If $f:[a,b]\to \mathbb{R}$ is continuous and nonnegative and $\int_a^b{f}=0$, then $f(x)=0$ for all $x\in [a,b]$

Assume $f(x_0) > \epsilon > 0$ for some $x_0 \in (a, b)$. By continuity, we can find $\delta > 0$ so that: $$ \forall x \in (x_0 - \delta, x_0 + \delta)\subset [a, b] : \left|f(x) - f(x_0)\right| < \frac{\epsilon}{2} $$

By the reverse triangle inequality, $\left|f(x)\right| > \left|f(x_0)\right| - \dfrac{\epsilon}{2} > \dfrac{\epsilon}{2}$.

Thus: $$ \int_a^b f(x) \,dx \ge \int_{x_0 - \delta}^{x_0 + \delta} f(x) \,dx \ge 2\delta \frac{\epsilon}{2} = \delta \epsilon > 0 $$

Hint: If you suppose $f(x)\neq 0$ for all $x$, then there is some $x$ such that $f(x)>0$. This means there is some interval around $f(x)$, say $(-\varepsilon,\varepsilon)$ such that $f(x)>\delta>0$. This means $$\int_a^bf(x)\,dx\geq\int_{-\varepsilon}^\varepsilon f(x)\,dx\geq\delta\int_{-\varepsilon}^\varepsilon\,dx=2\varepsilon\delta>0,$$a contradiction. I'll leave justifying the details to you.