Original subtitles

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Today we're going to talk about how to use properties of integrals to simplify a definite integral to

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complete this problem.

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Identify the properties we need.

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Use the properties to simplify the div an integral and then evaluate a definite integral.

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In this particular problem we've been asked to use properties of integrals to evaluate the definite

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integral of four plus three x squared from zero to 1.

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So in order to complete this problem we're going to be looking at a couple of different properties of

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integrals.

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The first one being that if you have an integral that is the sum or difference of multiple functions

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you can split them apart into separate integrals.

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So for example we have two different pieces of our integral here.

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One function which is four and the other function which is 3 x squared because they are added together

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if they were multiplied or divided one by the other we wouldn't be able to do this but because they're

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added together and if they were subtracted together from one another we could do the same thing.

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We can split them apart into two separate integrals.

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So the integral from zero to one of four plus three squared VX is the same thing as the integral from

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zero to one of four X plus the integral from 0 to 1 of 3 x squared x.

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So we just pulled them apart.

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And again the goal of using properties of integrals is to simplify integrals as much as possible.

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So that's our first property our second property is that we can pull constants out in front of our integral.

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So for example whenever everything is multiplied together inside an integral as we have here right we

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have three times x squared times x.

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All three of these are multiplied together whenever everything's multiplied together or divided.

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We can pull a constant out in front of the integral so we have a constant coefficient of three that

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can come out in front.

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Here we have the same thing in our first integral with four times dx because we have the constant coefficient

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of 4.

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It can come out in front.

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That's to be distinguished from what we had in our first step here we have four plus three x squared

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in this form four plus three x squared.

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We can't pull a four out in front or pull a three out in front because these two terms are added together

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only when everything is multiplied together.

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Can we pull out the constant coefficient.

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So to simplify will pull four out in front and we'll get four times the integral from 0 to 1 of X plus

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pulling our 3 out from three times the integral from 0 to 1 of x squared dx.

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And that's as far as we can simplify our integrals using the properties of integrals.

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So now we need to go ahead and evaluate our definite integrals which means we're going to be evaluating

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these on the range 0 to 1.

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So to take the end a girl of Dayaks the integral of the X is just X so we'll get four times X and we'll

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be evaluating X on the range 0 to 1.

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And over here we get plus three.

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The integral of x squared when we have a power like this x squared.

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We just add 1 to the exponent so two plus one gives us three and then we divide the coefficient which

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right now happens to be won by the new exponent 3.

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So we get ONE-THIRD like that and we're going to be evaluating one third x cubed on the range 0 to 1.

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Now before we evaluate either of these and the range 0 to 1 we can just go ahead and simplify as much

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as possible.

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So we'll get 4 x evaluated on 0 to 1.

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We'll get as you may notice three and three Here is a Cancel from the numerator and denominator so be

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left with plus x cubed from 0 to 1.

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And here's another property we can use because we're evaluating both of these terms on the range 0 to

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1 for x and x cube.

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We can do them separately or we can combine them so we can say actually for X plus x cubed all evaluated

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from 0 to 1 and if we want to remember that we're evaluating the whole thing we can put parentheses

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around it like this.

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So now what this tells us to do so when we have a function here and then we're saying we're evaluating

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from 0 to 1.

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We're going to plug the top number one in first.

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We're going to plug one in for x the 4 x plus x cubed and we're going to subtract whatever we get when

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we plug in 0 to a plug in one first will get four times one plus one cubed and then we always subtract

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whatever we get when we plug in the bottom number.

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In this case zero.

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So four times throw plus zero cubed like this and now we've plugged in one and zero.

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We just simplify as much as possible.

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So we'll get four plus one every year minus zero plus zero.

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And obviously this is all going to go away and we'll just be left with five and that's it.

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Five is the integral of four plus three x squared on the range 0 to 1.