Find an antiderivative of 3(3x + 1)5.
First we find the indefinite integral of 3(3x + 1)5. Let u = 3x + 1 and u' = 3. Then
This is the family of all antiderivatives of 3(3x + 1)5. To get one antiderivative, we pick a value of C. The simplest choice is C = 0. This gives us the antiderivative
First we use integration by substitution to find the corresponding indefinite integral. We did the work for this in a previous example:
This means is an antiderivative of 3(3x + 1)5.
First we use substitution to evaluate the indefinite integral
u = 4x2 + 1
u' = 8x.
We need to introduce a factor of 8 to the integrand, so we multiply the integrand by 8 and the integral by .
Letting C = 0, the simplest antiderivative of the integrand is
We use this antiderivative in the FTC.
If we're going to give an exact answer (which is advisable), it doesn't get any nicer than this.
Since the variable of integration is x, the limits of integration are values of x. The integral
is read "the integral of 3(3x + 1)5 as x goes from 0 to 1." To be explicit, we could write the integral as
If we take
u = 3x + 1
du = 3dx
and start the substitution, we run into a problem:
This could be read " the integral of u5 as x goes from 0 to 1," which doesn't make sense!
The problem is that the limits of integration are values of x, but now the variable of integration is u. We can fix this by changing the limits of integration to values of u. We know that
u = 3x + 1.
So when x = 0, the corresponding value of u is
u = 3(0) + 1 = 1.
This lets us change the lower limit of integration from a value of x to a value of u:
When x = 1 we have
u = 3(1) + 1 = 4.
This lets us change the upper limit of integration:
Now that we have everything in terms of u, there's no reason to go back to x. We can evaluate the integral we have, with no x in sight:
It's reassuring that we got the same answer as when we did this problem the other way!
Use Way 2 to evaluate
u = 4 – x
du = (-1)dx
Let's label the limits of integration as x-values so we don't mess up.
We're not done with the substitution yet. We still have to change the limits of integration so we have u-values instead of x-values. When x = 2,
u = 4 – (2) = 2.
When x = -1,
u = 4 – (-1) = 5.
Here's what the integral looks like after the limits of integration are changed:
Now that we've changed the limits of integration, we're done with the substitution. From here, we can use the FTC to integrate.
We conclude that
Be Careful: When using substitution to evaluate a definite integral, you aren't done with the substitution part until you've changed the limits of integration.
Make it rain.