Weighted mean value theorem for integrals (${\int }_a^{b}fg=f(\alpha ){\int }_a^{b}g$)

For real continuous functions $f,g$ on a compact interval with $g\ge 0$, some $\alpha$ in the interval satisfies $\int fg=f(\alpha )\int g$. If $\int g>0$, the weighted average lies between the minimum and maximum of $f$, and the intermediate value theorem supplies $\alpha$. If $\int g=0$, the bound $|\int fg|\le \sup |f|\int g$ makes both sides zero. Nonnegativity of the weight is essential to this argument.

If $f,g:[a,b]\to \mathbb{R}$ are continuous functions, $g\ge 0$, and $f$ is a nondecreasing monotonic function, there is $d\in [a,b]$ satisfying the identity in the title. It expresses the weighted integral as a mixture of the two endpoint values of $f$, with the split occurring in the accumulated weight.

Put $G={\int }_a^{b}g$ and $I={\int }_a^{b}fg$. The monotonicity of the Riemann integral gives $f(a)G\le I\le f(b)G$. The function $H(x)={\int }_a^{x}g$ is continuous; indeed, $|H(y)-H(x)|\le {\max }_{[a,b]}|g||y-x|$ by the extreme value theorem and the triangle inequality for the Riemann integral. Thereforeis continuous, with $K(a)=f(b)G$ and $K(b)=f(a)G$. The intermediate value theorem gives $K(d)=I$. If the endpoint values coincide, the bounds force $I$ to have that common value and an endpoint works. The proof covers zero total weight and a degenerate interval without division.