Original subtitles

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In this video we're talking about how to use you substitution indefinite integrals.

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And then this particular problem we've been given the integral from 0 to 4 of X times the squared of

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x squared plus nine.

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So we need to evaluate this integral and we're going to need to use substitution in order to do it.

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If we use your substitution we have to set you equal to x squared plus nine the value underneath the

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square root.

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And then d you if we take the derivative of you with respect to x we're going to get D U is equal to

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2 x plus 0.

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So just to X and then we multiply by d X. If we divide both sides by 2 x to solve for dx we get d x

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is equal to d u over 2 x.

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And I'm not going to focus so much on the substitution part in particular or why we picked x squared

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plus nine for you.

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What I want to focus on is two different ways to handle the limits of integration because we have a

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definite integral.

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So we can do this one of two ways.

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We can leave the limits of integration as they are these limits of integration are in terms of x.

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You can even think of them as X equals for four and X equals zero.

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So these limits of integration refer to the variable x.

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When we make a substitution and we change this function so that in terms of you instead of in terms

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of x we can leave the limits of integration as they are.

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But we need to remember that there in terms of x so we should write them as X equals four and X equals

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zero.

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And then at the end of the problem we'll have to back substitute to get the function back in terms of

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x so we can plug in these limits of integration that refer to X or alternately when we change the function

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so that it's in terms of you instead of X we can also change the limits of integration so that they

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are in terms of you instead of in terms of x and then we'll be able to evaluate over the interval directly

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because our limits of integration and our function will both refer to you.

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And now become more clear as we work through this and going to work through the problem in two ways.

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The first way is when we leave the limits of integration in terms of x.

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The second way is when we leave the limits of integration in terms of you.

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So in both cases we're going to make the substitution for x squared plus nine equals you and DXi equals

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Do you over 2 x into the integral.

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So in this first integral here what we would need to say is instead of just 0 to 4 we would need to

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say X equals zero to X equals 4 we'll say X times the square root of you since we said you as equals

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x squared plus 9 and X we know is d u over 2 x so we get you over to X. If we simplify this integral

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what we see is that we get X to cancel here and the numerator with X and the denominator.

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This 2 in the denominator is a one half that we can move outside of the integral.

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So we end up with 1 1/2 times the integral from X equals zero to X equals four of the squared of you

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which is the same as you to the 1 1/2 power D.

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You then reduce that our integral our function inside the integral here.

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Everything's in terms of you we have you to the one half do you.

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That's why we had to say X equals zero and exit was for.

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If we're going to leave our limits of integration as 0 and 4 we have to indicate that they refer to

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the variable x and not to the variable you.

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So now we would do is just evaluate this integral we know that the integral of you you're the one half

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is you to the three.

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Hobson we add one to the exponent 1 1/2 plus 1 gives us three halves.

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Then we divide it by the new exponent dividing by three halves is the same as multiplying by two thirds

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so we multiply by two thirds.

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We still have our one half out in front here and then we can say we're evaluating this whole thing over

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the interval X equals zero to X equals 4.

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Because this is still in terms of you we have to still indicate that our limits of integration are with

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respect to x and not you.

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Here we can get 2 in the denominator to cancel with two in the numerator so that's going to leave us

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with one third you to the three halves power over the interval X equals zero to X equals four.

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And now because of limits of integration are still in terms of x but our function is in terms of you

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because we didn't change those limits of integration.

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We need to back substitute to get you in terms of x where we know that you as equal to x squared plus

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nine.

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So we're going to plug back in for you and we're going to say one third time's the quantity x squared

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plus nine raised to the three halves power.

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And now because this function is back in terms of x we can drop the X equals here and just say we're

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evaluating over the interval 0 4 because it's implied that since we have our function here in terms

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of x if we just written 0 to four then these limits of integration are with respect to x in the same

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way that we just had limits of integration zero to 4 on the original integral which was in terms of

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x and then we would just evaluate over the interval so we would plug in for first and we would get one

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third times four squared which is 16.

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Sixteen plus nine is 25.

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So we would get 25 to the three halves power minus whatever we get when we plug in zero sum minus one

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third times zero squared is zero plus nine is nine.

