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Let $f : [a,b] \rightarrow \mathbb{R}$ be continuous, such that $f(x) ≥ 0$, for all $x \in \mathbb{R}$. Show that:

if $\int^b_a{f(x)dx} = 0$ then $f(x) =0$ for all $x \in \mathbb{R}$

Is the idea, schematically, that as x > $0$, we find the graph of x equaling the line y = $0$?

user381164
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    I think that the contrapositive is easier to prove: Assume $f(x) \neq 0$ for some $x \in [a,b]$. Show that $\int_a^b f(x)dx \neq 0$. – Ken Duna Nov 17 '16 at 22:57
  • To continue on the idea of @KenDuna, if you assume $f(z)\neq 0$ for some $z\in [a,b]$, then you can lower bound the lower Darboux sum quite easily (by considering a partition containing a neighborhood of $z$). – Hello Nov 17 '16 at 23:04
  • This might also be relevant: http://math.stackexchange.com/questions/635874/is-f-0-if-the-integral-is-zero In particular the answer of @HagenvonEitzen which develops the idea proposed by KenDuna. Along the same lines, there is also this post: http://math.stackexchange.com/questions/82839/prove-the-integral-of-f-is-positive-if-f-%E2%89%A5-0-f-continuous-at-x-0-and-f Well, this is to say that it might have been good to search for the question before asking it... – Hello Nov 17 '16 at 23:12

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$F(x)=\int_a^xf(y)dy$ is an increasing function, $F(b)=F(a)=0$ implies that $F$ is constant and its derivative $f=0$.

Tsemo Aristide
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