The Integral Mean Value Theorem 1   If $f$ and $g$ are continuous real functions on an interval $[a,b]$ , and $g$ is additionally non-negative on $(a,b)$ , then there exists a $\zeta\in(a,b)$ such that$$ \integral{a}{b}{f(x)g(x)}{x}=f(\zeta)\integral{a}{b}{g(x)}{x} $$

Proof. Since $f$ is continuous on a closed bounded set, $f$ is bounded and attains its bounds, say $f\left(x_0\right)\leq f(x)\leq f\left(x_1\right)$ for all $x\in[a,b]$ . Thus, since $g$ is non-negative for all $x\in[a,b]$ $$ f\left(x_0\right)g(x)\leq f(x)g(x)\leq f\left(x_1\right)g(x) $$ Integrating both sides gives$$ f\left(x_0\right)\integral{a}{b}{g(x)}{x}\leq\integral{a}{b}{f(x)g(x)}{x}\leq f\left(x_1\right)\integral{a}{b}{g(x)}{x} $$ If $\integral{a}{b}{g(x)}{x}=0$ , then $g(x)$ is identically zero, and the result follows trivially. Otherwise,$$ f\left(x_0\right)\leq\frac{\integral{a}{b}{f(x)g(x)}{x}}{\integral{a}{b}{g(x)}{x}}\leq f\left(x_1\right) $$ and the result follows from the intermediate value theorem.