In supplemental notes 21, I stated the Fundamental Theorem of Calculus, Part 2, because it dealt with the evaluation of definite integrals. But, I have not stated the first part of this theorem. The first part deals with how the derivative of a function is related to the integral. I will now state this part of the theorem.

FACT:

If f is continuous on [a, b], then

has a derivative at every point of [a, b] and

Now let us use this part of the theorem to find some derivatives of integrals.

EXAMPLE 1:

SOLUTION: To find the derivative of the above integral, you will first take the integrand and replace t with the upper limit of integration (in this case x). That will be the first part of the derivative. Next, you will take the derivative of the upper limit of integration with respect to x. Multiply this to the integrand.

EXAMPLE 2:

SOLUTION: In the integrand, I will replace t with sin x. Remember that you replace t with whatever the upper limit of integration is. Then remember to take the derivative of sin x.

EXAMPLE 3:

SOLUTION:

Remember that 1 + tan ² x = sec ² x.

EXAMPLE 4:

SOLUTION:

Remember that when you take the square root of a square it is like saying you want the absolute value.

Now let me do some examples of evaluating definite integrals.

EXAMPLE 5:

SOLUTION:

EXAMPLE 6:

SOLUTION:

EXAMPLE 7:

SOLUTION:

EXAMPLE 8:

SOLUTION:

EXAMPLE 9:

SOLUTION:

EXAMPLE 10:

SOLUTION:

EXAMPLE 11:

SOLUTION:

EXAMPLE 12:

SOLUTION:

EXAMPLE 13:

Find the total area of

on the interval [-2, 2].

SOLUTION:

Since the region is symmetric about the origin, I will find the area under the curve over the interval [0, 2]. Then I will take this area times 2 to find the total area. I will first set up the integral

Notice that I will have to use a u-substitution, so I will now do that.