If f is continuous on [a,b] and $f(x) \geq 0$, for $x \in [a,b]$, but $f$ is not the zero function, prove that $\int_{a}^{b} f(x) dx > 0$

Question

If f is continuous on [a,b] and $f(x) \geq 0$, for $x \in [a,b]$, but $f$ is not the zero function, prove that $\int_{a}^{b} f(x) dx > 0$

Could anyone give me a hint for this proof please?

Answer 1

If $f(c) >0$ then there exists $r>0$ such that $f(x) \geq \frac {f(c)} 2$ for $|x-c| \leq r$. Hence $\int_a^{b} f \geq \int_{c-r}^{c+r} f(x) dx \geq \frac {f(c)} 2 (2r) >0$.

Answer 2

f is not the zero function, and f ≥ 0, so f >0 somewhere. Since f is continuous, f>0 in a tiny interval, then the integral is positive in this interval.