Is it correct to take 0 out of the integral sign? Why is the definite integral of 0 equal to 0 but its indefinite integral equal to an arbitrary constant C? Is it correct to take 0 out of the integral sign? But that way, the indefinite integral of 0 ends up being only 0 while it should be C.
 A: The definite integral $\int_a^b0\,\mathrm dx$ can be computed as limit of Riemann sums, and already the individual Riemann sums are all zero, hence so is their limit.
The indefinite integral $\int0\,\mathrm dx$ is the general form of a function² whose derivative is the zero function. Apparently, every constant function fits.

² Strictly speaking, this is a set of functions and the common notation $\int f(x)\,\mathrm dx=F(x)+C$ might be considered an abuse of notation. A formally more correct notation would be something like $\int f(x)\,\mathrm dx=\{\,F(x)+C\mid C\in\Bbb R\,\}$, but that is cumbersome (and confusing). On the other hand, the common notation may also turn out to be confusing as soon as several integration constants come into play ...
A: The antiderivative
$$\int f(x)dx  = F(x) \iff F'(x) = f(x)$$
so it isn't unique. Per Newton-Leibniz formula ($f$ assumed to be continuous in $[a,b]$)
$$\int _a^b f(x) dx = F(b) - F(a) $$
As the antiderivative of $0$ is any constant (constant function)  we would have
$$\int _a^b 0dx = C(b) - C(a) = 0. $$
