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If $\int_{a}^{b} f(x)\,\mathrm dx=0$ considering Riemann integral with no condition on $f$ not even continuity, does this implies that there exist some $c \in (a,b)$ such that $f(c)=0$. With continuity i am ok and can prove this using mean value theorem, Is this still true and what's the proof (if so)?

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Though not an answer,with continuity u dont even need any theorem... :) The only way a defenite integral can be 0 is when the area below x axis and area above x axis cancells out,for this to happen the graph has to meet x axis at some point. Thus explained –  Sreekanth Karumanaghat Apr 13 '13 at 8:03

2 Answers 2

Let $f(x) = \begin{cases} -1 & \text{if $x < 0$} \\ 1 & \text{if $x \ge 0$}\end{cases}$. Now calculate $\displaystyle \int_{-1}^1 f(x) \,\mathrm dx$.

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No, if you cannot verify continuity, then you cannot verify Rolle's theorem. In other words, at the point where one might expect a zero, the function may be discontinuous.

Thus, if the function is discontinuous, there is no guarantee for any point equaling zero.

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You correctly state that you cannot apply certain theorems, but what is really needed is an example such as $f:[-1,1]\mapsto\mathbb{R}$ $$ f(x)=\left\{\begin{array}{} x&\text{if }x\ne0\\ 1&\text{if }x=0 \end{array}\right. $$ –  robjohn Apr 13 '13 at 8:33

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