quantity and plus one divided\n

sub N term, which is just negative four to\n

I've just copied from the formula here. Now\n

And now I'm going to rearrange terms so that\n

So I'm going to write negative four to the\n

to quantity n plus one, that's the same thing\n

two n. And then I'll write the N and N plus\n

terms, I get the limit of negative four times\n

one. Now, as n goes to infinity, n over n\n

value of negative four is just four. And the\n

just going to be x minus eight squared, since\n

have our limit. And the ratio test says the\n

than one. So next, let's set four times x\n

solve for x. In other words, x minus a squared\n

quadratic inequality by taking the square\n

like x minus eight is less than plus or minus\n

make any sense. And it's not true. What we\n

x minus eight squared is equal to 1/4. And\n

So now that we have an equation sign, we can\n

x minus eight is is equal to plus or minus\n

minus eight is plus or minus one half. So\n

half, or eight minus one half. That's either\n

to the inequality that we're interested in.\n

equal to 1/4. At x values of 15 halves and\n

eight squared is bigger or smaller than 1/4.\n

so when x is less than 15, half just by plugging\n

x minus eight squared will be bigger than\n

17 halves, say the value of eight, the plug\n

eight squared, which is zero, is going to\n

a value of x over here, maybe something like\n

minus eight squared, that is, again, bigger\n

see that x minus eight squared is less than\n

So by the ratio test, our series converges.\n

ratio test also tells us the series diverges.\n

order other words, x minus eight squared is\n

less than 15 halves are greater than 17 halves.\n

out is what happens at the endpoints of the\n

our series was given by this formula. So when\n

four to the n 15 halves minus eight to the\n

to the n, negative one half to the two n over\n

the n times four to the n times negative one\n

by n, which is the same thing as negative\n

the two n divided by n times two squared to\n

one to the two, n is always equal to one.\n

the n on the denominator cancels with the\n

with the alternating harmonic series, which\n

for x equal to 15 halves. Now at x equals\n

just using 17 halves in place of 15 halves,\n

a positive one half. And now we have a positive\n

everything else works the same. And so we\n

up to the top, we know that the series actually\n

halves and less than or equal to 17 halves.\n

bracket 50 and a half to 17 halves close bracket.\n

length one, because 17 halves minus 15 halves,\n

halves, which is one also, the interval of\n

average of the endpoints 17 halves plus 15\nhalves

is equal to eight. This should come as no\n

centered at eight. So if we draw our interval\n

at eight, and extends out a total distance\n

by half a unit on either side. And so the\n

interval divided by two, or one half. So we\n

the interval of convergence, which was this\n

the problem. In this video, I'll prove some\n

series. My ultimate goal is to prove that\n

convergence, a power series could converge\n

all real numbers. And if these two options\n

such that the power series converges for all\n

and the power series diverges.

his distance from a is greater than our first\n

with this one. If a power series converges\n

B, then it also converges for any x whose\n

of b. To prove this fact, let's assume that\n

to b, that is, the sum of c sub n times beats\n

If a series converges, then the limit of its\n

of the terms is not equal to zero, the series\n

Therefore, by the definition of limit, for\n

N, such that c sub n times b to the n is between\n

for little n bigger than or equal to capital\n

to one, this says there exists a capital n\n

n times b to the n is less than one for little\n

rewrite this statement as the absolute value\n

for little n bigger than or equal to capital\n

value less than the absolute value of b, we\n

x to the n as the absolute value of c sub\n

n, just using algebra, I can rewrite this\n

the n times the absolute value of x over b

For little n bigger than or equal to capital\n

n times b to the n is less than one.

