From the problems plus in Stewart Calculus 6e, it asks if $f$ is a differentiable function such that $f(x)$ is never $0$ and for any $x$, $\int_0^xf(t)dt=[f(x)]^2$, then what is $f(x)$?
I figured since it's differentiable I could take the derivative of both sides to get:
$$f(x)=2f(x)f'(x)$$ Since $f(x)$ is never $0$, then $f'(x)=1/2$. But that means that $f(x)=x/2+c$, which will equal $0$ for $x=-2c$. I can't just take $0$ out of the function, since it has to be differentiable everywhere. So how do I solve this problem?