Why does the following integral starts from $0$? Consider $$f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} x^{3n+1}$$
It's a power series with a radius, $R=1$. at $x=1$ it converges. Hence, by Abel's thorem:
$$\lim_{x\to 1^-} f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1}$$
Evaluating the derivative
$$f'(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} (3n+1)x^{3n+1} = \ldots = \frac{1}{1+x^3}$$
Now, consider this claim: "Since $f(0) = 0$: $$f(x) = \int_0^x \frac{1}{1+t^3} \ dt$$
I am familiar with the fundamental theorem of calculus, yet not sure why this claim true, More presicely; Why is the integral starts at $0$?
Thanks.

Note the following corollary:

$$\int_a^b f(t)\, dt = F(b)-F(a)$$

You are free to choose the points $a$ and $b$ as long the function $f$ is defined in $[a,b]$. But choosing $a$ such that $F(a)=0$ simplifies things.
