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I used two different theorems to come to my conclusion:

1. The definite integral with limits [-a,a] of a function f(x) will always be equal to zero if the function is odd.

[itex]\int[/itex][itex]^{a}_{-a}[/itex]f(x)dx = 0

2. Suppose a function F(x) is T-periodic function then for any real number a we have:

[itex]\int[/itex][itex]^{T}_{0}[/itex]f(x)dx = [itex]\int[/itex][itex]^{T+a}_{a}[/itex]f(x)dx

I have a problem where I need to find the Fourier Series of the function f(x) = 2x

^{2}-1

on the interval 0[itex]\leq[/itex]x[itex]\prec[/itex]2L and it tells me that the function has a period of 2L

now when I use the equation to find the b

_{n}fourier coefficient I get:

b

_{n}=[itex]\frac{1}{L}[/itex][itex]\int[/itex][itex]^{2L}_{0}[/itex](2x

^{2}-1)sin([itex]\frac{xnπ}{L}[/itex])

When I multiply sine by the 2x

^{2}-1 you can easily see that I get two integrals of odd functions sin(xnπ/L)(2x

^{2}) and sin(xnπ/L)

If we plug in -x to both functions we see that we get the negative of the function therefore we know both are odd functions. Now because the problem tells us that the function is of period 2L and we're integrating from 2L I apply my second theorem and subtract L from the limits which now gives an integral that satisfy the first theorem and by which I then conclude that both integrals should be equal to 0. But after calculating the integral I see that it doesn't and I'm not sure where I'm making a false assumption.