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I have qutoted that the absolute value of an integral is less than or equal to the integral of an absolute value of a function.

I have also said $|-g(x)| \le g(x) \le |g(x)|$ implies the integral g(x) is between the negative and positive absolute value integrals. I believe this implies the integral of g(x)=0?

I am struggling to put together a logical proof, any help would be great.

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I would love to answer but the way the OP, on this other page, apparently silently downvoted the two answers posted and never answered the queries for clarifications, is making me shy. –  Did Apr 26 '14 at 7:28

2 Answers 2

Another approach is based on the fact that, since $g$ is continuous, the existence of a point at which $|g(x_0)|>0$ implies the existence of a little interval $(x_0-\delta,x_0+\delta)$ where $|g|>0$. Therefore $\int_a^b |g| \geq \int_{x_0-\delta}^{x_0+\delta} |g|>0$.

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(a) Take a look of Mean Value Theorem for integration.

(b) Haow about $f(x)=x$, for $x\in[-1,1]$?

(c) You should try $f(x)=\lfloor x\rfloor$, for $x\in[0,1]$.

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I have set $m=min g(x)=0, M=max g(x)=0$, where $g(x)=0$ I have quoted the theorem saying: $0 = m \le 1/(b-a)\int_a^b |g(x)| dx \le M=0$ so that the integral is equal to zero. However, how do I show this in the other direction? –  user127700 Apr 24 '14 at 14:41

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