Mean Value Theorem for Integrals

Theorem
Let $f$ be a continuous 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
From Continuous Function is Riemann Integrable, $f$ is Riemann integrable on $\left[{a \,.\,.\, b}\right]$.

By the Extreme Value Theorem, there exist $m, M \in \left[{a \,.\,.\, b}\right]$ such that:


 * $\displaystyle f \left({m}\right) = \min_{x \in \left[{a \,.\,.\, b}\right]} f \left({x}\right)$
 * $\displaystyle f \left({M}\right) = \max_{x \in \left[{a \,.\,.\, b}\right]} f \left({x}\right)$

From Relative Sizes of Definite Integrals:


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

These bounds can be computed by Integral of Constant:


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

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


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

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

Also see

 * Definition:Average Value of Function


 * Upper and Lower Bounds of Integral