Areas, Integrals, and Antiderivatives
Using Antiderivatives to Evaluate
Now we can put the ideas of areas and antiderivatives together to get a way of evaluating definite integrals that is exact and often easy.
If , and is an antiderivative of , .
We also know that if is any antiderivative of , then and have the same derivative so and are "parallel" and differ by a constant, for all .
To evaluate a definite integral , we can find any antiderivative of and evaluate .
This result is a special case of Part 2 of the Fundamental Theorem of Calculus, and it will be used hundreds of times in the next several chapters. The Fundamental Theorem is stated and proved in Section 4.5.
Antiderivatives and Definite Integrals
If is a continuous, nonnegative function and and is any antiderivative of on the interval ,
The problem of finding the exact value of a definite integral reduces to finding some (any) antiderivative of the integrand and then evaluating . Even finding one antiderivative can be difficult, and, for now, we will stick to functions which have easy antiderivatives. Later we will explore some methods for finding antiderivatives of more difficult functions.
The evaluation is represented by the symbol .
Example 3: Evaluate in two ways:
By sketching the graph of and geometrically finding the area.
By finding an antiderivative of of and evaluating .
Solution:
(i) The graph of is shown in Fig. 6, and the shaded region has area 4.
(ii) One antiderivative of is (check that ), and which agrees with (i).
If someone chose another antiderivative of , say ( check that ), then . No matter which antiderivative is chosen, equals .
Practice 2: Evaluate in the two ways of the previous example.
This antiderivative method is an extremely powerful way to evaluate some definite integrals, and it is used often. However, it can only be used to evaluate a definite integral of a function defined by a formula.
Example 4: Find the area between the graph of the cosine and the horizontal axis for between and .
Solution: The area we want (Fig. 7) is so we need an antiderivative of .
is one antiderivative of . (Check that ). Then .
Practice 3: Find the area between the graph of and the horizontal axis for between 1 and 2.