When $f(0) = 0,$ is it always true that $G(0) = 0,$ where $G$ is the antiderivative of $f$? I have a hunch that it is, but it would be nice if somebody could confirm / disprove it for me.

Thank you.

Edit
Is it when the constant of integration is equal to zero?
No, this is not true in general. Let $F(x)$ be a anti-derivative of $f(x)$ then $$\int f(x) \, \mathrm{d}x = F(x) + k$$ now, even if $F(0) = 0$ there's still the $+ k$ constant to contend with. 

An example would be $f(x) = x\exp x$ then we have $$F(x) = \int x\exp x \, \mathrm{d}x = e^x (x-1) + k$$ Now, even if $k = 0$, then $F(0) = -1$ whilst $f(0) = 0$.
No. The simplest example I can think of?
$$f(x) = x$$
Then $G(x) = \frac{x^2}{2} + C$, which is not guaranteed to be always $0$.