So how would do below Fourier?
where $$f(x)= \begin{cases} 1-x, &x = [0,4]\\ x-3, &x = [4,8] \end{cases}$$
Should is this way to solve it?
$$b_n = \frac{1}{8}\int_{0}^{4} (1-x) \cdot \cos(\frac{2\cdot \pi\cdot n\cdot x}{4})dx +\frac{1}{8}\int_{4}^{8} (x-3) \cdot \cos(\frac{2\cdot \pi\cdot n\cdot x}{4})dx$$
It is in fact$$b_n = \frac{2}{8}\int_{0}^{4} (1-x) \cdot \cos(\frac{\pi\cdot n\cdot x}{4})dx +\frac{2}{8}\int_{4}^{8} (x-3) \cdot \cos(\frac{\pi\cdot n\cdot x}{4})dx$$since the period is $T=8$ and $L={T\over 2}=4$.