Question: Let $f(x)=x-x^2$ in $[0,1]$. Define the Fourier Serie in that interval.
My Try: I define the Fourier Serie to $g(x)=x$ in $[0,1]$ as $\mathcal{F(g)}$. As $f$ is continuous, differentiable by parts and $f(0)=f(1)$, so I intend to use the theorem of uniform converge for Fourier Serie of $f$ so that
$$\int_{0}^{x} f'(t) dt =\int_{0}^{x} g'(t)dt - \int_{0}^{x} 2g(t)dt$$
My question is: Can I do the integration using the the fundamental theorem of calculus for $f'(t)$ and $g'(t)$ (and replace by $\mathcal{F(g)}$ of $g$, again because the uniform convergence of this) and for $2g(t)$ integration term-by-term of the $2\mathcal{F(g)}$?
Although, I know that can do the problem in a straightforward way using definitions, I want to learn the theorems to make easier the calculations.
Suppose you have the Fourier series for $g(x)=x$ on $[0,1]$: $$ x=\sum_{n=-\infty}^\infty a_ne^{2\pi i nx}\tag{1} $$
To get the Fourier series of $x^2$ from (1), you need to do integration term by term, which would be justified by uniform convergence of the Fourier series.
If you don't know how to deal with the exponential functions, then just rewrite (1) by using the Euler formula $e^{ix}=\cos x+i\sin x$.