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 $\int_a^b f(x)\,dx>0$.
Show that $\int_a^b f(x)\,dx>0$.
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calculus
integration
proof-writing
epsilon-delta
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1Let $f(x_0)>0$. The integral on $[a,b]$ is $\ge$ the integral on a suitably small neighbourhood of $x_0$. Now... – 2017-02-09
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0And [this](http://math.stackexchange.com/questions/1223367/if-f-geq-0-is-continuous-and-int-ab-fx-dx-0-then-f-0?rq=1). – 2017-02-09
1 Answers
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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 $\int_{a}^b f(y) dy \ge \int_{x - \delta}^{x+\delta} f(y) dy \ge \delta c > 0$