Read this section to see the connection between derivatives and integrals. Work through practice problems 1-5.
Every continuous function has an antiderivative, even those nondifferentiable functions with "corners" such as absolute value.
The Fundamental Theorem of Calculus (Part 1)
then . is an antiderivative of .
Proof: Assume is a continuous function and let . By the definition of derivative of ,
By Property 6 of definite integrals (Section 4.3), for
Dividing each part of the inequality by , we have that is between the minimum and the maximum of on the interval . The function is continuous (by the hypothesis) and the interval is shrinking (since h approaches 0), so and . Therefore, is stuck between two quantities (Fig. 2) which both approach .
Then must also approach , and .
Example 1: for in Fig. 3. Evaluate and for .
Practice 1: for in Fig. 4. Evaluate and for and .
Example 2: for the function shown in Fig. 5.
For which value of is maximum?
For which is the rate of change of maximum?
Solution: Since is differentiable, the only critical points are where or at endpoints. and has a maximum at . Notice that the values of as goes from 0 to 3 and then the values decrease. The rate of change of is , and appears to have a maximum at so the rate of change of is maximum when . Near , a slight increase in the value of yields the maximum increase in the value of .