Here's a question I can't figure out =[

Suppose f(x) is a 2L-periodic function: $\displaystyle f(x+2L) = f(x)$. Show:

$\displaystyle \int_{-L}^{L} f(x) dx = \int_{a}^{a+2L} f(x) dx $

for any value of a.

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- Feb 2nd 2010, 11:25 AMCreebePeriodic Functions and Fourier Series
Here's a question I can't figure out =[

Suppose f(x) is a 2L-periodic function: $\displaystyle f(x+2L) = f(x)$. Show:

$\displaystyle \int_{-L}^{L} f(x) dx = \int_{a}^{a+2L} f(x) dx $

for any value of a. - Feb 4th 2010, 10:50 AMAryth
You just need to show that the integral yields the same value when the period is used as the interval. You can see that [-L,L] is 2L in length, and then [a, a+2L] is also 2L in length, and since f(x) is a 2L periodic function, then it is the same over all periods, regardless of the starting point.