Prove that if $f(x) = \int_{0}^x f(t)\,dt$, then $f(x) = 0$ 
Prove that if $f(x) = \displaystyle\int_{0}^x f(t)\, dt$ for all $x$, then $f(x) = 0$.

I first differentiated to get $f'(x) = f(x) - f(0)$. Then by the mean value theorem there exists a $c$ in $(0,x)$ such that $f'(c)=\dfrac{f(x)-f(0)}{x}$. Thus, $f'(x) = xf'(c)$. What do I do from here?
A summary of the deliberation in the comments about the necessity of assumptions on $f$: 
There was worry that we needed assumptions on $f$ such as continuity in order to exploit FTC.

Thanks to Aloizio and Clark for pointing out that no assumptions need be placed on $f$ as the integral of a Riemann integrable function is continuous. This gives us that $f$ is continuous (since by assumption we have that $f$ is integrable) and thus the Fundamental Theorem of Calculus applies.


Suppose $f$ is continuous. The fundamental theorem of calculus tells us that it is actually differentiable, and moreover by differentiating both sides we get
$$f'(x) = f(x)$$
So $f(x)$ is a function which is its own derivative, and hence is of the form $f(x) = ce^{x}$. What is the constant $c$? Simply compute
$$f(0) = \int_{0}^{0} f(x) = 0$$
to see that $c = 0$, and we are done.
