Chapter 5 - Applications of Integration - 5.5 Average Value of a Function - 5.5 Exercises - Page 392: 23

See proof

Let $F(x)$ = $\int_a^x{f(t)dt}$ for $x$ in $[a,b]$. Then $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$, so by the Mean Value Theorem there is a number $c$ in $(a,b)$ such that $F(b)-F(a)$ = $F'(c)(b-a)$ But $F'(x)$ = $f(x)$ by the Fundamental Theorem of Calculus. Therefore $F(b)-F(a)$ = $f(c)(b-a)$ $\int_a^b{f(t)dt}-\int_a^a{f(t)dt}$ = $f(c)(b-a)$ $\int_a^b{f(t)dt}-0$ = $f(c)(b-a)$ $\int_a^b{f(t)dt}$ = $f(c)(b-a)$ which is the Mean Value Theorem for Integrals.