Answer
See explanation
Work Step by Step
Mean Value Theorem for Integrals:
If $f$ is continuous on $[a,b]$, then there exists a number $c \in [a,b]$ such that:
$f(c)=\frac{1}{b-a}\int_a^b f(x)dx$
Mean Value Theorem for Derivatives:
If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a point $c \in (a,b)$ such that:
$f'(c)=\frac{f(b)-f(a)}{b-a}$
We are provided with a function:
$F(x) = \int_a^xf(t) dt$
The Fundamental Theorem of Calculus tells us three significant facts:
1. $F(x)$ is differentiable over $(a,b)$
2. $F(x)$ is continuous over $[a,b]$
2. $F'(x) = f(x)$
Let's use the Mean Value Theorem for Derivatives, since $F(x)$ is continuous over $[a,b]$ and differentiable over $(a,b)$.
$F'(c) = \frac{F(b) - F(a)}{b-a} = \frac{1}{b-a} F(b) - F(a)$
Approach 1: Apply the Net Change Theorem
$F'(c) = \frac{1}{b-a} \int_a^bF'(x)dx = \frac{1}{b-a}\int_a^bf(x)dx = f(c)$
Approach 2: Directly substitute $F(b)$ and $F(a)$
$F(b) = \int_a^bf(x)dx$
$F(a) = \int_a^af(x)dx = 0$
$F'(c) = \frac{1}{b-a} \int_a^bf(x)dx = f(c)$
Since $F'(c) = f(c)$, this proves the Mean Value Theorem for Integrals using the Mean Value Theorem for Derivatives.
