Suppose that the function $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous and that $$f(x)=\int_{0}^{x}f(t)dt\qquad\text{for all $x$}$$ Prove that $f(x)=0$ for all $x$.
Attempt Since $f(x)=\int_{0}^{x}f(t)dt$, by the First Fundamental Theorem of Calculus, this gives $f(x)=F(x)-F(0)$. So, $$F(x)=\int_{0}^{x}f(t)dt+F(0)\Longrightarrow F'(x)=f(x)=F(x)-F(0)=\int_{0}^{x}f(t)dt$$
At this step, I don't know how to keep going to work on it. I think my proof doesn't right. Can someone give me a hint or suggestion to the right direction? Thanks
By the fundamental theorem you have $$ f'(x)=f(x) $$ Since $f$ is defined on $\mathbb{R}$, this implies that $f(x)=ke^x$, for some $k\in\mathbb{R}$. But, since $f(0)=0$, we have $k=0$.
Hint for proving $f(x)=ke^x$: consider $g(x)=e^{-x}f(x)$ and compute $g'(x)$.