We have $g:RR->RR$ a continous function with property that $int_(-x)^xg(t)dt=0$ for each $x inRR$ .How to demonstrate that g is an odd function?

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Answer 1 · 2017-04-20T14:52:53

Define $f(x) = int_0^x g(t) dt$ and use the fact that if $f$ is even then $f'$ is odd.

Suppose that $g$ is continuous on $RR$ and for every real $x$ , we have $int_-x^x g(t) dt = 0$ .

Define $f(x) = int_0^x g(t) dt$ .

Then for every $x$ , $f'(x) = g(x)$

Then $int_-x^x g(t) dt = [ f(t) ]_-x^x = f(x) - f(-x) = 0$

Therefore, $f(-x) = f(x)$ so $f$ is even.

This implies that $f'$ is odd, and since $f' = g$ , we see that $g$ is odd.

To see that $f$ even implies $f'$ odd, apply the chain rule to $f(-x)$

Answer 2 · 2017-04-20T15:05:15

See below.

$int_(-x)^x g(t)dt=int_(-x)^x g(-t)(-dt)=-int_(-x)^xg(-t)dt$

so

$int_(-x)^x (g(t)+g(-t))dt=0 forall x$ then $g(t) = -g(-t)$ so $g(t)$ is odd.

We have g:RR->RRg:RR->RR a continous function with property that int_(-x)^xg(t)dt=0int_(-x)^xg(t)dt=0 for each x inRRx inRR.How to demonstrate that g is an odd function?