Original subtitles

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In this video we're talking about how to use tabular integration to evaluate an integral and tabular

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integration is similar to integration by parts.

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It's just an alternate method.

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The tricky thing is that integration by parts.

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If it's applicable to a certain integral it'll work every time.

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Tabular integration is a more specific case and it doesn't work every time like integration by parts

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well.

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So you can't always use it but where you can use it.

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Oftentimes it can be a lot faster than integration by parts.

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So what I want to do here is take this integral we have the integral of x squared either the X x.

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And we've been asked to use tabular integration to evaluate this integral.

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Now we could use traditional integration by parts which is what I've done here.

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What I want to show you is how to use the tabular integration method to get to the exact same answer

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and you'll see that we end up with the same answer.

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So before we get tabular integration Let's review really quickly what we would do if we used integration

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by parts to evaluate this integral.

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First of all we would notice that we have the product of two functions so we have x squared multiplied

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by either the X..

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So those are two separate functions they're multiplied together.

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So this is a good candidate for integration by parts.

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And what we want to do first is identify you and divi inside of our integral.

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So in this case we would say you equal to x squared which means DV has to be everything else in the

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integral which would be either the X X so we say us x squared.

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So DV is either the X D x then we take the derivative of you get d you.

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So we would say D-New is equal to the derivative x squared which is 2 x and we have that x and then

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we take the integral of DV to get V.

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So the integral of each of the x x is just either the X then we use this formula here this is the integration

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by parts formula it tells us that if we have the integral of u times divi in a row we assigned you and

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divi to values and our integral.

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So we do have the integral of U time Stevie when we have that integral it's equal to u times V minus

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the integral of VDU.

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So when we Bilour and roll this on the left hand side here is our original integral on the right hand

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side.

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Is this right hand side of the integration by parts formula.

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So we take you Times V or x squared times either the X and we get X squared times the X minus the integral

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of VDU from our formula.

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So we take V E to the X and D u to X X and we put that inside our integral.

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But now we're at a point where we need to use integration by parts again to evaluate this integral.

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The remaining integral here is a little simpler than what we started with.

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Because instead of x squared we have a first degree x variable but we still have to use integration

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by parts a second time.

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So when we do that we say you is 2 x.

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So DV has to be everything else either the X so we have u equals 2 x D-B equals either the X X we take

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the derivative of you and we do is two times d x we take the integral of DV and we get v is equal to

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x and then we use these values here.

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And the right hand side of our integration by parts formula we replace just this integral here.

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So we replaced just this integral right here with the right hand side from our integration by parts

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formula.

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And that's where this comes in here because we left in our answer we left the X square either the X

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right here X square either the X we left or minus sign.

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But then this integral gets replaced by the right hand side of our integration by parts formula.

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So we have u times V or 2 x times each of the X and then minus the integral of the times.

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D u so V and D U.

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We get to either the X and X inside of our integral.

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Then we distribute this negative sign across these two terms.

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We distribute the negative sign here and we distribute the negative sign here and we end up with a minus

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2 x either the X.

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And then this negative cancels with this negative.

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We have a positive.

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We pull that two out in front of the integral.

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So we just have the integral of either the X X and then we just have to take this integral and the integral

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of each of the X is either the X so we end up with this plus two times either the X and we had C to

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account for our constant of integration.

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So that's our final answer using integration by parts.

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And it took us a little while and it was a little complicated.

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So this is a perfect candidate for tabular integration.

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You only want to use tabular integration when ever one of the functions inside of your integral So in

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this case we have x squared and you do the X when one of the functions if you take its derivative over

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and over and over again the derivative will eventually go to zero.

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And we can see that that would be the case with x squared because if we take the derivative x squared

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we get to X.

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If we take the derivative again we get two.

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If we take the derivative again we get zero.

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So the derivatives of x squared eventually go to zero which means that we could probably use tabular

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integration to evaluate this integral.

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So the first thing we want to do in the same way that with integration by parts we assign you and D.V.

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to values in our integral The first thing we want to do is assign f of x and g of x to the functions

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inside of our integral.

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So essentially with tabular integration instead of the integral of you DV we're looking at the integral

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of f of x x x.

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So what we want to do is figure out whether f of x is x squared or the X and then whether G of x is

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x squared or either the X you want to sign f of x to the function whose derivatives go to zero which

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means we would say that f of x is going to be equal to x squared.

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That means the other function is going to be g of X will say g of x is equal to E to the X and now from

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this point the steps might seem a little foreign but they're actually really simple.

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So we just write those functions right underneath these values here.

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We're going to create a little table so forever x we're always going to take successive derivatives

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until we get to zero.

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So we're going to say the derivative x squared is to X the derivative to x is to the derivative of two

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is zero.

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So we always take successive derivatives of f of x.

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We're going to take successive integrals of each of the X so what we have here we have either the X

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if we take the integral we get either the X..

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If we take the integral again we get either the X..

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If we take the integral again we get either the X and we just drop this down all the way until we hit

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this last row here where we go 0 4 F of X..

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Now we're going to do something a little strange.

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We're just going to take the value from the first row in the 5 x column and we're going to hit with

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a line.

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Connect it to the value in the second row in the G of X column and then we're just going to draw parallel

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lines here so this 2 x is going to get connected to this value.

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The twos going to get connected to this value and then we're done.

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When you hit that zero term you don't have to connect that to anything in the G of X column.

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So you just start with the value in the first row in the X column and you connect to the next lowest

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value in the G of X column until you're done here.

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Then the last thing you do is you add signs to this f of x column and you always start with a positive

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sign of new alt..

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So we're going to do a positive sign a negative sign and a positive sign and you want to keep doing

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that all the way down until you get here to zero.

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You don't have to assign a sign to zero just to the non-zero term.

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So you always start with positive and you just alternate positive negative positive negative positive

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negative.

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As long as it takes for you to get down to this zero value and now this is actually all we need to do

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to get directly to our final answer because here's what we do we start with this value in the first

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row for f of x and we just multiply it by the value it's connected to.

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So we say x squared times each of the X and the sign is positive.

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So in other words positive x squared either.

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So we start with positive x squared B to the X..

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Then we go to the next row we have a negative 2 x either the X so minus 2 x either the X then we have

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a positive so plus two times either the X or plus two times either the X and then we can't forget to

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add our constant of integration.

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See but that's our final answer.

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And if you notice here this is exactly the same answer that we got when we used integration by parts.

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They match exactly.

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So tabular integration is going to work for you in place of integration by parts.

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Whenever you have one function inside of your integral whose derivatives go to zero if that's the case

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you can use tabular integration to find the correct answer.

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And especially for a problem like this one where you have to use integration by parts multiple times

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to get to the answer tabular integration can sometimes be a lot faster.