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  "body": "\n## Proof\n\nWe trap the average value of $f$ between its minimum and maximum and apply the intermediate value theorem.\n\n1. Let $a \u003c b$ 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.\n2. 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 $x \\in [a, b]$.\n3. 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 $k (b - a)$.\n4. By step 2, $m \\le f \\le M$ on $[a, b]$. By the [[monotonicity of the integral|0MBWQKM8QR]] and step 3, $m (b - a) \\le I \\le M (b - a)$.\n5. Let $\\mu = \\frac{I}{b - a}$. Dividing step 4 by $b - a \u003e 0$ gives $f(p) = m \\le \\mu \\le M = f(q)$.\n6. If $p = q$, then $m = M = \\mu$, and $c = p$ satisfies $f(c) = \\mu$.\n7. 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 $f(s)$.\n   By the [[intermediate value theorem|00KAFBR32E]], there is $c \\in [r, s] \\subseteq [a, b]$ with $f(c) = \\mu$.\n8. In both cases $c \\in [a, b]$ and $\\int_a^b f(x) \\, dx = \\mu (b - a) = f(c)(b - a)$.\n\n## Idea\n\nThe 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.\n",
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