$\int f(x)\,\mathrm{d}x$ is an antiderivative. It represents any function whose derivative with respect to $x$ is $f(x)$.
$\int_0^af(x)\,\mathrm{d}x$ is a definite integral, and for any of the antiderivatives $g(x)=\int f(x)\,\mathrm{d}x$ (which incorporate a constant of integration),
$$
\int_0^af(x)\,\mathrm{d}x=g(a)-g(0)
$$
For example,
$$
\int x^3\,\mathrm{d}x=\frac14x^4+C
$$
for some constant $C$, and
$$
\begin{align}
\int_0^ax^3\,\mathrm{d}x
&=\left(\frac14a^4+C\right)-\left(\frac140^4+C\right)\\
&=\frac14a^4
\end{align}
$$
no matter which $C$ is chosen.
In the case above, $\int_0^xf(x)\,\mathrm{d}x$, there is confusion because the same variable is used inside the integration as in the bounds. The bound variable $x$ inside the integral is not the same as the free variable $x$ in the limit. To reduce the confusion, your integral can also be written as $\int_0^xf(t)\,\mathrm{d}t$ by renaming the bound variable. In any case, this is a definite integral.