- Use antiderivatives to evaluate definite integrals
- Use the Mean Value Theorem for integrals to solve problems
- Use general rules of integrals to solve problems
In the Lesson on Definite Integrals, we evaluated definite integrals using the limit definition. This process was long and tedious. In this lesson we will learn some practical ways to evaluate definite integrals. We begin with a theorem that provides an easier method for evaluating definite integrals. Newton discovered this method that uses antiderivatives to calculate definite integrals.
If is continuous on the closed interval then
where is any antiderivative of
We sometimes use the following shorthand notation to indicate
The proof of this theorem is included at the end of this lesson. Theorem 4.1 is usually stated as a part of the Fundamental Theorem of Calculus, a theorem that we will present in the Lesson on the Fundamental Theorem of Calculus. For now, the result provides a useful and efficient way to compute definite integrals. We need only find an antiderivative of the given function in order to compute its integral over the closed interval. It also gives us a result with which we can now state and prove a version of the Mean Value Theorem for integrals. But first let’s look at a couple of examples.
Compute the following definite integral:
Using the limit definition we found that We now can verify this using the theorem as follows:
We first note that is an antiderivative of Hence we have
We conclude the lesson by stating the rules for definite integrals, most of which parallel the rules we stated for the general indefinite integrals.
Given these rules together with Theorem 4.1, we will be able to solve a great variety of definite integrals.
- We used antiderivatives to evaluate definite integrals.
- We used the Mean Value Theorem for integrals to solve problems.
- We used general rules of integrals to solve problems.
Proof of Theorem 4.1
We first need to divide into sub-intervals of length . We let be the endpoints of these sub-intervals.
Let be any antiderivative of
We will now employ a method that will express the right side of this equation as a Riemann Sum. In particular,
Note that is continuous. Hence, by the Mean Value Theorem, there exist
Taking the limit of each side as we have
We note that the left side is a constant and the right side is our definition for .
Proof of Theorem 4.2
By the Mean Value Theorem for derivatives, there exists such that
From Theorem 4.1 we have that is an antiderivative of Hence, and in particular, Hence, by substitution we have
Note that Hence we have
and by our definition of we have
This theorem allows us to find for positive functions a rectangle that has base and height such that the area of the rectangle is the same as the area given by In other words, is the average function value over
In problems #1–8, use antiderivatives to compute the definite integral.
- Find the average value of over
- If is continuous and show that takes on the value at least once on the interval
- Your friend states that there is no area under the curve of on since he computed Is he correct? Explain your answer.
- Apply the Mean Value Theorem for integrals.
- He is partially correct. The definite integral computes the net area under the curve. However, the area between the curve and the -axis is given by