I am given that $$c_0 + \sum^{\infty}_{k=1}b_k\sin{\frac{k\pi t}{2}}$$ is the fourier series of $g(t)$ as given in this image:
I notice that the period is $T=4$, and $c_0 = 1$.
I am now asked to calculate the sum: $$\sum^{\infty}_{k=1}|b_k|^2.$$
Parsevals theorem states that: $$\frac{1}{T}\int_{period}|g(t)|^2dt = |c_0|^2 + \frac{1}{2}\sum^{\infty}_{k=1}(|a_k|^2 + |b_k|^2),$$ so in my case, with $a_k = 0, c_0 = 1, T=4$, I get: $$\sum^{\infty}_{k=1}|b_k|^2 = \frac{2}{4}\int_{-2}^{2}|g(t)^2| dt - 2 = \frac{2}{4}\left( \int_{-2}^{-1} 0 dt + \int_{-1}^{1}|t^2|dt + \int_{1}^{2}2^2 dt\right) - 2 = \frac{1}{3}$$
The answer is however supposed to be $\frac{4}{3}$. Am I using Parseval's theorem wrong or what is going on?