Expanding Fourier Series of $f(x)=x^2$ where $0<x<1$ (even and odd)

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I tried to solve Fourier series (which appeared on title) and ended up to below solution :

on even state :

$ \phi(x)= \begin{cases} x^2 & 0<x<1 \\ x^2 & -1<x<0 \end{cases} $

$a_{0}=\dfrac{2}{3}$ and also $a_{n}=\dfrac{4(-1)^n}{n^2\pi^2}$

on odd state I reached below answer :

$ \phi(x)= \begin{cases} x^2 & 0<x<1 \\ -x^2 & -1<x<0 \end{cases} $

$a_{0}=0$ and $b_{n}=\begin{cases} \dfrac{-2}{n\pi}& n \text{ is even} \\ \dfrac{-2}{n\pi}-\dfrac{8}{n^3\pi^3}& n \text{ is odd} \end{cases}$

I would appreciate any help to confirm I this answer is true or not.