Fourier Coefficient

Question

I have to compute the coefficient $b_3$ of the odd Fourier Series associated with the function $y=2-x$ in the interval $(0,1)$, period $2$. By using the formula $$ b_k = \frac{1}{T}\int_{-T}^{T} f(t)\sin\left(\frac{k\pi t}{T}\right)dt $$ I get $$ b_3 = \frac{1}{2}\int_{-2}^{2} (2-x)\sin\left(\frac{2\pi 3}{2}\right)dt = -\frac{4}{3\pi} $$ but the text of the exercise tells me it should be 2/$\pi$. Is my answer wrong?

Answer 1

Thanks to Aaron up there for pointing me to the right direction.

Being the period $2$, the formula becomes $$ \frac{1}{1}\int_{-1}^1 (2-t)\sin(3\pi t)dt = 2\int_0^1(2-t)\sin(3\pi t)dt = \frac{2}{\pi} $$