Uploaded by TheIntegralCALC on 25.07.2011

Transcript:

Hi, everyone! Welcome back to integralcalc.com. Today, we’re going to talk about how we’re

going to use u-substitution in definite integrals. So this is a definite integral because we’ve

got limits of integration from 0 to 4. And we’ve got a problem that we need to use

u-substitution for. And we can treat this a little bit differently than a regular u-substitution

problem where we don’t have a definite integral. We don’t have limits of integration and

we’re going to talk about how to do that. So first of all, let’s identify the u-substitution

that we need to make. We’re going to go ahead and substitute u for x^2 + 9. So I’m

going to assume that you’re somewhat familiar with u-substitution but basically, you identify

u and then you find the derivative of u which is du and taking the derivative of x^2 + 9,

we just get 2x. We go ahead and add dx to this because we just took the derivative.

So we have to add dx. And then we have to divide both sides by 2x because we want to

solve for dx. So doing that, we’ll get that dx is equal to du over 2x. So now, we can

go ahead and substitute these things back into our integral here. I’m going to leave

out the limits of integration for one second but we’ll have x and then the square root

of u because we said that x^2 + 9 was going to be equal to u. And we’re also going to

plug in for dx. So plugging in for dx, we’ll get du/2x. So at this point, we can do one

of two things. You can either use the same limits of integration that you had in your

original problem. You could put in zero and 4 as your limits of integration but what that

means is that you’re going to end up having to plug back in for u. You’ll end up having

to substitute x^2 + 9 back in for u at the end of the problem to make sure that that

matches the limits of integration 0 to 4. Or you can transform your limits of integration

to be related to u instead of x and you won’t have to plug back in for u later.

So let’s actually go ahead and solve it both ways. I’m going to go ahead and put

zero and 4 here and then we’re going to do another integral over here where we change

or limits of integration to be related to u instead. So we’ll have the square root

of u times du/2x. So in order to change our limits of integration with respect to u, what

we need to do is go ahead and plug each of our limits of integration 0 and 4 into our

equation here for u. So when we do that, we’re going to plug it in for x. So when we plug

in zero, our lower limit of integration in for x, we get u = 0^2 + 9 which of course

will just be 9. When we plug in 4 to the equation for u, we’ll get u = 16 + 9 which is going

to be 25. So those are our new limits of integration. So we can go ahead and write those in, 9 to

25. So this is going to be x and this is going to be u. So we can either change the limits

of integration or not. And this is what it’s going to look like. In both cases, x is going

to cancel because we’ve got x in the numerator and denominator. And we can bring the 2 here

out in front of the integral because it’s a constant so we’ll end up with 1/2 times

the integral from 0 to 4 of the square root of u du. And over here we’ll end up with

1/2 times the integral from 9 to 25 of the square root of u du.

So now, we can go ahead and integrate the square root of u. Remember that the square

root of u is the same thing as saying u to the 1/2 power. So when we integrate, remember

that we add 1 to the exponent and then we divide by the new exponent so 1/2 + 1 gives

us 3/2. So we’ll end up with [1/(3/2)][u^(3/2)]. And when we divide 1 by 3/2, we’ll just

flip that fraction on the bottom and we’ll get (2/3)[u^(3/2)]. So that’s going to be

our integral. So we can go ahead and write that in now so we get (2/3)[u^(3/2)] and we’re

going to evaluate that on the range 0 to 4. And then over here, we’ll get (1/2)(2/3)[u^(3/2)]

and we’ll be evaluating that on the range 9 to 25. So we can simplify this somewhat.