So this expression has to be less than the\n

the absolute value of x is less than the absolute\n

of x over b is less than one. So the series\n

the n is a geometric series, whose ratio has\n

series. Now, the ordinary comparison test\n

of c sub n, x to the n also converges, because\n

less than the terms of our convergent geometric\n

sum of c sub n, x to the n converges absolutely,\n

first statement. The second statement says\n

equal to d for some nonzero number D, then\n

absolute value is greater than the absolute\n

from the first statement, because suppose\n

n diverges. If the absolute value of x is\n

sum of c sub n x to the n converged, then\n

n would have to converge, since the absolute\n

of x. But this contradicts the assumption\n

And therefore, we know that the sum of c sub\n

That's all for the proof of facts about convergence\n

power series as functions. Consider the function\n

n equals zero to infinity of x to the n. This\n

and so on. When we think of this as a function,\n

or input variable. So to figure out the values\n

Please pause the video and calculate F of\n

plus a third squared, and so on, this is a\n

term one divided by one minus the ratio of\n

question, what's the domain of f of x, we\n

values of the input variable x, that give\n

video to write down your answer for the domain\n

n converges when x is between negative one\n

a finite real number as our answer for f of\n

of x. So we don't get a real number answer\n

to negative one or greater than or equal to\n

set of X values for which negative one is\n

notation, we can write this as the interval\n

And in general, the domain of a power series\n

By a closed form expression for f of x, I\n

I can write the sum of X to the N, without\n

series formula, first term is one divided\n

for all x values between negative one and\n

Therefore, I can write f of x as one over\n

Notice that if I just looked at the function,\n

context, it would have domain spanning from\n

interval one to infinity. Because this function\n

not equal to one. So f of x and g of x are\n

domains, but they are exactly equal on the\n

say that the function g of x is represented\n

zero to infinity of x to the n. If we want\n

by this power series 4x between negative one\n

of this series sum of x to the n as a way\n

minus x with polynomials. Please pause the\n

sums, your answers should have Xs. S sub zero\n

one plus x, and so on. s sub n is one plus\n

an nth degree polynomial. In this figure,\n

x in blue. And I've drawn the first partial\n

Notice that these two functions are close\n

farther away from each other when x is far\n

partial sum s two, which is a degree two polynomial,\n

s four. And here I've got partial sums through\n

one minus x is here in blue on mark over

And you can see that these partial sums are\n

original function on the interval from negative\n

example, for x values below negative one

our partial sums deviate wildly from our original\n

this graph, the function one over one minus\n

series, the sum of X to the N is shown in\norange

the blue function is actually obscured by\n

are identical for values of x between negative\n

two functions, as discussed before, is that\n

for all x values except for the x value of\n

even when x values are less than negative\n

series has domain in between negative one\n

those x values. In this video, I talked about\n

x with the power series, the sum from n equals\n

progressions are equal for x values between\n

I also used a graph to give an idea of what\n

in various colors give excellent approximations\n

the interval of X values between negative\n

this gives us a way to approximate this rational\n

The idea of approximating functions with polynomials\n

and again. This video is about rewriting functions\n

that we'll do in this section will be based\n

fact that the sum from n equals zero to infinity\n

x. For x between negative one and one. We\n

three as a power series. And we want to do\n

trick here is going to be to rewrite to over\n

one minus something, then we can treat whatever\n

to get a power series. So that's the idea.\n

And I don't really like the X minus three,\n

x because that reminds me more of one minus\n

two expressions now are equal. This one's\n

by just sticking a negative sign out in front.\n

I've just multiplied my first expression by\n

expression, but I still don't really like\n

x, it'd be nice if I could just divide the\n

leave the expression on change, I'm going\n

the top and the bottom. This gives me negative\n

three, which I can also write as negative\n

if I bring the negative two thirds out front,\n

one minus x over three. Using my geometric\n

two thirds times the sum from n equals zero\n

just plugging in x over three 4x. In this\n

But I'm going to clean it up a little bit\n

negative two thirds into the summation sign\n

n over three to the N. And now I can rewrite\n

one, times x to the n. To figure out the interval\n

are two different approaches that I could\n

using the ratio test. I'll let you work out\n

the radius of convergence is three, and the\n

A second approach to finding the interval\n

how we made the power series.

