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?

Only polyinomials that can be periodic are constant polynomials.
Say you have one nonconstant polynomial. Then it has a root (by the fundamental theorem of algebra). But then it has infinite number of roots which is impossible.
So there is no solution to your problem.