These 2s here are going to cancel. Same with this over here. So we end up with (1/3)[u^(3/2)]

on the range zero to 4 and over here, we’ll end up with (2/3)[u^(3/2)] on the range 9

to 25. So you can see that they’re exactly the same, which makes sense. That’s exactly

what we wanted. The difference is that because we didn’t change our limits of integration

over here for x, before we would evaluate on the range 0 to 4, we would need to plug

back in for u. So we can, taking this one over here, plug back in for u and say 1/3

and we said that u is x^2 + 9 so x^2 + 9 and that is raised to the 3/2 power. We can evaluate

that on zero to 4. That would be the exact same thing as this over here, evaluating this

one on the range 9 to 25. Because we already changed our limits of integration to be with

respect to u, we don’t have to plug back in for x.

So you can do it either way. Whatever is easier for you. You can either leave your limits

of integration as they are and then plug back in for u at the end and then evaluate with

those original units of integration or you can change your limits of integration when

you make that substitution for u and then you don’t have to back substitute. You can

just leave it as u and plug in your limits of integration 9 and 25 in for u. Because

I’ve got a whole section of definite integrals about this method here where you use the original

units of integration, I’m going to use this method where we don’t have to back substitute.

We changed our limits of integration. So in the case of all definite integrals, you plug

in this top number here first. So we’ve got 1/3 times 25 to the 3/2 and then you subtract

from that what you get when you plug in the bottom number so 1/3 times 9 to the 3/2. And

we can do this kind of simply here. Since we’ve got basically, 25 to the 3/2 here,

that’s the same thing as saying 25 to the 1/2 raised to the third. So we can break apart

that exponent 3/2 into 1/2 and 3 and 25 to the 1/2 is the same thing as the square root

of 25 so we’re going to do this in two parts. We’re going to deal with the denominator

of our exponent here and take the square root of each of these and then cube them. So the

square root of 25 is 5 but then we still have to cube that. And then taking the square root

of 9, we get 3 but we still have to cube it. Now, 5^3 = 125 so we get 125/3 – 3^3 = 27

so 27/3. And when we subtract 27 from 125, we get 98. So our final answer is 98/3. And

that’s it. You could have plugged x^2 + 9 back in for

u as we did here. We would have plugged 4 first and then subtract it from that what

we got when we plugged in zero and you still would have ended up with 98/3. So it works

either way but that’s it. I hope this video helped you guys and I’ll see you in the

next one. Bye!

going to use u-substitution in definite integrals. So this is a definite integral because we’ve

got limits of integration from 0 to 4. And we’ve got a problem that we need to use

u-substitution for. And we can treat this a little bit differently than a regular u-substitution

problem where we don’t have a definite integral. We don’t have limits of integration and

we’re going to talk about how to do that. So first of all, let’s identify the u-substitution

that we need to make. We’re going to go ahead and substitute u for x^2 + 9. So I’m

going to assume that you’re somewhat familiar with u-substitution but basically, you identify

u and then you find the derivative of u which is du and taking the derivative of x^2 + 9,

we just get 2x. We go ahead and add dx to this because we just took the derivative.

So we have to add dx. And then we have to divide both sides by 2x because we want to

solve for dx. So doing that, we’ll get that dx is equal to du over 2x. So now, we can

go ahead and substitute these things back into our integral here. I’m going to leave

out the limits of integration for one second but we’ll have x and then the square root

of u because we said that x^2 + 9 was going to be equal to u. And we’re also going to

plug in for dx. So plugging in for dx, we’ll get du/2x. So at this point, we can do one

of two things. You can either use the same limits of integration that you had in your

original problem. You could put in zero and 4 as your limits of integration but what that

means is that you’re going to end up having to plug back in for u. You’ll end up having

to substitute x^2 + 9 back in for u at the end of the problem to make sure that that

matches the limits of integration 0 to 4. Or you can transform your limits of integration

to be related to u instead of x and you won’t have to plug back in for u later.

