I plan it as follows:
$a0=\frac{1}{\pi 2}\int _{-2}^2x\cdot dx\:$
$an=\frac{1}{\pi }\int _{-2}^2x\cdot \:cos\left(nx\right)dx\:$
$bn=\frac{1}{\pi }\int _{-2}^2x\cdot \:sin\left(nx\right)dx\:$
But my teacher says that the argument (nx) is wrong in an. I do not understand why? I think that being odd an=0. But the integral as it is correct?. Thanks for the help I have exam tomorrow
Note $L=2$ thus $$a_0=\frac{1}{4}\int_{-2}^{2}xdx=0$$ $$a_n=\frac{1}{2}\int_{-2}^{2}x\cos\left(\frac{n\pi x}{2}\right) dx$$ $$b_n=\frac{1}{2}\int_{-2}^{2}x\sin\left(\frac{n\pi x}{2}\right) dx$$