Mean Value Theorem for Integrals

Theorem
Let $f$ be a continuous real function on the closed interval $\closedint a b$.

Then there exists a real number $k \in \closedint a b$ such that:


 * $\displaystyle \int_a^b \map f x \rd x = \map f k \paren {b - a}$

Proof
From Continuous Real Function is Darboux Integrable, $f$ is Riemann integrable on $\closedint a b$.

By the Extreme Value Theorem, there exist $m, M \in \closedint a b$ such that:


 * $\displaystyle \map f m = \min_{x \mathop \in \closedint a b} \map f x$
 * $\displaystyle \map f M = \max_{x \mathop \in \closedint a b} \map f x$

From Relative Sizes of Definite Integrals:


 * $\displaystyle \int_a^b \map f m \rd x \le \int_a^b \map f x \rd x \le \int_a^b \map f M \rd x$

These bounds can be computed by Integral of Constant:


 * $\displaystyle \map f m \paren {b - a} \le \int_a^b \map f x \rd x \le \map f M \paren {b - a}$

Dividing all terms by $\paren {b - a}$ gives:


 * $\displaystyle \map f m \le \frac 1 {b - a}\int_a^b \map f x \rd x \le \map f M$

By the Intermediate Value Theorem, there exists some $k \in \openint a b$ such that:

Also see

 * Definition:Average Value of Function


 * Upper and Lower Bounds of Integral


 * Mean Value Theorem