twice differentiable function (question from exam) $f$ is twice differentiable, $f(0)=f(1)=0$ and $f''$ is continuous. Prove that there exists $c\in[0,1]$ such that $$\int_0^1f(x)dx=-\frac1{12}f''(c).$$
I haven't progressed much on this problem. A lot of ideas came up to my mind but none seems to work. Obviously, this has something to do with mean value theorem. In fact, we only need to show that there exist $a,b\in[0,1]$ such that $\int_0^1f(x)dx=\frac{f'(a)-f'(b)}{a-b}$. By mean value theorem, there exists $c\in[a,b]$ such that $f''(c)=\frac{f'(a)-f'(b)}{a-b}$. But this does not seem to be the correct path, because we haven't used that $f''$ is continuous.
This leads to the second idea to show that $f''(x)$ attains some values below and above $\int_0^1f(x)dx$. So I think we need to work out some inequalities, which I don't have any idea.
Anyway, I just started learning calculus for a few weeks. This question is from the previous exam paper, it's the only question I can't solve. Any help is appreciated, thanks.
 A: Hint:
$$
\int_0^1x(1-x)f''(x)\mathrm dx=-2\int_0^1f(x)\mathrm dx.
$$
A: Here is how I continue from Andrew's hint. Please check it:
Firstly we prove that $\int_0^1x(1-x)f''(x)dx=-2\int_0^1f(x)dx$: $$\begin{align*}-2\int_0^1f(x)dx&=-2\left(xf(x)\Big|^1_0-\int_0^1xf'(x)dx\right)\\&=2\left(\frac12x^2f'(x)\Big|^1_0-\int_0^1\frac12x^2f''(x)dx\right)\\&=f'(1)-\int_0^1x^2f''(x)dx\\&=xf'(x)\Big|^1_0-\int_0^1x^2f''(x)dx\\&=xf'(x)\Big|^1_0 - \int_0^1f'(x)dx -\int_0^1x^2f''(x)dx\\&=\int_0^1xf''(x)dx-\int_0^1x^2f''(x)dx\\&=\int_0^1x(1-x)f''(x)dx\end{align*}$$ So we will prove that there exists $c\in[0,1]$ such that $\int_0^1x(1-x)f''(x)dx=\frac16f''(c)$. Note that $x(1-x)\ge0$ for $x\in[0,1]$. Since $f''$ is continuous, by EVT it has a maximum $M$ and minimum $m$ on $[0,1]$. Hence $$\frac16m=\int_0^1x(1-x)m dx\le\int_0^1x(1-x)f''(x)dx\le \int_0^1x(1-x)M dx=\frac16M.$$ Since $f''$ is continuous, by IVT we know that there exists $c$ such that $\frac16f''(c)=\int_0^1x(1-x)f''(x)dx$, as desired.
QED
A: Let be $$f(x)= x^2-x$$ Clearly, $f(0)=f(1)=0$. And 
$$\int_0^1 f(x) \ dx= \frac{1}{3}-\frac{1}{2}= \frac{-1}{6}$$ In the other hand, 
$$f''(x)= 0$$for all $x \in \mathbb{R}$. Therefore, on this case, do not exists $c \in \mathbb{R}$ such that $ \int_0^1 f(x) \ dx = f''(c)$.
