For which lower bound of integration $a$ does a definite integral of $f(x)$ from $a$ to $x$ equal its antiderivative $F(x)$ with $C=0$?

Question

For an arbitrary antidifferentiable function $f(x)$, my goal is to construct a definite integral of $f(x)$: $$ \int_a^x f(t) dt $$

which is equal to one of the infinitely-many antiderivatives of $f(x)$, $F(x)+C$. Specifically, the one with a constant of integration $C=0$.

I've come up with the following relation between $a$ and $C$ which I think implies the existence of such an $a$: $$ \int f(x) dx = F(x) + C $$

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

$$ F(x) - F(a) = F(x) + C $$

$$ F(a) = -C $$

But, I don't know where to go from here.

Answer 1

If $F(x)$ is an antiderivative of $f(x)$, then for every real number $C$ the function $F(x) + C$ is also a valid antiderivative. Then, when you take the definite integral, you will get

$$\begin{eqnarray} \int_a^x f(x) dx & = & (F(x) + C) - (F(a) + C) \\ & = & F(x) - F(a) \\ \implies F(x) & = & F(a) + \int_a^x f(x) dx \end{eqnarray}$$

So what you have found is that your choice of $C$ is equivalent to fixing the value at $F(a)$.