So let’s actually go ahead and solve it both ways. I’m going to go ahead and put

zero and 4 here and then we’re going to do another integral over here where we change

or limits of integration to be related to u instead. So we’ll have the square root

of u times du/2x. So in order to change our limits of integration with respect to u, what

we need to do is go ahead and plug each of our limits of integration 0 and 4 into our

equation here for u. So when we do that, we’re going to plug it in for x. So when we plug

in zero, our lower limit of integration in for x, we get u = 0^2 + 9 which of course

will just be 9. When we plug in 4 to the equation for u, we’ll get u = 16 + 9 which is going

to be 25. So those are our new limits of integration. So we can go ahead and write those in, 9 to

25. So this is going to be x and this is going to be u. So we can either change the limits

of integration or not. And this is what it’s going to look like. In both cases, x is going

to cancel because we’ve got x in the numerator and denominator. And we can bring the 2 here

out in front of the integral because it’s a constant so we’ll end up with 1/2 times

the integral from 0 to 4 of the square root of u du. And over here we’ll end up with

1/2 times the integral from 9 to 25 of the square root of u du.

So now, we can go ahead and integrate the square root of u. Remember that the square

root of u is the same thing as saying u to the 1/2 power. So when we integrate, remember

that we add 1 to the exponent and then we divide by the new exponent so 1/2 + 1 gives

us 3/2. So we’ll end up with [1/(3/2)][u^(3/2)]. And when we divide 1 by 3/2, we’ll just

flip that fraction on the bottom and we’ll get (2/3)[u^(3/2)]. So that’s going to be

our integral. So we can go ahead and write that in now so we get (2/3)[u^(3/2)] and we’re

going to evaluate that on the range 0 to 4. And then over here, we’ll get (1/2)(2/3)[u^(3/2)]

and we’ll be evaluating that on the range 9 to 25. So we can simplify this somewhat.

These 2s here are going to cancel. Same with this over here. So we end up with (1/3)[u^(3/2)]

on the range zero to 4 and over here, we’ll end up with (2/3)[u^(3/2)] on the range 9

to 25. So you can see that they’re exactly the same, which makes sense. That’s exactly

what we wanted. The difference is that because we didn’t change our limits of integration

over here for x, before we would evaluate on the range 0 to 4, we would need to plug

back in for u. So we can, taking this one over here, plug back in for u and say 1/3

and we said that u is x^2 + 9 so x^2 + 9 and that is raised to the 3/2 power. We can evaluate

that on zero to 4. That would be the exact same thing as this over here, evaluating this

one on the range 9 to 25. Because we already changed our limits of integration to be with

respect to u, we don’t have to plug back in for x.

So you can do it either way. Whatever is easier for you. You can either leave your limits

of integration as they are and then plug back in for u at the end and then evaluate with

those original units of integration or you can change your limits of integration when

you make that substitution for u and then you don’t have to back substitute. You can

just leave it as u and plug in your limits of integration 9 and 25 in for u. Because

I’ve got a whole section of definite integrals about this method here where you use the original

units of integration, I’m going to use this method where we don’t have to back substitute.

We changed our limits of integration. So in the case of all definite integrals, you plug

in this top number here first. So we’ve got 1/3 times 25 to the 3/2 and then you subtract

from that what you get when you plug in the bottom number so 1/3 times 9 to the 3/2. And

we can do this kind of simply here. Since we’ve got basically, 25 to the 3/2 here,

that’s the same thing as saying 25 to the 1/2 raised to the third. So we can break apart

that exponent 3/2 into 1/2 and 3 and 25 to the 1/2 is the same thing as the square root

of 25 so we’re going to do this in two parts. We’re going to deal with the denominator

of our exponent here and take the square root of each of these and then cube them. So the

square root of 25 is 5 but then we still have to cube that. And then taking the square root

of 9, we get 3 but we still have to cube it. Now, 5^3 = 125 so we get 125/3 – 3^3 = 27

so 27/3. And when we subtract 27 from 125, we get 98. So our final answer is 98/3. And

that’s it. You could have plugged x^2 + 9 back in for

u as we did here. We would have plugged 4 first and then subtract it from that what

we got when we plugged in zero and you still would have ended up with 98/3. So it works

either way but that’s it. I hope this video helped you guys and I’ll see you in the

next one. Bye!