Solution:

Let, $ f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos n \pi x+\sum_{n=1}^{\infty} b_{n} \sin u \pi x \\ $

Then, $ a_{0}=\int_{0}^{2} f(x) d x=\int_{0}^{1} \pi x d x+\int_{1}^{2} \pi(2-x) d x=\pi\left[\frac{x^{2}}{2}\right]{0}^{1}+\pi\left[2 x-\frac{x^{2}}{2}\right]{1}^{2} \\ $

$$ =\pi\left(\frac{1}{2}\right)+\pi\left[(4-2)-\left(2-\frac{1}{2}\right)\right]=\pi \\ $$

$$ \begin{aligned} a_{n} &=\int_{0}^{2} f(x) \cos n \pi x d x=\int_{0}^{1} \pi x \cos …