Our basic template power series was the sum\n

n, which converges when x is between negative\n

for x. Well, this good should converge when\n

In other words, when x is between three and\n

series by negative two thirds. This doesn't\n

interval of convergence for our final power\n

and three, just like you could have gotten\n

let's find a power series representation of\n

to use the geometric series summation formula.\n

like one over one minus something. Well, one\n

minus 5x squared. So if I just wanted a power\n

could do that easily by using the geometric\n

zero to infinity of minus 5x squared to the\n

5x squared for x in this formula. Since I\n

squared, instead, I can just multiply everything\n

make this expression look more like a standard\n

summation sign, I can do that because the\n

do with x. Now I can use my laws of exponents\n

to the n times x to the two n. Now, x times\n

plus one. And so this gives me a good power\n

the problem didn't explicitly ask for it,\n

convergence to see for what values of x

this equation actually holds. I'll use the\n

familiar power series, which converges when\n

with plugged in net Get a 5x squared for x.\n

than negative 5x squared is less than one,\n

is greater than x squared, which is greater\n

around the inequality signs when i divided\n

equals x squared, I can see that x squared\n

corresponding to this section of the graph\n

of X values on the x axis that I'm drawing\n

interval, I just need to find where x squared\n

equal to plus or minus the square root of\n

are the x values in between these two values,\n

less than x is less than the square root of\n

I multiplied everything by x, this doesn't\n

this interval here. In this video, I represented\n

the geometric series summation formula. Although\n

using this formula, some of the techniques\n

my whole power series by x or plugging in\n

can be used in a much broader context to represent\n

series. As we'll see. Up to now, we've only\n

specific series like geometric series and\n

see how to use Taylor series to find the sums\n

Taylor series for arc tan of x centered at\n

to the integral of one over one plus x squared,\n

arc tan of x is to build it up, starting with\n

one minus x is the sum from n equals zero\n

plus x squared is equal to one over one minus\n

x squared in for x in this power series. And\n

of negative x squared to the n, which simplifies\n

negative one to the n x to the two n. Therefore,\n

over one plus x squared dx is going to be\n

at least up to a constant, I can integrate\n

sum of negative one to the n, x to the two\n

a constant. To figure out the constant C,\n

my equation. Since all of my powers of x involve\n

zero, I've got x to the one, so there's always\n

in x equals zero, all of these terms go to\n

plugging in x equals zero gives me zero equals\n

words, the constant is zero. Therefore, this\n

representation of arc tan. Let's take a moment\n

is actually true for. We know that the geometric\n

negative one and one not including the endpoints.\n

for x, I get In the equation that holds for\n

one, which is equivalent to saying that x\n

And when I take the integral of both sides,\n

x between negative one and one. So I'm guaranteed\n

between negative one and one. But in fact,\n

converges at the endpoints of negative one\n

series test sets when we plug in x equals\n

we get an alternating series that converges.\n

from negative one to one, and it's equal to\n

turns out that is equal to our tan

even on the closed interval. In particular,\n

x is equal to one, that I plug into the equation\n

sum from n equals zero to infinity, negative\n

that's just one over two n plus one

I can rewrite that. Now, arc tan of one is\n

of one is going to be pi over four. In other\n

over four, let's write out the first few terms\n

The first term is one, the next term minus\n

a ninth, and so on. In other words, multiplying\n

to four minus four thirds plus four fifths\n

4/11, and so on, if you've ever wondered how\n

we found the sum of kind of a natural series\n

For the next example, let's start by finding\n

at x equals one, we can write out the pattern\n

soon notice that the nth derivative will have\n

one factorial times x to the n. Since we're\n

and get the nth derivative of f at one is\n

n minus one factorial. Therefore, the Taylor\n

equals zero to infinity of f to the n at one\n

n. Since this pattern for the nth derivative\n

first derivative, not with the zeroeth derivative,\n

is just going to be ln of one, which is actually\n

the same pattern. And we have negative one\n

divided by the n factorial times x minus one\n

to the sum from n equals one to infinity of\n

one to the n over n. Since the n minus one\n

leaving just the factor and in the denominator,\n

for ln of x. It's easy to check using the\n

of a convergence of one and so it converges\n

one. It other words when x is between zero,\n

it turns out that this Taylor series really\n

in fact, it converges to ln x on the interval\n

