Show that $\int_a^b f(x)\,dx>0$.

Question

Suppose $f(x)$ is continuous on $[a, b]$ with $f(x)\geq 0$ and such that $f(x) > 0$ for some $x\in\mathbb [a,b]$. Show that $\int_a^b f(x)\,dx>0$.

Answer 1

Let $f(x) = c > 0$ for some $x$. Choose $\delta > 0$ so that if $y \in (x-\delta, x+\delta)$ then $f(y) > \frac{c}{2}$. Then $\int_{a}^b f(y) dy \ge \int_{x - \delta}^{x+\delta} f(y) dy \ge \delta c > 0$