Why is $f(4)$ the area under $f'(x)$ specifically from $0$ to $4$ and not for ex from $1$ to $4$ or $2$ to $4$? I've seen the geometric argument for why any differentiable function $f(x)$ gives the rate of change of the area under its own curve to $x$ for a specific input $x$, and it makes sense to me.
It also makes sense that if $f(x)$ gives you $\frac{dA}{dx}$, then $F(x)$, its antiderivative, should give you the actual area under the curve of $f(x)$.
What I don't understand however, is why does $F(x)$ give you the area from $\textbf{0}$ specifically to $x$ given a single input $x$? In other words, where does that $0$ come from?
Because $f(x)$ gives you $\frac{dA}{dx}$ at $x$ regardless of how much area comes "before" (for smaller values of $x$). 
How do we decide that $F(x)$ starts computing the area under the curve of $f(x)$ from $0$ and not some other starting point? Obviously it's very convenient but it seems pretty arbitrary to me.
 A: 
Obviously it's very convenient but it seems pretty arbitrary to me.

People choose $F(x)$ to start from $0$ because it's convenient. There's no reason why the area has to start from $0$ other than that people like to define $F(x)=\int_0^x f(x) \ dx$. However, $F(x)$ can be anything as long as $F'(x)=f(x)$ and by the Fundamental Theorem of Calculus, all such $F(x)$ differ by a constant. This means you could also use $G(x)=\int_1^x f(x) \ dx=F(x)-F(1)$, which starts from $1$ instead of $0$. Because of the Fundamental Theorem of Calculus, really, you can make $F(x)$ start calculating the area from any value you want since all such $F(x)$ will have a derivative of $f(x)$.
However, this can be a problem for functions where this would become an improper integral, like $f(x)=\frac 1 x$. In most cases, most people would probably just let $F(x)=\ln x$ because that's the easiest way to express the indefinite integral, but this function certainly does not start giving you the area from $0$ because $F(0)$ is undefined. There are other cases like this, too, so make sure you don't just assume that $F(x)$ starts calculating the area from $0$ because that's not always true. If you always stick with the identity $\int f(x) \ dx=F(x)\bf{+C}$ and $\int_a^b f(x) \ dx=F(b)\bf{-F(a)}$, even when $a=F(a)=0$, you will always avoid the mistake of assumption.
A: It is simply
$$
A = \int\limits_a^b f(x) dx = F(b) - F(a)
$$
if $f$ is non-negative on $[a,b]$ and $F' = f$.
The case you mention seems to be a special case with $F(a) = 0$, which must not be true in general.
A: It depends on the function you are using. Looking at an arbitrary function, you can see that we can move it up or down changing its integral, but maintaining the derivative constant, since the slope does not change. The primitive is only the function $F(x)$ that when taken the derivative will yield the main function $f(x)$ without any terms that do not contribute to $f(x)$, these terms would only move the integral up or down without affecting it's derivative. This means that the antiderivative of a function is the integral in which the integration constant is $0$. For many functions this means the initial point of the integral that defines the antiderivative is $x=0$. For example:
$$\int_{a}^{x}x^2dx=\frac{x^3}{3}-\frac{a^3}{3}$$
The second term is a constant, and does not contribute to the derivative, so it must be zero for this integral to define an antiderivative, and that yields $a=0$ for the antiderivative.
Another example is:
$$\int_{a}^{x}e^{Ax}dx=\frac{e^{Ax}-e^{Aa}}{A}$$
For the constant to be zero, we must start the integral from $-\infty$ in irder for the integral above to define the antiderivative of $e^{Ax}$.
This comes from the fundamental theorem in this form:
$$\int_{a}^{x}f(x)dx=F(x)-F(a)$$
The antiderivative that we where looking to find is obviously $F(x)$ so this shows that $F(a)=0$ so that the integral in the left represents just $F(x)$.
