Rolles theorem question involving integrals 
If $f(x)$ and $g(x)$ are continuous in $[a,b]$ and $g(x)\ge0$ for all $x\in[a,b]$ then
  $$\int_a^bf(x)g(x)\,\mathrm dx$$
  is
  $$
\begin{array}{lll}
(a)\quad&g(\xi)\int_a^bf(x)\,\mathrm dx\quad&\text{for exactly one }\xi\in(a,b)\\
(b)&f(\xi)\int_a^bg(x)\,\mathrm dx&\text{for all }\xi\in(a,b)\\
(c)\quad&f(\xi)\int_a^bg(x)\,\mathrm dx\quad&\text{for some }\xi\in(a,b)\\
(d)\quad&g(\xi)\int_a^bf(x)\,\mathrm dx\quad&\text{for all }\xi\in(a,b)\\
\end{array}
$$

I am sensing rolles theorem but i haven't found  way yet .Thanks
 A: Since 
$$\min_y f(y) \int g(x)dx \le \int f(x)g(x) dx\le \max_y f(y)\int g(x) dx$$
(here it is used that $g\ge 0$)
and $\min$ and $\max$ are attained you can apply the mean value theorem. Since $f$ is continuos and the domain of definition is compact, there are values $c, d$ such that $f(c)=\min_y f(y)$ an $f(d) =\max_y f(y)$. 
This implies that for any value $\eta$ sucht that $$\min_y f(y) \int g(x)dx=f(c)\int g(x) dx\le \eta\le f(d)\int g(x)dx=\max_y f(y)\int g(x) dx$$ there is $\xi \in [c,d]$ such that $f(\xi)\int g(x) dx= \eta$. Now chose $\eta = \int f(x)g(x) dx$
A: While I'm thinking over (a) and (c)  I do think that (b) and (d) are wrong.
Just check with an easy example:
Let $ f(x) = x $ and $ g(x) = x $ where $ [a,b] = [0,1] $. Then $ \int\limits_a^b f(x)g(x) dx = \int\limits_a^b x^2 dx = \frac 1 3 (b^3 - a^3) = \frac 1 3 $.
Obviously (b) and (d) won't hold for $ \int\limits_a^b f(x)dx = \frac 1 2 $ or $ \int\limits_a^b g(x)dx  = \frac 1 2 $.
Edit: This is only to disprove (b) and (d) by counter example.
