 Solution

Write $G={\int }_a^{b}g(t)dt\ge 0$ and $I={\int }_a^{b}f(t)g(t)dt$. By the extreme value theorem, $f$ attains a minimum $m$ and a maximum $M$ on $[a,b]$. Because $g\ge 0$, the monotonicity of the Riemann integral givesIf $G=0$, this forces $I=0$, and every $c\in [a,b]$ satisfies the required identity. This includes $a=b$.

If $G>0$, divide to obtain $m\le I/G\le M$. The intermediate value theorem, applied between points where the continuous function $f$ attains $m$ and $M$, supplies $c\in [a,b]$ with $f(c)=I/G$. Hence the weighted mean value theorem for integrals gives