- #1
bigplanet401
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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].