Something about Fourier expansion Suppose f is continuity on $[-\pi,+\pi],period ~T=2\pi$
$$f(x)\sim \frac{a_0}{2}+\sum_{n=1}^{+\infty}a_n \cos nx +b_n\sin nx$$
$$F(x)=\int_0^x[f(t)-\frac{a_0}{2}]dt$$
1)Calculate the F(x) Fourier expansion
2)proof
$$\int_a^bf(t)dt=\int_a^b \frac{a_0}{2}dt +\sum_{n=1}^{\infty}\int_a^b a_n \cos nx +b_n\sin nx dt$$
My attempt 
1)$a_n=\int_{-\pi}^{\pi}f(t)\cos ntdt$
$b_n=\int_{-\pi}^{\pi}f(t)\sin nt dt$
Then I don’t know what to do next
2)
I think it’s a little like uniformly convergence’character
But I don’t have idea to proof this
 A: Let $$F(x) \sim \frac{c_0}{2}+\sum_{n=1}^{\infty}c_n \cos nx +d_n\sin nx.$$ Since $F$ is $2\pi$-periodic $C^1$, note that the Fourier series of $F$ converges uniformly to $F$ and therefore equality holds ($\alpha$-Holder continuity is one of the sufficient conditions for uniform convergence, see e.g. this earlier post). Now, we can find $c_n$ and $d_n$ as follows.
$$
\pi c_n = \int_{-\pi}^\pi F(t)\cos nt dt = \frac{\sin nt}{n}F(t)|^{\pi}_{-\pi}-\int_{-\pi}^\pi (f(t)-\frac{a_0}{2})\frac{\sin nt}{n}dt = -\pi \frac{b_n}{n}, \quad n\neq 0,
$$
$$
\pi d_n = \int_{-\pi}^\pi F(t)\sin nt dt = \frac{-\cos nt}{n}F(t)|^{\pi}_{-\pi}+\int_{-\pi}^\pi (f(t)-\frac{a_0}{2})\frac{\cos nt}{n}dt = \pi \frac{a_n}{n}.
$$ This gives
$$
F(x) = \frac{c_0}{2}+\sum_{n=1}^{\infty}\left(-\frac{b_n}{n}\right)\cos nx +\frac{a_n}{n}\sin nx\tag{*}.
$$ where $\frac{c_0}{2} = \sum_{n=1}^\infty \frac{b_n}{n}$ from $F(0) = 0$. It can be also calculated explicitly:
$$\begin{eqnarray}
\pi c_0 = \int_{0}^{2\pi} F(t) dt& =& (t-\pi)F(t)|^{2\pi}_{0}-\int_{0}^{2\pi}(t-\pi) (f(t)-\frac{a_0}{2})dt \\&=&\int_{0}^{2\pi}\left(\sum_{n=1}^\infty \frac{2}{n}\sin nt\right) (f(t)-\frac{a_0}{2})dt \\
&=&2\pi\sum_{n=1}^\infty \frac{b_n}{n}.
\end{eqnarray}$$
For (b), note that the stament is equivalent to
$$
F(b) - F(a) = \sum_{n=1}^\infty \int_a^b \left(a_n \cos nx +b_n\sin nx\right)dx.
$$ But this is obvious from $(*)$:
$$\begin{eqnarray}
\sum_{n=1}^\infty \int_a^b \left(a_n \cos nx +b_n\sin nx\right)dx &=&\sum_{n=1}^{\infty}\frac{a_n}{n}(\sin nb-\sin na)+\left(-\frac{b_n}{n}\right)(\cos nb-\cos na)\\& =& F(b) - F(a). 
\end{eqnarray}$$
