= Solution
The <function> is an <odd function>, so $a_0=a_n=0$. Here $L=1$, and the <Fourier coefficients> are
$$
b_n=2\int_0^1x(1-x)\sin(n\pi x)\,dx.
$$
Write $k=n\pi$. One <integration by parts> has zero endpoint terms because $x(1-x)$ vanishes at both endpoints:
$$
b_n=\frac2k\int_0^1(1-2x)\cos(kx)\,dx.
$$
A second <integration by parts> gives
$$
\int_0^1(1-2x)\cos(kx)\,dx
=\frac2k\int_0^1\sin(kx)\,dx
=\frac{2(1-\cos k)}{k^2}.
$$
Therefore $b_n=4(1-(-1)^n)/(n^3\pi^3)$, zero for even $n$ and $8/(n^3\pi^3)$ for odd $n$. The <Fourier series> is
$$
\boxed{f(x)=\frac8{\pi^3}\sum_{m=0}^\infty\frac{\sin((2m+1)\pi x)}{(2m+1)^3}}.
$$
The coefficients are absolutely summable, and the periodic <function> is continuous even at the joins $x=\pm1$, so the displayed <Fourier series> represents it everywhere.
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