So I have been trying to solve this question for some time now but I couldn't find much information on how to solve it. There is much information in the case of for (-π,π) over the period of 2π but nothing in the case of (0,2π).
Problem:
Let f be a $2π$-periodic function given in the $[0,2π[$ as: $$f(x)=1 , x ∈ [0,\frac{7π}{16}[$$ $$f(x)=5 , x ∈ [\frac{11π}{16},\frac{7π}{16}[$$ $$f(x)=4 , x ∈ [\frac{11π}{16},2π[$$
We need to find $a_0$ , $a_n$,$b_n$ and finally find the value of the series for $x =\frac{7π}{16}$
Hint: It's actually just a matter of applying the definition
\begin{align*}a_n&=\frac{1}{\pi}\int_0^{2\pi}f(x)\cos(nx)\,dx\\ &=\frac{1}{\pi}\int_0^{7\pi/16}1 \cdot \cos(nx)\,dx+\frac{1}{\pi}\int_{7\pi/16}^{11\pi/16}5 \cdot \cos(nx)\,dx+\frac{1}{\pi}\int_{11\pi/16}^{2\pi}4 \cdot \cos(nx)\,dx\\ &=\frac{1}{\pi}\int_0^{7\pi/16}\cos(nx)\,dx+\frac{5}{\pi}\int_{7\pi/16}^{11\pi/16}\cos(nx)\,dx+\frac{4}{\pi}\int_{11\pi/16}^{2\pi}\cos(nx)\,dx\\ &= \frac{\sin(7n\pi/16)-\sin(0)}{\pi}+\frac{5\sin(11n\pi/16)-5\sin(7n\pi/16)}{\pi}+\frac{4\sin(2n\pi)-4\sin(11n\pi/16)}{\pi}\\ &= \frac{\sin(11n\pi/16)-4\sin(7n\pi/16)}{\pi} \end{align*}
You can take care of the rest! Let me know if you encounter any trouble.
This of course works in general for any stepwise defined function $f(x)=\sum c_n\chi_{I_n}(x)$