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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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.
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.
