Is $f=0$ if the integral is zero If $f: [0,1]\to \mathbb R$ is continuous and for all $s\in [0,1]$ 
$$ \int_0^s f(t)dt = 0$$
does it then follows that $f=0$? I can show it for $f \ge 0$ but I am wondering if it is also true if $f$ not positive.

Use the fundamental theorem of calculus to show it is constant.
imagine the next line with a strike through:
Assume that $f=c$ and $c\ne 0$ and reach a contradiction. Thus proving f=0.
Fleshing this out:
$\frac{d}{ds}(\int^s_0f(t)dt)=f(s)$
You know that $f(s)=\frac{d}{ds}(\int^s_0f(t)dt)=\frac{d}{ds}(0) = 0$ for $s\in[0,1]$
So you can ignore the contradiction bit, it doesn't apply. I panicked because the prospect of answer a question first made me rush :p
