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.


Firstly, we know $f$ is continuous, since $\int_0^x f$ is a continuous function when $f$ is Riemann-integrable, or even Lebesgue-integrable (which I'm assuming is the case since otherwise this doesn't make sense). Now, we then have by the FTC that $f$ is differentiable, and it follows that $f'(x)=f(x)$ for all $x$ also due to the FTC. Since $f(0)=0$, we are done.

"Order of information": We look at the right side, and see that it is a continuous function on $x$ due to the fact that $f$ is integrable. Therefore, the left side is continuous. Looking at the right side now, we have the integration up to $x$ of a continuous function, which is differentiable by the FTC.
