- #1

bigplanet401

- 104

- 0

## Homework Statement

Prove the Mean Value Theorem for integrals by applying the Mean Value Theorem for derivatives to the function

[tex]

F(x) = \int_a^x \, f(t) \, dt

[/tex]

## Homework Equations

[/B]

Mean Value Theorem for integrals: If f is continuous on [a, b], then there exists a number c in [a, b] such that

[tex]

\int_a^b \, f(x) \, dx = f(c) (b - a) \, .

[/tex]

Mean Value Theorem for derivatives, Fundamental Theorem of Calculus.

## The Attempt at a Solution

Since the function f is continuous on [a, b], F(x) is continuous on [a, b] and differentiable on (a, b) by the FTC. The MVT for derivatives says

[tex]

F(b) - F(a) = F^\prime (c) (b - a) \quad \text{for} \, c \in (a, b) \, .

[/tex]

By FTC, F'(c) = f(c). Then f(c) (b - a) = F(b) - F(a). So

[tex]

f(c) = \frac{F(b) - F(a)}{b - a} = \frac{1}{b - a} \int_a^b \, f(x) \, dx

[/tex]

again by FTC.

What I don't understand is why the book says c can be equal to a or b...F is differentiable on the open interval (a, b), not [a, b]. But the book says the MVT for integrals permits c to lie anywhere in the

*closed*interval [a, b].