---
tag: F89W27S5QT
kind: proof
name: Proof of the mean value theorem for integrals
msc: [26A42]
proves: 0N91N98EG5
uses: [06SESV7GQY, 0WS1A8E2P9, 0HBY48G8G1, 0N1JQXZXHC, 0MBWQKM8QR, 00KAFBR32E]
sources:
  - url: https://www.math.ucdavis.edu/~hunter/m125b/ch1.pdf
    locator: Theorem 1.29
---

## Proof

We trap the average value of $f$ between its minimum and maximum and apply the intermediate value theorem.

1. Let $a < b$ and let $f \colon [a, b] \to \mathbb{R}$ be continuous. Since [[continuous functions are Riemann integrable|06SESV7GQY]], the integral $I = \int_a^b f(x) \, dx$ exists.
2. By the [[extreme value theorem|0WS1A8E2P9]], there are $p, q \in [a, b]$ with $m = f(p) \le f(x) \le f(q) = M$ for all $x \in [a, b]$.
3. Let $k$ be a real constant. Every [[Riemann sum|0HBY48G8G1]] of the constant function $k$ is $\sum_{i} k (x_i - x_{i-1}) = k (b - a)$. So the constant $k$ is [[Riemann integrable|0N1JQXZXHC]] with integral $k (b - a)$.
4. By step 2, $m \le f \le M$ on $[a, b]$. By the [[monotonicity of the integral|0MBWQKM8QR]] and step 3, $m (b - a) \le I \le M (b - a)$.
5. Let $\mu = \frac{I}{b - a}$. Dividing step 4 by $b - a > 0$ gives $f(p) = m \le \mu \le M = f(q)$.
6. If $p = q$, then $m = M = \mu$, and $c = p$ satisfies $f(c) = \mu$.
7. If $p \ne q$, let $[r, s]$ be the closed interval with endpoints $p$ and $q$. The restriction of $f$ to $[r, s]$ is continuous, and $\mu$ lies between $f(r)$ and $f(s)$.
   By the [[intermediate value theorem|00KAFBR32E]], there is $c \in [r, s] \subseteq [a, b]$ with $f(c) = \mu$.
8. In both cases $c \in [a, b]$ and $\int_a^b f(x) \, dx = \mu (b - a) = f(c)(b - a)$.

## Idea

The average value $\frac{1}{b - a} \int_a^b f$ lies between the smallest and the largest value of $f$. A continuous function on an interval takes every value between two of its values. So it takes its average value somewhere.