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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.

Every real-valued monotone function on a real interval is differentiable with a finite derivative at almost every point.

This strengthens the regularity known for a monotonic function beyond the restriction on its jump discontinuities.

A real-valued function $f:D\to \mathbb{R}$, with domain of a function $D\subseteq \mathbb{R}$ in the usual order of the real numbers, is a monotonic function if it is either a nondecreasing function throughout $D$ or a nonincreasing function throughout $D$. The same direction must work for every pair $x<y$ in $D$. Being a strictly increasing function on one part and a strictly decreasing function on another does not make the whole function monotone: for example, $x^2$ on the real line changes direction at zero. A constant function is both a nondecreasing function and a nonincreasing function.

The word "increasing" is ambiguous across textbooks: some allow equal outputs and some require unequal outputs. Use nondecreasing function for the weak inequality and strictly increasing function for the strict inequality; use nonincreasing function and strictly decreasing function in the other direction. Neither continuity nor differentiability is part of the definition of a monotonic function.

Strict monotonicity gives injectivity, so an equation $f(x)=b$ has at most one solution and an inverse function exists on the image of a function. Weak monotonicity permits repeated outputs and jumps. In probability theory, a cumulative distribution function is a nondecreasing function; its quantile function uses a generalized inverse to handle those flat parts and jumps. Monotonicity alone does not guarantee continuity, even for a strictly increasing function.

Again let $G={\int }_a^{b}g(t)dt\ge 0$ and $I={\int }_a^{b}f(t)g(t)dt$. Since $f$ is a nondecreasing monotonic function, $f(a)\le f(t)\le f(b)$ on $[a,b]$. The condition $g\ge 0$ and the monotonicity of the Riemann integral giveSet $H(x)={\int }_a^{x}g(t)dt$. By part (a), $H$ is continuous. Therefore the functionis continuous on $[a,b]$. Its endpoint values are $K(a)=f(b)G$ and $K(b)=f(a)G$, which enclose $I$. The intermediate value theorem provides $d\in [a,b]$ with $K(d)=I$, soIf the two endpoint values coincide, the preceding bounds force $I$ to equal that common value and either endpoint works. Thus the proof also covers $G=0$, $f(a)=f(b)$ and $a=b$, without dividing by any possibly zero quantity.

Solving $\mu x(1-x)=1$ by the quadratic formula givesFor $\mu >4$, $0<x_{-}<1/2<x_{+}<1$. On $L=[0,x_{-}]$ the derivative $f^{\prime }(x)=\mu (1-2x)$ is strictly positive, while on $R=[x_{+},1]$ it is strictly negative. Thus the restrictions are strictly monotonic functions. Their endpoint values areBy continuity and strict monotonicity, each restriction maps its interval bijectively onto $[0,1]$, with opposite orientations. The corresponding inverse functions are