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
Example 2
Evaluate .
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.
We're shooting for a definite, though. We need to the bounds into this antiderivative and then take the difference.
Example 3
Evaluate .
First we use substitution to evaluate the indefinite integral
.
Take
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.
Example 4
Use Way 2 to evaluate
We take
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, we aren't done with the substitution part until we've changed the limits of integration.