Generlized IVT

Suppose $f,g$ are continuous functions on $[a,b]$ and $g \ge 0$. Then $\exists x \in (a,b)$:

$$f(x) \int_a^b g = \int_a^b fg$$

Proof Hint

Consider this and proof is similar to IVT.

$$mg \le fg \le Mg$$