$c=-\pi{}\\[2ex]c+2l=\pi{}\\[2ex] i.e\ 0+2l=2\pi{}\ \\[2ex] \therefore{}l=\pi{}\\[2ex] Now,\\[2ex] \displaystyle a_0=\frac{1}{l}\int_c^{c+2l}f\left(x\right)dx\\[2ex] \displaystyle a_0=\frac{1}{\pi{}}\int_0^{\pi{}}x^2dx\ \\[2ex] \displaystyle a_0=\frac{1}{\pi{}}{\left[\frac{x^3}{3}\right]}_0^{\pi{}}\\[2ex] a_0 \displaystyle =\frac{1}{3\pi{}}[{\pi{}}^3-0] \\[2ex] \displaystyle a_0=\frac{{\pi{}}^2}{3}\ \ \\[2ex] $

$\displaystyle a_n=\ \frac{1}{l}\int_c^{c+2l}f\left(x\right)\cos\frac{n\pi{}x}{l}dx\ \\[2ex] \displaystyle a_n=\frac{1}{\pi{}}\int_0^{\pi{}}\ \ x^2\cos{nx}\ dx \\[2ex] \displaystyle a_n=\frac{1}{\pi{}}{\left[x^2\ \left(\frac{\sin{nx}}{n}\right)-\left(2x\right)\left(-\frac{\cos{nx}}{n^2}\right)+2\ \left(-\frac{\sin{nx}}{n^3}\right)\right]}_0^{\pi{}} \\[2ex] \displaystyle a_n=\frac{1}{\pi{}}\left[{\pi{}}^2\times{}0-\frac{2\pi{}\left(-{\left(-1\right)}^n\right)}{n^2}+2\left(0\right)-0+0+0\right] \\[2ex] \displaystyle a_n=\frac{2{\left(-1\right)}^n}{n^2}\ \ \\[2ex] $

$\displaystyle b_n=\ \frac{1}{l}\int_c^{c+2l}f\left(x\right)\sin\frac{n\pi{}x}{l}dx \\[2ex] \displaystyle b_n=\frac{1}{\pi{}}\int_0^{\pi{}}\ \ x^2\sin{nx}\ dx \\[2ex] \displaystyle b_n=\frac{1}{\pi{}}{\left[x^2\ \left(\frac{-\cos\ {nx}}{n}\right)-\left(2x\right)\left(-\frac{\sin{nx}}{n^2}\right)+2\ \left(\frac{\cos{nx}}{n^3}\right)\right]}_0^{\pi{}}\ \\[2ex] \displaystyle b_n=\frac{1}{\pi{}}\ \left[{\pi{}}^2\left(-\frac{{\left(-1\right)}^n}{n}\right)-\left(2\pi{}\right)\times{}0+2\frac{{\left(-1\right)}^n}{n^3}-0-0-\frac{2}{n^3}\right] \\[2ex] \displaystyle b_n=\frac{1}{\pi{}n^3}\ \left[\left(2-{\pi{}}^2n^2\right){\left(-1\right)}^n-2\right] \\[2ex] $

$\displaystyle In\ fourier\ series, \\[2ex] \displaystyle \\[2ex] \displaystyle f(x)=\ \frac{a_0}{2}+\sum_{n=1}^{\infty{}}a_n\ \cos\frac{n\pi{}x}{l}+\sum_{n=1}^{\infty{}}b_n\sin\frac{n\pi{}x}{l} \\[2ex] \displaystyle f\left(x\right)=\frac{{\pi{}}^2}{6}\ \ +\sum_{n=1}^{\infty{}}\frac{{2\left(-1\right)}^n}{n^2}\ \cos nx+\sum_{n=1}^{\infty{}}\frac{1}{\pi{}n^3}\ \left[\left(2-{\pi{}}^2n^2\right){\left(-1\right)}^n-2\right]\sin nx \\[2ex] $