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Let $f$ be continuous on $[a,b]$ and suppose that $f(x)\ge0$ for all $x \in\ [a,b].$

Prove that if there exists a point $c \in\ [a,b]$ such that $f(c)>0$, then $$\int_{a}^b f(x)\,dx > 0 .$$

I feel like by proving this function is uniformly continuous, which is trivial, that it somehow shows me that the integral is greater than $0$, but I don't quite know how to get there. Can someone help?

Thanks! On a side note, I don't think the mathtex stuff came out right, let me know how to fix that!

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The function is continuous - what does this imply about the set $\{x: f(x)>0\}$? – Brad Dec 7 '12 at 4:15

let $c\in [a,b]$ : $f(c)>0$ then

from continuous we have

$\exists \delta >0 : \forall x \in (c-\delta, c+\delta)$ $\qquad f(x)>0$

now, we have

$\int_a^b f(x)dx=\int_a^{c-\delta} f(x)+\int_{c-\delta}^{c+\delta} f(x)dx+\int_{c+\delta}^b f(x)d(x)\geq \int_{c-\delta}^{c+\delta} f(x)dx >0$

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Hint: Let $f$ be positive at the point $p$. Then by the definition of continuity, there is an interval about $p$ (one-sided if $p$ is an endpoint) such that $f(x)\gt f(p)/2$ in that interval. Thus the integral over that interval is positive. The integral(s) over the rest of $[a,b]$ are $\ge 0$. So the full integral over $[a,b]$ is positive.

The above sketch assumes that you have already proved that if $f\ge g$ on some interval $[c,d]$, then $\int_c^d f(x)\,dx \ge \int_c^d g(x)\,dx$. It also assumes the result that if $a\le c\le b$, then $\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx$.

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