The Mean Value Theorem for Integrals is $\int_Sf(x)g(x)dx$ =$f(c)\int_Sg(x)dx$ I am asked to use this to prove the generalized MVT which is $$\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$$
how can I link the integral MVT to the generalized MVT? I was thinking to use the FUndamental Calculus theorem which is F(x)=$\int_a^xf(t)dt$ and F'(x)=f(x).
On
Consider the function $h(x) = f(x)*(g(b) - g(a)) - g(x)*(f(b) - f(a))$ on $[a, b]$. Then: $$ h(a) = f(a)g(b) - f(a)g(a) - g(a)f(b) + g(a)f(a) = f(a)g(b) - f(b)g(a), $$ and $$ h(b) = f(b)*(g(b) - g(a)) - g(b)*(f(b) - f(a)) = f(a)g(b) - f(b)g(a). $$ We see that $h(a) = h(b)$, so by Rolle's theorem: there is a $c \in (a, b)$ such that $h'(c) = 0$. So: $$ f'(c)*(g(b) - g(a)) - g'(c)*(f(b) - f(a)) = 0 , $$ and we have: $$ \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}. $$
One approach is like this: $$\int_{a}^{b}f'(x)\,dx = \int_{a}^{b}\frac{f'(x)}{g'(x)}\cdot g'(x)\,dx = \frac{f'(c)}{g'(c)}\int_{a}^{b}g'(x)\,dx$$ so that $$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}$$ The conditions required for mean value theorem of integrals is that both $f'(x),g'(x)$ are continuous and $g'(x) \neq 0$.
So this approach does establish the generalized mean value theorem also called by the name "Cauchy's MVT" but the above proof requires more conditions on $f, g$.