Suppose that $f$ is a continuous function on $[a,b]$ and that $f(x)\geq0$ for all $x\in [a,b]$. Show that if $\int_a^bf(x)=0$, then $f(x)=0$ for all $x\in[a,b]$.
Let $F(x)=\int_a^xf(x)$. Since $\int_a^xf(x)+\int_x^bf(x)=\int_a^bf(x)=0$ for every $x\in[a,b]$, we have $\int_a^xf(x)=-\int_x^bf(x)$. But since $f(x)\geq0$ for all $x\in[a,b]$, both $\int_a^xf(x)$ and $\int_x^bf(x)$ are positive. So $F(x)=0$ for all $x\in[a,b]$ and by the fundamental theorem of calculus, $f(x)=F'(x)=0$ for all $x\in[a,b]$.
Is this a valid proof?
The proof is correct, though invoking the fundamental theorem for this result is a bit of an over-kill. If you like, think of how to approach it quite directly. If $f(x)> 0$ for some $x$, then by continuity you can find an interval about $x$ where the function is greater than $f(x)/2$ on that interval. What can you now say about the integral of the function on $[a,b]$?