 Solution

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.