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!