Using the Fourier Cosine Series for Integral Calculation

[libelec]

Homework Statement

Using the Fourier cosine series for $$\[f(x) = \left\{ \begin{array}{l}<br /> 1,x = 0 \\ <br /> 10,x = \pi \\ <br /> x,x \in (0,2\pi ) - \left\{ {0,\pi } \right\} \\ <br /> \end{array} \right.\]$$, find a series that converges to $$\[\int\limits_0^{2\pi } {{x^2}dx} \]$$

The Attempt at a Solution

For the Fourier cosine series, I need the even extention of f(x), that is, $$\[f(x) = \left\{ \begin{array}{l}<br /> - x,x \in ( - 2\pi ,0) - \left\{ {0, - \pi } \right\} \\ <br /> 10,x = - \pi \\ <br /> 1,x = 0 \\ <br /> 10,x = \pi \\ <br /> x,x \in (0,2\pi ) - \left\{ {0,\pi } \right\} \\ <br /> \end{array} \right.\]$$. Now, $$\[\int\limits_0^{2\pi } {{x^2}dx} \]$$ = $$\[2\left\| {f(x)} \right\|_2^2\]$$, so I can use Parseval's equality, right?

But if that's correct, I'm unable to find a series that converges to that defined integral, since it has constant terms: $$\[\frac{8}{3}{\pi ^3} = \frac{{{\pi ^3}}}{4} + \frac{8}{\pi }\sum\limits_{n = 1}^\infty {\frac{1}{{{{(2n - 1)}^4}}}} \]$$

Evidently, I'm doing something wrong, but I don't know what. Is it the cosine series? Is it the convergence of the series? Or is it that I can't use Parseval's?

Thanks.

 

[libelec]

OK, I just realized that the even extention is wrong. But that would solve anything if I still have the a₀ term in the series.

 

[libelec]

Anybody? I still can't solve it. And the even extension is OK.