Mean Value Theorem for Integrals

Mean Value Theorem for Integrals
Let $f$ be a continuous and integrable real function on the closed interval $[a..b]$, where $a<b$. Then there exists a number $k$ in the interval such that


 * $\displaystyle \int_a^b f \left({x}\right) \ \mathrm d x = f(k)(b-a)$

Proof
Case 1: $f$ is a constant function.


 * $\displaystyle \int_a^b c \ \mathrm d x = cb - ca$

$ = c(b-a)$

In which case $c = f(k) = k$.

Case 2:

$f$ is not a constant function.

By the Extreme Value Theorem and from Relative Sizes of Definite Integrals it follows that


 * $\displaystyle \int_a^b f \left({m}\right) \ \mathrm d x \leq \displaystyle \int_a^b f \left({x}\right) \ \mathrm d x \leq \displaystyle \int_a^b f \left({M}\right) \ \mathrm d x \quad \forall x \in [a..b]$

From Case 1,


 * $f(m)(b-a) \leq \displaystyle \int_a^b f \left({x}\right) \ \mathrm d x \leq f(M) (b-a) \quad $

Dividing all terms by $(b-a)$ gives us


 * $f(m) \leq \displaystyle \frac 1 {b-a}\int_a^b f \left({x}\right) \ \mathrm d x \leq f(M) \quad$

By the Intermediate Value Theorem, there exists some $f(c)$ in $(a..b)$ such that

$\displaystyle \frac 1 {b-a}\int_a^b f \left({x}\right) \ \mathrm d x = f(c)$

$\implies$

$ \displaystyle \int_a^b f \left({x}\right) \ \mathrm d x = f(c)(b-a)$