Original subtitles

1

Get a room to talk about how to evaluate the integral of an even function to complete this problem will

2

confirm their function is even and then simplify and evaluate the integral.

3

In this particular problem we've been asked to evaluate the integral of x to the 6 plus one on the range

4

negative to the two.

5

And the first thing we should notice with this problem is that we have the definite integral range here

6

negative to the positive too which means that we're being asked to evaluate a symmetric area about the

7

y axis so we have here the x axis and the y axis and we want to evaluate from negative to here to positive

8

2.

9

So the same distance from negative to the y axis as it is from the y axis the positive 2.

10

So that's always our first give away when it comes to simplifying this as an even function.

11

The next thing that we want to look for is to test whether or not our function actually is even if our

12

function is in fact even and we're being asked to evaluate the same distance to the left of the y axis

13

as we are to the right of the y axis and we can take advantage of a specific kind of simplification

14

when it comes to evaluating the definite integrals of even functions.

15

So we want to test to see whether or not our function is even.

16

So first of all let's call our function f of x so we just take everything that's inside our integral

17

here and say F X Ziegel to x to the sixth plus one.

18

Now in order to test whether or not the function is even we want to go ahead and plug in negative x

19

for x into our function.

20

So everywhere we see X and our function will plug in negative X will get negative x to the sixth plus

21

one.

22

And when we simplify this we'll see that we actually get X to the sixth plus one whenever we raise negative

23

x to an even power.

24

We're going to end up with a positive sign here and just x to the sixth.

25

So what we see is that we actually end up with the exact same function as we did when we started with

26

our original function.

27

So because these two are the same.

28

The original function and the function we got after plugging in negative X because those two are the

29

same.

30

That means that f x is even.

31

So we can go ahead and say F X is an even function.

32

So now that we've proven that f of BAC's is even we can go ahead and evaluate the integral and take

33

advantage of the simplification that we were talking about.

34

So because the function is even and because we're evaluating from negative to the positive to the same

35

distance on the left of the y axis as on the right of the y axis instead of evaluating from negative

36

to the positive to we can go ahead and evaluate from 0 to 2 and just multiply by 2.

37

And the reason is because since we know that the function is even that means that we know that the function

38

is symmetrical about the y axis and if it's symmetric about the y axis that means we're going to have

39

the same area in this range as we are in this range.

40

And so we can just take the area on this range here and multiply it by two and it's the same thing as

41

taking the area from negative to the positive two.

42

So we'll simplify the integral and call it 0 to 2 multiplied by 2 x to the sixth plus 1.

43

Now we can just go ahead and integrate.

44

So we'll get 1 7 using power rule here.

45

1 7 x the seventh plus X will be our integral and we're going to evaluate that on the range 0 to two.

46

Remember that with definite integrals we always want to plug in the top number first and then subtract

47

whatever we get when we plug in the bottom number.

48

So let's go ahead and plug two in.

49

So we'll get one seven times to the seventh power plus two minus two times 1 7 times 0 to the seventh

50

plus 0.

51

And obviously we can see here this entire second term will go away because we're just going to have

52

zero.

53

And in the first term here we'll have 2 times one seven times two to the seventh which is 128 plus two.

54

Simplifying further will get 2 times 128 over seven plus and we can find a common denominator here and

55

called to 14 over 7.

56

That gives us 2 times.

57

Over 7 which of course just simplifies to to 80 for all over seven and that's it.

58

That is the value of the integral of x 6 plus one from negative to positive to.