to two. Now, if I plug in x equal to two into\n

I get that ln of two is equal to the sum from\n

the n minus one of two minus one to the n,\n

one. And so I get ln f two is equal to the\n

n. That should be looking familiar to you.\nAnd yes

it's true. This is just the alternating harmonic\n

1/4, and so on. So Taylor series has given\n

and it is ln of two. In this video, we use\n

We also used Taylor series for arc tangent,\n

is actually equal to pi. As you get more familiar\n

calculate the sum of other series by recognizing\n

in a certain value of x into the Taylor series\n

you see the series one, plus one over two,\n

four factorial, and so on, you might recognize\n

for the Taylor series of either dx, in other\n

one which is e. for any function f of x, whose\n

Taylor series for f of x centered at x equals\n

we can write the Taylor series down, doesn't\n

converges to the function we started with.\n

of When can we be sure that the Taylor series\n

series does converge? How good is the approximation\n

other words, how big is the remainder? The\n

always converge to the function that's made\n

of convergence is zero. And sometimes, even\n

or even infinite, the Taylor series converges,\n

an example of the second situation. If we\n

of x is defined as e to the minus one over\n

so that it's continuous at zero to be zero,\n

out the value of g prime is zero, using the\n

zero is the limit as h goes to zero of g of\n

is the limit as h goes to zero

of e to the minus one over h squared minus\n

as h goes to zero of one over h

divided by e to the one over h squared. as\n

is an infinity over infinity indeterminate\n

side, is a negative infinity over infinity\n

use loopy talls rule to replace this limit\n

to a zero over infinity kind of limit

In a similar way, it's possible to prove that\n

and so is the third derivative, and so are\n

if we write out The Taylor series is just\n

Certainly this Taylor series converges for\n

function. And that's different from the function\n

that we started with g of x is not zero for\nany x

except x equals zero. So the Taylor series\n

x equals zero and nowhere else. We found an\n

but not to its function g of x. Fortunately,\n

functions that we typically deal with. To\n

guaranteed to converge to their functions,\n

For a function f of x as Taylor series T of\n

f of x minus T sub n of x, where T sub n of\n

can be expanded out as follows. Previously,\n

we wrote that the remainder was the infinite\n

nth partial sum. The analogous expression\n

series, minus the first terms up through the\n

the remainder to be for Taylor series. Instead,\n

between the function and the first terms up\n

a little bit differently is because for Taylor\n

series converging to its function. And it's\n

series happens to converge to its infinite\n

difference between the function and its Taylor\n

series for f of x converges to f of x and\n

of the remainders is zero in this interval,\n

goes to infinity of the Taylor series equals\n

f of x. If and only if f of x minus this limit\n

limit as n goes to infinity of f of x minus\n

x equals zero. Since the limit as n goes to\n

no ends in this expression. Using limit laws,\n

the quantity f of x minus t and f of x equals\n

that the limit of the remainders is zero by\n

question about when does a Taylor series converge\n

does the limit of the remainders

tailors in equality gives us a bound on these\n

of when the remainders limit to zero. This\n

question of how close is the approximation\n

a function. Here's some details about when\n

capital N, such that the n plus one derivative\n

N, for all X's within a distance d of the\n

a number capital M. And for all X's within\n

lies between negative m and m. So if such\n

n of x of the Taylor series satisfies the\n

x is less than or equal to this bound capital\n

the absolute value of x minus a to the n plus\n

of length two D that we're talking about.\n

the number m can be chosen just to work for\n

can happen if we are able to choose the same\n

our values of little n. In fact, if all derivatives\n

we can guarantee that the Taylor series converges\n

convergence criterion that I'll show you on\n

condition says that if there's a number capital\n

at x is less than capital M, for all numbers,\n

And for all numbers, and then the Taylor series\n

interval. I represent the convergence condition\n

the derivatives are within this bound. So\n

and its derivative lies within this bound,\n

bound. These are not necessarily accurate\n

the idea. So as long as the bound holds for\n

converges to the function. And it's not too\n

From Taylor's inequality. Remember that Taylor's\n

the nth remainder is bounded by M over n plus\n

x minus a to the n plus one. But it's a fact\n

over n plus one factorial, times the absolute\n

to zero. And it's not hard to prove this fact,\n

And using the ratio test, to show that the\n

for the viewer. Therefore, by the divergence\n

has to be zero, which is what we want.

