$f(x)=x^2\qquad\qquad 0\le x\le 2\pi\qquad\qquad T=2\pi$
I was watching a youtube video about complex form of Fourier Series. For this problem (I wrote exactly what was presented in the video) the teacher said we first find $c_0$ from the formula $$c_n=\frac{1}{2l}\int_{-l}^{l} f(x)e^{\tfrac{-in\pi x}l}$$ And said since $f(x)$ is periodic function it doesn't matter if you integrate over $[-\pi,\pi]$ or $[0,2\pi]$. Then he evaluated $$c_0=\frac1{2\pi}\int_{0}^{2\pi} x^2 dx$$ But I think this is wrong, since from the graph of $y=x^2$ we can intuitively see the area under the curve for $x\in[0,2\pi]$ is more than area for $x\in[-\pi,\pi]$.
Hence I think according the formula, the correct integral should be $$c_0=\frac1{2\pi}\int_{-\pi}^{\pi} x^2 dx$$Am I right?