Mean Value Theorem for integrals

[bigplanet401]

Homework Statement

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

<br /> F(x) = \int_a^x \, f(t) \, dt<br />

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

<br /> \int_a^b \, f(x) \, dx = f(c) (b - a) \, .<br />

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

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

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

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

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].

 

[mfb]

You derived a slightly stronger result - there has to be a "c" in (a,b). That implies the weaker result of the existence of a "c" in [a,b].