U-Substitution in Definite Integrals
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!