Let's say for simplicity $F(x) = 3 x^2$, just so we have an example to talk about. As you wrote $$f(x) = \int F(x) \mathrm d x + C = x^3 +C\,.$$ And now the initial conditions come into play (the $x_0$, $v_0$ and so on) or in some cases boundary conditions. Those are always needed to find a specific solution to a differential equation. So if I also know, that $f(0) = f_0$, then we can just plug that in: $$f(0) = 0^3 +C = C = f_0$$ So we see, if $f(x)$ should also fulfill the condition $f(0) = f_0$, $C$ can not have any value, but needs to be $f_0$.

Edit:

A lot of times, when you solve a differential equation you try to find all or a whole bunch of solutions (that's the $x^3 +C$ above) and then pick one of them, that fits your initial/boundary conditions.

If you have higher order DGLs (differential equations) like $F = m \ddot x$ (where each dot means a derivative with respect to time), then you need more than one initial condition to solve this for $x(t)$ (the initial position and initial velocity).

Let's say for simplicity $F(x) = 3 x^2$, just so we have an example to talk about. As you wrote $$f(x) = \int F(x) \mathrm d x + C = x^3 +C\,.$$ And now the initial conditions come into play (the $x_0$, $v_0$ and so on) or in some cases boundary conditions. Those are always needed to find a specific solution to a differential equation. So if I also know, that $f(0) = f_0$, then we can just plug that in: $$f(0) = 0^3 +C = C = f_0$$ So we see, if $f(x)$ should also fulfill the condition $f(0) = f_0$, $C$ can not have any value, but needs to be $f_0$ in this specific example - in general (for other $f(x)$'s) you always need to calculate the $C$ that you need to fulfill your initial conditions.

Here some more notes:

A lot of times, when you solve a differential equation you try to find all or a whole bunch of solutions (that's the $x^3 +C$ above) and then pick one of them, that fits your initial/boundary conditions.

If you have higher order DGLs (differential equations) like $F = m \ddot x$ (where each dot means a derivative with respect to time), then you need more than one initial condition to solve this for $x(t)$ (the initial position and initial velocity).

Let's say for simplicity $F(x) = 3 x^2$, just so we have an example to talk about. As you wrote $$f(x) = \int F(x) \mathrm d x + C = x^3 +C\,.$$ And now the initial conditions come into play (the $x_0$, $v_0$ and so on) or in some cases boundary conditions. Those are always needed to find a specific solution to a differential equation. So if I also know, that $f(0) = f_0$, then we can just plug that in: $$f(0) = 0^3 +C = C = f_0$$ So we see, if $f(x)$ should also fulfill the condition $f(0) = f_0$, $C$ can not have any value, but needs to be $f_0$.

Let's say for simplicity $F(x) = 3 x^2$, just so we have an example to talk about. As you wrote $$f(x) = \int F(x) \mathrm d x + C = x^3 +C\,.$$ And now the initial conditions come into play (the $x_0$, $v_0$ and so on) or in some cases boundary conditions. Those are always needed to find a specific solution to a differential equation. So if I also know, that $f(0) = f_0$, then we can just plug that in: $$f(0) = 0^3 +C = C = f_0$$ So we see, if $f(x)$ should also fulfill the condition $f(0) = f_0$, $C$ can not have any value, but needs to be $f_0$.

Edit:

A lot of times, when you solve a differential equation you try to find all or a whole bunch of solutions (that's the $x^3 +C$ above) and then pick one of them, that fits your initial/boundary conditions.

If you have higher order DGLs (differential equations) like $F = m \ddot x$ (where each dot means a derivative with respect to time), then you need more than one initial condition to solve this for $x(t)$ (the initial position and initial velocity).