The Fourier coefficients for a function with period 2a over a symmetric interval is obtained by integrating over $(-a,a)$. If the interval however, is not symmetric, say $(0,2a)$, one can still integrate over $(-a,a)$ if the period is 2a and get the same Fourier Series.
I will attach pictures with full solutions, but here's a summary of my conundrum:
I thought this was pretty straightforward and I calculated the F-series for $f(x)=x~~$ on $(0,1)$ in part a) by integrating the coefficients on $(-\frac{1}{2},\frac{1}{2})$. This will be a sine series, so I shifted the sine function by using $sin(x-\frac{1}{2})$ in the series representation and it gave the correct series.
In part b) for the function $f(x) = ax^2+bx+c$ on $(0,2\pi)$ I used the same method and it did not work. The solution is close but the correct solution has an extra term of $2a\pi ~sin(nx)$ which I can't explain. (see top of third pic) What am I missing?
Showing what I did in LaTex will take a lot of time to write, hence the pictures.