Mean Value Theorem for Integrals

Mean Value Theorem for Integrals
Let $f$ be a continuous and integrable real function on the closed interval $\left[{a..b}\right]$.

Then there exists a real number $k$ in the interval such that:


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

Proof

 * Case 1:

$f$ is a constant function:
 * $\forall x \in \left[{a..b}\right]: f \left({x}\right) = c$

Then:

In this case:
 * $\forall x \in \left[{a..b}\right]: f \left({x}\right) = c$

... so in this case $k$ is fulfilled by any real number in the interval.


 * Case 2:

$f$ is not a constant function.

By the Extreme Value Theorem and from Relative Sizes of Definite Integrals it follows that:
 * $\displaystyle \forall x \in \left[{a..b}\right]: \int_a^b f \left({m}\right) \ \mathrm d x \leq \int_a^b f \left({x}\right) \ \mathrm d x \leq \int_a^b f \left({M}\right) \ \mathrm d x$

where:
 * $m = \inf f \left({x}\right)$
 * $M = \sup f \left({x}\right)$

for $x \in \left[{a..b}\right]$.

From Case 1:


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

Dividing all terms by $\left({b - a}\right)$ gives us


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

By the Intermediate Value Theorem, there exists some $k$ in $\left({a..b}\right)$ such that:

Also see

 * Average Value of a Function