Original subtitles

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Today we're going to talk about how to use integration by parts three times in order to find the antiderivative

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of a function to complete this problem will understand our integration by parts formula and then use

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it three times in a row to simplify and evaluate the integral.

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In this particular problem we've been asked to evaluate the integral of x cube times either the x dx.

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The first thing that we should recognize is that we have essentially two functions inside of arm integral.

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Here we have x cubed and we have e to the X and they are multiplied together.

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So the first thing that we should think is to try to use integration by parts to evaluate this integral.

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Whenever you have two functions like this and they're multiplied together integration by parts should

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be your first thought.

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So I'd gone ahead and written the integration by parts formula over here on the right and what it tells

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us is that when we have a a when we have an integral when we are when we're asked to take the integral

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of two pieces that are multiplied together you and DVH when we want to take the integral of something

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in that form the formula will use to do it is you Times V minus the integral of V times d u.

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So essentially what we need to do is we need to identify which part of our integral is you and which

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part is DVH.

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And then once we've identified you MTV will take the derivative of you to find the you and the integral

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of DVH to find V and then we have u d u V and D and those four components we can use to plug into our

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formula and build out the right hand side over here.

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So let's go ahead first and identify you and DVH you.

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Is the part you want to focus in on first.

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So to find you you should be looking for something that becomes simpler when you take its derivative.

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So in this case we have X to the third and E to the x which are two candidates.

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If we take the derivative of each of the X we're going to get it the X the exact same thing and we really

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haven't made any progress which should immediately rule it out as a candidate for you.

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That means that you will be equal to x to the third.

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The other option.

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So we set you equal to x cubed.

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That means that the rest of what's inside the integral here has to be set equal to dvh because we have

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two components here we have you and we have DVH.

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So because we already set you equal to x cube that means that divi has to be equal to everything else

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including the X.

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So we'll set divi equal to each of the X Dyaks.

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Now as I mentioned before now that we've identified these two pieces we need to go ahead and take the

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derivative of you to get the U.

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So the derivative of x cubed is 3 x squared.

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And it's important then that we add a D X to this since we took the derivative.

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We're going to take the integral of DV to find V and the integral of the x is simply E to the X..

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The same thing.

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So here with you you're going in this direction taking the derivative with divi you're going in this

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direction and taking the integral.

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Now that we have those four components we can go ahead and plug them into our integration by parts formula.

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So you'll notice that we have first you Times V.

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So going to be multiplying you and B together.

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So go ahead and say that are integral x cubed either the x dx our original integral is equal to this

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right hand side here.

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So you times being we know to be x cubed times each of the X minus the integral of V times you saw say

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the integral of V times d you the times c you will give us 3 x squared E to the x dx and now you can

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see we still have an integral that's too complicated to integrate on its own.

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But it is different than what we originally started with instead of having executed.

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We've got 3 x squared.

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So notice that the degree of that term went down by 1.

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Right we had to do a third degree power function here and now we have a second degree power function.

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So what we're going to have to do is use integration by parts again to try to attempt to reduce our

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integral to something simpler.

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So we'll again say this time that you is equal to 3 x squared we'll say that DVM is equal to everything

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that's left over which will be e to the X.

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The x will take the derivative of you to get d u.

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And that will give us 6 x times DX right we just used power rule to find the derivative of 3 x squared.

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So 6 x x and we'll take the integral of DVH the integral of each of the x is simply E to the X and now

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we'll plug these four components into our formula again for this integrals so what we're left with here

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is is x cubed either the X Kari's down so x cubed into the X minus.

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And now this integration by parts operation we're going to perform represents this integral right here.

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So what we'll do is we carry down our minus sign here.

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We're going to draw a big parentheses and plug these four components into the right hand side of our

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integration by parts formula and put that in parentheses here.

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So you at times V will be 3 x squared each to the X minus the integral V times d u which will give us

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6 x to the x x.

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Right we have the 6 x here.

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The either the X here and the X here are just reordered the terms like that.

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So that's our entire integration by parts formula is right hand side over here that replaced this integral

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right here.

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So when we simplify this we'll get x cubed into the X minus 3 x squared E to the X the negative sign

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here and here will cancel out and we'll get a positive.

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Now we can go ahead and pull this 6 here out in front of our integral because it's a constant coefficient

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so we can say plus six times the integral of x E to the x dx.

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It's not really crucial that we pull that 6 out in front but we can do it just a simple fire integral

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a little bit.

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So now notice that we still have a function that is too complicated for us to integrate as is will in

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fact need to use integration by parts.

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A third time.

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But the trend that we should take note of is that we started out with a third degree power function

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here.

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We use integration by parts and we reduce it to a second degree power function and now essentially we

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have X to the first power which is of course a first degree power function.

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So we've gone from three to two to one.

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So if the trend holds and we use integration by parts again this term here should drop out and we should

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be left with something that we can integrate.

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So if we use integration by parts again and we say that you is equal to x and that DVN is equal to each

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of the x dx and we take the derivative of you the derivative of X is just one.

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So we would get 1 times DX which of course would just be the X and then we integrate divi to find.

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So we get we equals.

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It is the X.

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Now plug those four components into our integration by parts formula to replace the integral that we

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had left over here.

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So what we get for our integral function will carry over this excuse me the x this whole thing here

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will bring up so x cubed you to x minus 3 x squared each x plus 6.

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And now here's where we draw our big parentheses and in place of this entire integral right here or

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place that with our the right hand side here of our integration by parts formula plugging in these components

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that we have.

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So remember you Times V.

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So you Times V will give us X E to the X minus the integral of V times d u.

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So v times d you here will just give us e to the x dx and notice now that our x term dropped away and

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we're just left with either the ex insider integral so it looks like we're finally going to be able

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to integrate after using integration by parts for the third time.

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So we'll go out and simplify and we'll get x cubed to the X minus 3 x Square-D to the X plus 6 x 8 to

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the X..

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Notice now we have to distribute 6.

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So we'll get minus six times the integral of either the X dx.

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We know that the integral of E to the x is simply E to the X. So I'll go ahead and take the integral

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and we'll get 6 x either the X minus six times each to the X.

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At this point because we finished our final integral we want to make sure to add the constant of integration

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plus C to account for that that constant that could have dropped off.

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So we have the constant integration there and now we could leave this as our final answer.

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But why don't we go ahead and factor out in either the X because we have an either the X and each one

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of our terms here and we'll simplify if we factor out the either the X so our final answer for the integral

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will be either the X times x cubed minus 3 x squared.

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Plus 6 X minus 6 plus C.