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} \\ $

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

$$ \begin{aligned} a_{n} &=\int_{0}^{2} f(x) \cos n \pi x d x=\int_{0}^{1} \pi x \cos n \pi x d x+\int_{1}^{2} \pi(2-x) \cos n \pi x d x \\\\ &=\left[\pi x \cdot \frac{\sin n \pi x}{n \pi}-\pi\left(-\frac{\cos n \pi x}{n^{2} \pi^{2}}\right)\right]{0}^{1}+\left[\pi(2-x) \cdot \frac{\sin n \pi x}{n \pi}-(-\pi)\left(-\frac{\cos n \pi x}{n^{2} \pi^{2}}\right)\right]{1}^{2} \\\\ &=\left[\frac{\cos n \pi}{n^{2} \pi}-\frac{1}{n^{2} \pi}\right]+\left[-\frac{\cos 2 n \pi}{n^{2} \pi}+\frac{\cos n \pi}{n^{2} \pi}\right]=\frac{2}{n^{2} \pi}(\cos n \pi-1)=\frac{2}{n^{2} \pi}\left[(-1)^{n}-1\right] \\ \end{aligned} $$

$=0$ or $-\frac{4}{n^{2} \pi}$ according as $n$ is even or odd. $$ \begin{aligned} b_{n} &=\int_{0}^{2} f(x) \sin n \pi x d x=\int_{0}^{1} \pi x \sin n \pi x d x+\int_{1}^{2} \pi(2-x) \sin n \pi x d x \\ &=\left[\pi x\left(-\frac{\cos n \pi x}{n \pi}\right)-\pi\left(-\frac{\sin n \pi x}{n^{2} \pi^{2}}\right)\right]{0}^{1}+\left[\pi(2-x)\left(-\frac{\cos n \pi x}{n \pi}\right)-(-\pi)\left(-\frac{\sin n \pi x}{n^{2} \pi^{2}}\right)\right]{1}^{2} \end{aligned}\ $$

$$ \begin{aligned} &=\left[-\frac{\cos n \pi}{n}\right]+\left[\frac{\cos n \pi}{n}\right]=0 \\\\ \therefore \quad f(x) &=\frac{\pi}{2}-\frac{4}{\pi}\left(\frac{\cos \pi x}{1^{2}}+\frac{\cos 3 \pi x}{3^{2}}+\frac{\cos 5 \pi x}{5^{2}}+\ldots . .\right) . \end{aligned}\\ $$

Note. Putting $x=0$, we have, $ f(0)=\frac{\pi}{2}-\frac{4}{\pi}\left(\frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\ldots \ldots\right) \\ $

$$ \begin{aligned} &0=\frac{\pi}{2}-\frac{4}{\pi}\left(\frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\ldots \ldots\right) \\\\ &\frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\ldots \ldots=\frac{\pi^{2}}{8} . \end{aligned} \\ $$