How to show that $f(x) = x^{1/3}$ on [-1,1] is the indefinite integral of $g$ over $[-1,1]$?
I want to show that $f$ is an indefinite integral. I know that $f'(x)=\frac{1}{3}x^{-2/3}$and it is integrable so I can take it as $g.$ Also, I know that the definition of $f$ being indefinite integral of $g$ over $[a,b]$ provided $g$ is Lebesgue integrable over $[a,b]$ is that: $$f(x) = f(a) + \int_{a}^{x} g$$ for all $x \in [a,b].$
But how can I calculate $\int_{a}^{x} g,$ is it the ordinary integration which is equal to $\frac{1}{3}x^{-2/3} - \frac{1}{3}a^{-2/3}$, if so how will this give me the LHS, could anyone explains this for me please?
Also what about the point $0$ in the interval $[-1,1]$?
EDIT:
in the interval $[-1,1]$, the LHS will give me 2 while the RHS will give me 0.