Proving a statement with the mean value theorem If $f:[-2,2]\rightarrow R$ is a continuous function, there exists $c $ in $(-2,2)$ such that $$\int_{-2}^cf(x)dx=\left( \frac{4}{c^3}-\frac{c}{4}\right) f(c)$$.I tried to prove this using the mean value theorem.
So there exists $c$  in $(-2,2)$ such that  $$\int_{-2}^2f(x)dx=4f(c)$$.
Can somebody tell me what should I do next, please?

First, we define the function $g : [-2,2] \to \mathbb R$ by $g(x) = \int_{-2}^x f(t)\mbox{dt}$. By the fundamental theorem of calculus, $g$ is differentiable, with $g'(x) = f(x)$.
Consider the function $h(x) = 16 - x^4$ on $[-2,2]$. Then, note that:
$$
\frac{4}{c^3}- \frac c4 = \frac{16 - c^4}{4c^3} = -\frac{h(c)}{h'(c)}
$$ 
We are essentially asked to show that there is a $c$ such that $g(c) = \frac{-f(c)h(c)}{h'(c)}$, or $h'(c)g(c) = -g'(c)h(c)$, or $(gh)'(c) = 0$ by the product rule.
Note that $h(-2)  = h(2) = 0$. Conclude that $gh(-2) = gh(2) = 0$. Can you use some theorem now?
