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.
 A: HINT: If $\int_a^b f(x)\,dx\ne0$, then $\int_0^a f(x)\,dx\ne\int_0^b f(x)\,dx$.
A: Define $F(x) = \int_0^x f(s)\,ds$.  By the Fundamental Theorem of Calculus, $F'=f$. But by your assumption, $F(x) = 0$ for all $x \in [0,1]$.  So $f \equiv 0$.
A: Assume $a:=f(x_0)\ne0$ for some $x_0\in(0,1)$. Then $\frac1a{f(x)}>\frac12$ on some open neighbourhood $x_0\in (x_0-r,x_0+r)\subset(0,1)$ because $f$ is continous. Hence
$$ 0=\frac1a\int_0^{x_0+r}f(t)\,\mathrm dt-\frac1a\int_0^{x_0-r}f(t)\,\mathrm dt=\int_{x_0-r}^{x_0+r}\frac1af(t)\,\mathrm dt\ge \int_{x_0-r}^{x_0+r}\frac12\,\mathrm dt=r>0.$$
Again by continuity, $f(0)=f(1)=0$, too.
A: 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