Now, because the limit of this expression\n

of the our ends has to be zero as well, which\n

function. This practical convergence criterion\n

converges to their function. But even if it\n

Taylor series may converge to its function,\n

practical convergence condition to prove the\n

x. Recall the Taylor series for sine x is\n

any nth derivative of x for f of x equals\n

or negative sine x, or cosine of x, or negative\n

take repeated derivatives of sine and cosine,\n

possible answers. Now, since the absolute\n

equal to one and same thing for cosine, we\n

has to be bounded by one, so we'll let m be\n

all real numbers. Therefore, we know that\n

sine of x. for all values of x. We've used\n

equals one to prove this. In this video, we\n

as a difference between the function and its\n

on the size of the nth remainder. It's always\n

times the value of x minus a to the n plus\n

n plus ones. derivative of x for x within\n

for the remainder known as Taylor's inequality,\nwe can show that

nth derivative of x is always bounded by the\n

around a, and for all values of n, then the\n

video introduces the idea of parametric equations,\n

f of x, we can describe the x coordinates\n

third variable t, usually thought of as time.\n

as a separate function of t. This is especially\n

satisfy the vertical line test, and therefore\n

of y in terms of x. A Cartesian equation for\n

only. parametric equations for a curve give\n

usually T. The third variable is called the\n

graph the parametric equations given here\n

finding x and y coordinates that correspond\n

t is negative two, you can calculate that\n

five and y, when you plug in negative two\n

for a moment and fill in some additional values\n

t. Your chart should look like this. And when\n

we get something like this. It says this point\n

two. And this point over here corresponds\n

as time, we're traversing the curve in this\n

this curve, we need to eliminate the variable\n

is to solve for t and one equation, say the\n

x, which means that t is one half minus x\n

for t into the second equation and get y equals\n

which simplifies to the quadratic equation,\n

17 fourths. Let's try some more examples.\n

us draw the familiar graph of a circle of\n

since the equations x equals cosine t and\n

a way of describing the x and y coordinates\n

when t equals zero, our curve lies on the\n

to two pi, we traverse the curve once in the\n

for this unit circle is given by the equation\n

follows from the trig identity cosine squared\n

in X for cosine t, and y for sine t. Please\n

second curve and rewrite it as a Cartesian\n

you see that the graph is again a unit circle.\n

to two pi, we actually traverse the circle\n

this with a double arrow going clockwise The\n

x squared plus y squared equals one. And so\n

graph on the x, y axis. Let's take a look\n

value specified for t here. So let's just\n

Now as T ranges from negative infinity to\n

cosine t, oscillate between one and negative\n

our Y values. So the graph of this curve has\n

which is a sideways parabola. But a parametrically\n

Remember that y is given by cosine of t. So\n

one. And so we're only getting the portion\n

varies from say, zero to pi, I traverse this\n

pi to two pi, I go back again in the other\n

I traverse this parabola infinitely many times.\n

equation x equals y squared, with the restriction\n

seen several examples where we went from parametric\n

start with a Cartesian equation and rewrite\n

y is already given as a function of x. So\n

to just let x equal t. And then y is equal\n

in T for x, the domain restriction in terms\n

terms of t. I call this the copycat parameterization.\n

new variable t, but T just copies, whatever\n

setting x equal to t, then we get 25 t squared\n

for y, we'd have y squared equals 900 minus\n

the square root of this quantity. This is\n

why is that even a function of t here because\n

for a better way to parameterize this curve.\n

this equation is a good candidate for parameterizing\n

both sides of the equation by 900, we get\n

900 is equal to one, which simplifies to x\n

equal to one. If I rewrite this as x over\n

one, then I can set x over six equal to cosine\n

And I can see that for any value of t x over\n

simply because cosine squared plus sine squared\n

x equals six cosine of t, y equals five sine\n

ellipse. As a final example, let's describe\n

For any point, x, y on the circle, we know\n

