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 = 4x^{2} + 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.