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.

Take $f(x) = xe^{-x^2}$. Clearly, $f(0) = 0$.

The integral function of $f(x)$ is:

$$G(x) = \int_{-\infty}^{x} te^{-t^2} dt = \left.-\frac{1}{2}e^{-t^2}\right|_{t=-\infty}^{t=x} = -\frac{1}{2}e^{-x^2}$$

In this case, $G(0) = -\frac{1}{2} \
eq 0.$