center of the circle is equal to r. So using\n

root of x minus h squared plus y minus k squared\n

gives us the equation for the circle in Cartesian\n

has radius five, and has Center at the point\n

minus negative three, that's x plus three\n

25. One way to find the equation of a general\n

with the unit circle and work our way up.\n

centered at the origin is given by the equation\n

we want a circle of radius r centered around\n

everything by a factor of R. So we multiply\n

the center to be at HK instead of at the origin,\n

and add K to all our Y coordinates. This gives\n

equations. to match the Cartesian equation\nabove

our same example circle and parametric equations\n

five sine t plus 17. In this video, we translated\n

Cartesian equations and parametric equations\n

circles. This video is about finding the slopes\n

for curves to find parametrically. To find\n

y equals p of x, given an ordinary Cartesian\ncoordinates

we just take dydx or equivalently, we calculate\n

by the equations, x equals f of t, y equals\n

we still want to find dydx. But since our\n

ready access to the y dx. Instead, we'll need\n

are easy to get from our parametric equations.\n

need to use the chain rule. Recall that the\n

y dx times dx dt. So rearranging, we know\n

dx dt. And that's how we'll calculate the\n

formula equivalently as d y dx is equal to\n

use these formulas in an example. For the\n

and drawn below, let's find the slopes of\n

x&y coordinates of zero. And let's find the\n

of the tangent line is given by dy dx, which\n

going to be cosine of two t times two and\n

the ratio, we see that dydx is twice cosine\n

to calculate this slope not when t is zero,\n

01. cosine of t is zero, which is when t is\n

the only two values that work in the interval\n

it's easy to check that when T has these values,\n

to be zero. So we want to calculate d y DT\n

pi over two. plugging into our formula for\n

negative sine of pi over two, which simplifies\n

two, we get twice cosine of three Pi over\n

to negative two. So our tangent lines at the\n

two. Next, let's find where the tangent line\n

be four places. If we set d y dx equals zero,\n

sine of t needs to be zero, which means that\n

two t, two equal pi over two, plus some multiple\n

equal pi over four, plus some multiple of\n

of t in the interval from zero to two pi.\n

four, five pi over four, and seven pi over\n

of these points simply by plugging in these\n

are the XY coordinates of those four points.\n

by parametric equations, the slope of the\n

d y d t, divided by dx dt. In this video,\n

parametrically. Recall that the area under\n

coordinates is just the integral from x equals\n

the integral from a to b of p of x dx. If\n

equations, x equals f of t and y equals g\n

of y dx. But now y can be written as g of\n

Therefore, the area is going to be the integral\n

integrating with respect to t, now, our bounds\nof integration

have to also be t values, I'll still call\n

t values here. It's important to note that\n

words, in between the curve and the x axis.\n

by this list as you figure given by these\n

the area under one segment of the lisu curve,\n

a bye for now, a is equal to the integral\n

we know that y is sine of two t, and dx is\n

sine of t dt. the rightmost point of the section\n

here happens when x is one and y equals zero,\n

equals zero. The leftmost point of the section\n

setting both our equations equal to zero and\n

plus any multiple of pi. So the first time\n

zero is when t is simply pi over

So we'll set our bounds of integration as\n

plugging this information into our equation,\n

zero to pi over two of sine of two t times\n

sign out and use the double angle formula\n

t, multiply that by the sine t dt

we can pull the two out and rewrite this as\n

use substitution will allow us to compute\nthe integral

we get negative two times the integral from\n

integrates to negative two u cubed over three,\n

two thirds. Notice I get a negative answer\n

from right and point to the left endpoint\n

followed the t values in increasing order\n

order in order to make the x values in increasing\n

switching my bounds of integration, or more\n

front and changing my sign. Now I can figure\n

figure just by multiplying by four. In this\n

curve given in parametric equations, is given\n

t and y is g of t, this is just the integral\n

of integration need to be t values.

