Finding a constant to make this function periodic So I have the function $$ f(x) = x^2 + b $$ which is $ 2\pi $ periodic for $ 0 \leq x < 2\pi $ 
If $ F $ is the antiderivative of $ f $ with $ F(0) = 1 $, what value of $ b $ is $ F $ periodic?
So I found that $$ F(x) = \frac{1}{3}x^{3}+bx+1 $$ so in order to prove something is periodic I need to show that:
$$ F(x) = F(x+2\pi) $$
So I plugged in $ (x+2\pi) $ in $ F $:
$$ \frac{1}{3}x^{3}+bx+1 = \frac{1}{3}(x+2\pi)^{3}+b(x+2\pi)+1 $$
and then I solved for $ b $ to get:
$$ b = -x^{2}-2\pi x-\frac{4}{3}\pi^2 $$
$ b $ is a real constant, my question is how do I solve for this $ b $ to get a real constant?

Note that $f(x)=x^2+b$ only in the interval $0\leq x<2\pi$. For all other $x$ we have to compute first the "fractional part mod $2\pi$ of $x$".  In other words we have
$$f(x):=\left(x-2\pi\left\lfloor{x\over2\pi}\right\rfloor\right)^2+b\qquad(-\infty<x<\infty)\ .$$
The antiderivative $F$ is obtained by inserting this $f$ into
$$F(x):=1+\int_0^x f(t)\>dt\qquad(x\in{\mathbb R})\ ,$$
and $F$ is in a way staircase like again. For $F$ to be periodic we need $F(2\pi)=F(0)=1$, which amounts to the condition
$$\int_0^{2\pi}f(t)\>dt=\left({x^3\over3}+bx\right)\biggr|_0^{2\pi}=0\ ,$$
hence
$$b=-{4\pi^2\over3}\ .$$
This value of $b$ makes $F$ in fact periodic.