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So 9 raised to the three halves power we can rewrite this we can say ONE-THIRD 25 to the three halves

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is the same thing as 25 to one half then raised to the third because only of exponents like this we

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multiply the exponents together and one half times three gives us three halves which is our original

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exponent down here.

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So we can split up the xponent this way and that's useful because then 25 to the one half is the same

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as the square root of 25 so we can call this value 5.

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But anyway we're going to say minus one third and the nine to the three halves is the same as nine to

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the one half raised to the third.

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So then we say one third the square root of 25 is five five to the third power is 125.

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Here we have minus one third to the one half of the squared of nine is three three to the third is twenty

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seven.

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So we end up with 125 over three minus 27 over three which leaves us with ninety eight over three.

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So we can go ahead and say Here our answer is ninety eight over three.

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And now if we we're going to do this where we change our limits of integration so that they are in terms

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of you will be able to avoid this back substitution step that we did at the bottom here.

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So here's what that would look like.

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We would make the same substitution we would have the integral here we would keep our X and then we

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would say the square root of instead of x squared plus nine we said you is x squared plus 9 so we get

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the squared of you.

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And then we replace X with do you over to X so do you over to X the same thing that we had here in our

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first integral.

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But our limits of integration we want to change them so that they're in terms of you instead of in terms

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of x the way that we do that is we want to get a value for you.

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When x is equal to zero our lower limit of integration.

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So all we do is we take X equals zero and we plug it into our equation here for you.

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So we get you is equal to plugging zero and for x we get zero squared plus nine so zero squared plus

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nine which is of course equal to nine when we plug in X equals 4 we get you is equal to 4 squared plus

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9 or you is equal to 9:16 which is 25.

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So these limits of integration are with respect to the variable you.

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They're the same as our original limits of integration zero to four except they apply to the variable

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you instead of the variable x.

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So what we can say then is 9 to 25 because when we cancel here the X in the numerator with the X in

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the denominator and we pull the one half out in front.

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We have 1 1/2 times the integral from 9 to 25 of the squirt of you which is the same as you to the 1

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1/2 power times do you.

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So now notice here compared these two integrals that are side by side.

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They're exactly the same.

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The only thing that's different is the limits of integration because this function over here on the

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right is in terms of you and these limits of integration 9 to 25 are in terms of you we can just go

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out and state the limits of integration 9 and 25 because when there's no designation you equals or X

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equals.

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It implies that these limits of integration are with respect to the variable that's inside the integral.

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But here because these limits of integration are with respect to x and the function was in terms of

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view we had to say X equals to indicate that these limits of integration did not match the function

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the way we took care of that was by back substituting for you down here we took out you and we put back

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in x squared plus nine.

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So our limits of integration would match.

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But here because we've changed the function and the limits of integration we won't have to do the back

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substitution step.

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So we'll just integrate we'll keep the one half out in front and again we'll get you to the three halves

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when we add one to one half then we'll divide by the new exponent three has which is the same as multiplying

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by two thirds.

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And we're going to be evaluating this on the interval 9 to 25.

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So again we cancel and we end up with one third times you to the three halves over the interval nine

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to 25.

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Now if you'll notice here we're going to end up in the same situation we ended up in before we're going

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to play in our upper limit of integration first 25.

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So we'll get 25 to the three halves which is the same as 25 to the one half raised to the third power

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again.

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So 25 to one half is the squared of 25 which is 5 5 raised to the third power or five cubed is 125.

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So we end up with one 25 times one third or 125 over three.

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Then we want to subtract whatever we get.

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We plug in nine.

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So again nine of the three halves power is the same as the square root of 9 which is three raise to

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the third which is 27.

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So we end up with 27 times one third or 27 over three and you can see that again we end up with 98 over

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three which has the same value that we got when we did it the other way here when we didn't change the

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limits of integration.

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But we back substituted instead.

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So the conclusion is that it really doesn't matter which way you evaluate a definite well with your

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substitution you can leave the limits of integration in terms of the original variable as long as you

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back substitute for the original variable.

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At the end of the problem before you plug in your original limits of integration or if you prefer you

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can do this step where you find new limits of integration and then you just evaluate over those new

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limits of integration and you don't have to do the back substitution.

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So it's really up to you whatever you're more comfortable with.

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Whatever's easier is fine because they're both going to give you the same answer.