In this video, we'll calculate the length\n

a warm up, let's calculate the length of this\n

we can just calculate the length of each linear\n

For example, the length of the first segment\n

y one squared, that's three minus two squared\n

minus one squared, which gives us the square\n

length given by the square root of five minus\n

which is the square root of five, we can make\n

third segment and the fourth segment. adding\n

the square root of five plus the square root\n

to approximate the length of any curve by\n

each piece with a straight line and using\n

the line segments. If the curve is given by\n

and y equals g of t. Then we can write each\n

f and g. For example, P of i minus one, we\n

one and the y coordinate g of t minus one.\n

f of t AI, y coordinate g of t AI. If we think\n

T sub i is the time at which we get to point\n

us that the length of the line segment from\n

given by the square root of x two minus x\n

t sub i minus one squared plus y two minus\n

Now, the total length of the curve, which\n

equal to the sum of the length of these segments.\n

sum, but it's missing the delta t. So I'll\n

the denominator by delta t. If I suck the\n

root sign, it needs to become a delta t squared,\n

up my fractions, I can rewrite this. Now there's\n

here looks a lot like a slope. In fact, it's\n

secant line for the function f that we get\n

with respect to t. And similarly, this expression,\n

slope of the secant line, we'd get if we were\n

to t. Because these quotients here, are approximately\n

because of the mean value theorem, I can replace\n

and g prime of t star squared, where ti star\n

exact arc length is going to be the limit\n

goes to infinity. As usual, I can replace\n

where the bounds of integration are the t\n

to the end of the curve. This arc length formula\n

from a to b of the square root of dx dt squared\n

two versions of this very useful

formula for arc length. Now let's use this\n

arc length of this list as you figure

since dx dt is given by negative sine of t.\n

t, our Clank is given by the integral of the\n

cosine of two t squared dt, we do still need\n

terms of t. That will make us wrap around\n

check that when t equals 0x is equal to one\n

The next time that we get to this point, with\n

t to equal one. So the next time will be when\nt equals two pi.

Therefore, our bounds of integration are going\n

up This integral, but it would be very difficult\n

the case with our clients. But we could use\n

approximation of about 9.4. In this video,\n

video introduces the idea of polar coordinates.\n

of describing the location of points on the\n

of its x and y coordinates, those are the\n

polar coordinates, we instead describe the\n

r is the distance of the point from the origin,\n

with the positive x axis. Let's plot these\n

eight here is the value of the radius, and\n

the angle theta. The negative angle means\n

x axis, instead of counterclockwise like I\n

a negative two thirds pi means that I need\n

of for the radius means I need to go eight\n

be around right here. The next point has a\n

angle of positive three pi means that I go\n

x axis, here, I've gone around by two pi.\n

pi. Now the radius of five means I need to\n

puts me about right here. Notice that I could\n

of five Pi, there's more than one way to assign\n

The next point has an angle of pi over four,\n

radius means that I need to jump to the other\n

In other words, instead of plotting the point\n

12, which would be about right here, I go\n

at the same distance from the origin, but\n

over here. Now I could have also labeled this\n

an angle of pi over four plus pi, or five\n

polar coordinates of negative r theta means\n

r, theta plus pi. Adding pi just makes us\n

To convert between polar and Cartesian coordinates,\n

First, x is equal to r cosine theta, y is\n

x squared plus y squared, which means that\n

plus y squared. And tangent theta is equal\n

come from. If we draw a point with coordinates,\n

the height of that triangle is y. The length\n

length R. Theta is the measure of this interior\n

is equal to adjacent over hypotenuse, so that's\n

cosine theta. Similarly, sine theta is opposite\n

means that y is equal to r sine theta. That\n

orien theorem tells us that x squared plus\n

gives us the third equation. Finally, tangent\n

y over x, which is the fourth equation.

To convert five, negative pi over six, from\n

the fact that x equals r cosine theta, and\n

is equal to five times cosine of negative\n

of three over two, and y is equal to five,\n

to negative five halves, did convert negative\n

we know that negative one and negative one\n

fact that r squared is x squared plus y squared,\n

plus negative one squared, or two. Also, tangent\n

over negative one, or one. Now there's several\n

are could be squared of two, or negative the\n

over four, or five pi over four. Or we could\n

answers. But not all combinations of r and\n

with Cartesian coordinates negative one negative\n

use a theta value of say pi over four and\n

get us to the first quadrant. So instead,\n

and five pi over four. Or if we prefer, negative\n

also add any multiple of two pi to either\n

way of representing the point and polar coordinates.\n

and converting in between Cartesian coordinates\n