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I am an adult learner trying to learn calculus again. I have learnt that for an odd periodic function with period, say, T: $$ f\left(-t\right)=-f\left(t\right) $$ $$ \int_{-\frac{T}{2}}^{\frac{T}{2}}{f\left(t\right)dt=0} $$ But what about if the integration interval is from 0 to T, i.e.: $$ \int_{0}^{T}f\left(t\right)dt $$ Will the integral still sum to 0? If yes, could anyone show me the proof?

Thanks in advance!

lyl
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    Split the integral in two and use the relation between $f$ on $[-T/2,0]$ and $[T/2,T]$. – nicomezi Jul 31 '23 at 08:53
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    On a more general note, the integral of a $T$-periodic function on an interval of length $T$ is always the same (by using similar techniques as the one given above, one can bring any of these integral to the one over $[0,T]$). – nicomezi Jul 31 '23 at 08:55

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Firstly, the integral from $-a$ to $a$ for any odd function will always be zero because of the rotational symmetry of the graph...

Secondly, if $f(x)$ is periodic with period $T$ then $f(-\frac{T}{2})=f(\frac{T}{2})$ because $f(x-T)=f(x+T)=f(x)$.

Using both of these ideas, the answer is quite clearly, Yes.

Red Five
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