Question

Questionvvv

Answer 1

lol

Answer 2

\[\large \rm a_n = \frac{1}{2}\int\limits_{-L}^{L} x \cos(\frac {n \pi x}{L}) dx\]\[\large \rm a_n = \frac{1}{2}\int\limits_{-2}^{2} x \cos(\frac {n \pi x}{2}) dx\] \[\rm u= x ~~~,~~~du = dx \]\[\large\rm dv= \cos(\frac{ \pi n x}{2})~dx,~~~ v=\frac{2\sin(\frac{ \pi n x }{2})}{\pi n}\]\[\frac{1}{2} [x \cdot \frac{2\sin(\frac{ \pi n x}{2})}{\pi n } -\int\limits_{ }^{ } \frac{2\sin ( \frac{ \pi n x}{ \pi n })}{\pi n}]\]\[\frac{1}{2}[x \cdot \frac{2\sin(\frac{ \pi n x}{2})}{\pi n }-(\frac{2}{\pi n}-cos(\frac{\pi n x }{2}) \cdot \frac{2}{\pi n})]\]
intergrate from -2 to 0
\[\frac{1}{2}[(\frac{2}{ \pi n })^2 -[0 +(\frac{2}{\pi n })^2] =0\]

Answer 3

this is totally wrong.
i'll post the correct answer..later