Assume $f$ is continuous and nonnegative over $[a,b]$. Show that $f$ is identically zero if and only if $\displaystyle \int_{a}^b f(x)dx = 0$.
Proving the first direction is easy: If $f(x) = 0$, then obviously $\displaystyle \int_{a}^b f(x)dx = 0$. On the other hand if $\displaystyle \int_{a}^b f(x)dx = 0$, then since $f$ is continuous on $[a,b]$ and $f(x) \geq 0$ we must have $f(x) = 0$.
HINT: If $f(x)$ is not identically $0$ on $[a,b]$, there is a $c\in[a,b]$ such that $f(c)>0$. Use the continuity of $f$ to show that there is a non-degenerate interval $[u,v]\subseteq[a,b]$ such that $f(x)\ge\frac12f(c)$ for each $x\in[u,v]$. Now compare $\int_a^bf(x)\,dx$ with $\frac12f(c)(v-u)$.
Hint:
If $f(c) \neq 0$ for some $c$, then by continuity, there is a small interval around $c$ where $f$ is strictly positive...
This theorem is sometimes proved by contradiction, but a direct proof is possible if you know the Fundamental Theorem of Calculus. What you need to do is prove that for $x \in (a,b)$
$$f(x) = \lim_{h \to 0} \frac{1}{h}\int_{x}^{x+h} f(x) dx$$
You know that the right-hand side is identically zero, as $f(x) \geq 0$ implies
$$\int_{x}^{x+h}f(x) dx \leq \int_{a}^{b} f(x) = 0$$
since you know the right-hand-side is identically zero, you will be able to conclude that $f(x)$ is zero as well.
Assume that there exists $x_{0} \in [a,b]$ such that $f(x_{0}) >0$. Then, we can find $\delta>0$ such that for $x \in A=]x_{0}-\delta,x_{0}+\delta[$, $|f(x)-f(x_{0})|<\frac{f(x_{0})}{2}$, which implies that, for $x \in A$, $\frac{f(x_{0})}{2}<f(x)<\frac{3f(x_{0})}{2} $. Integrating sides from $a$ to $b$, will lead to a contradiction!
