Assume that $ h : [a,b] → (0,+∞)$ is a positive integrable function and $f : [a, b] → $ is continuous. Prove that there exists $c ∈ [a, b] $ such that
$\displaystyle\frac{\int_a^b f (x)h(x)dx}{\int_a^bh(x)dx} = f(c)$
assume $f(x)h(x)$ is integrable.
Im assuming I need a theorem such as MVT or IVT to prove this, i used integration by parts, and deduced the left hand side, but it didn't get me anywhere for the proof.
Consider $F,H:[a,b]→R$ as $F(x)=∫_a^x[f(t)h(t)dt]$ and $H(x)=∫_a^x[h(t)dt]$. Then both $F$ and $H$ satisfies Cauchy’s MVT, provided $h(x)≠0$ for all $x$ in $[a,b]$. Therefore we get at least one $ c $ in $(a,b)$ such that $$\frac{(F(b)-F(a))}{(H(b)-H(a))} =\frac{(F' (c))}{(H' (c))}$$ that is, $$\frac{(∫_a^b[f(t)h(t)dt])}{(∫_a^b[h(t)dt])}=f(c).$$