TITLE: Prove that $ f(\xi)=f'(\xi)\int_{0}^{\xi}{f(x)dx} $
QUESTION [6 upvotes]: Let $f:[0,1]\rightarrow\mathbb{R}$ be a differentiable function with continuous derivative such that
$$ \int_{0}^{1}{f(x)dx}=\int_{0}^{1}{xf(x)dx} $$
How can we prove that there exists $\xi\in(0,1)$ such that $$ f(\xi)=f'(\xi)\int_{0}^{\xi}{f(x)dx} $$
I tried to use 
$$ F(x)=\int_{0}^{x}{f(t)dt} $$
then the condition gives
$$ \int_{0}^{1}{F(x)dx}=0 $$
and I have to show there exists $\xi\in(0,1)$ such that 
$$ F'(\xi)=F''(\xi)F(\xi) $$
I was stuck here.

REPLY [5 votes]: Note that $F(0) = 0$.
Consider function $G(x) = e^{-F'(x)} F(x)$, $G'(x) = e^{-F'}(F'-F''F)$, thus $x=0$ is a zero for $G$, then

if $F(x)=0$ for all $x$, then it is done.
Otherwise, there are $x_1,x_2$ such that $F(x_1)>0$, $F(x_2)<0$. Then $G(x_1)>0$, $G(x_2)<0$, by Roll's theorem (Since we have another zero at $x=0$) we have there is a $\xi$ such that $G'(\xi) = 0$.

Done.