f(x)=∑bnsin(nπx/L) (for n∈[1,∞)), where bn=(2/L)∫f(x)sin(nπx/L)dx.
In this problem, since 0<x<2, L=2 and f(x)=-23x
bn=∫(-23x)sin(nπx/2)dx for x∈[0,2].
bn=-23∫xsin(nπx/2)dx.
Let u=x, du=dx; dv=sin(nπx/2)dx, v=-(2/(nπ))cos(nπx/2),
bn=-23(2x/(nπ))cos(nπx/2)+(2/(nπ))∫cos(nπx/2)dx,
bn=-23[2x/(nπ))cos(nπx/2)+(2/(nπ))2sin(nπx/2)]02,
bn=-23(4/(nπ))(-1)n=23(4/(nπ))(-1)n+1.
f(x)=∑(92/(nπ))(-1)n+1sin(nπx/2)=(92/π)∑{(-1)n+1/n}sin(nπx/2).
[Note: f(x+4)=f(x); so sin(nπ(x+4)/2)=sin(nπx/2+2πn)=sin(nπx/2).]