The Fourier series of $x^2$ is:
$$x^2 = \frac{4}{3} + 16\sum_{r=1}^{\infty}\frac{(-1)^r}{\pi^2r^2} \cos(\frac{\pi r x}{2}) \\ 0 \le x \le 2$$
To find the Fourier series of $x^3$, we can integrate term by term to obtain:
$$\frac{x^3}{3} = \frac{4x}{3}+32\sum_{r=1}^{\infty} \frac{(-1)^r}{\pi^3 r^3} \sin(\frac{\pi r x}{2}) + c$$
where c is the constant of integration.
However, I have noted from my textbook, that this is not the actual Fourier series and the $\frac{4x}{3}$ must written into a Fourier series by differentiating the series for $x^2$
Why is the differentiation required? Could we simply leave as it is? Would the deviation be great?
No , you cannot leave $\frac{4x}{3}$ , as a Fourier series contains a constant and the terms containing sine and cosines. So you have to expand the linear part also.