What is $a$ in $F(x)=\int_a^xf(t)\,dt$? I often see things like
$$
F(x)=\int_a^xf(t)\,dt.
$$
What is $a$? Is its value important? I ask this because I often get the feeling that $a$ could be any constant.
I also see $a$ sometimes be replaced by $x_0$.
Here is an example: please see this previous question of mine. Notice that the last expression of the chosen answer has a lower limit $t_0$; I have no clue what its significance is or where it came from.
 A: I am posting nothing more than what @Christian noted. Just note that if we set $$\Phi(x)=\int_a^x f(t)dt$$ then $f(x)$ has an indefinite integral as $\Phi(x)+C$ in $[a,b]$. To find why $x$ is important here look at the following plot:

We see that $\Phi(x)$ is the shaded area (here I assumed $f$ is positive) which varies continuously from $$\int_a^a f(t)dt$$ to $$\int_a^b f(t)dt$$
A: If a function $f$ has a primitive $F$ then it has infinitely many primitve since $F(x)+C$ is also a primitive for any constant $C$  but there's only one primitive of $f$ that vanishes at $a$ and this function is
$$F(x)=\int_a^x f(t)dt$$
A: The antiderivative to a function $f(x)$ is not uniquely defined by $F'(x) = f(x)$ since you can add an arbitrary constant $C$ to $F$ and the requirement would still be fulfilled:
$$\frac{d}{dx}\left(F(x) + C\right) = F'(x) = f(x)$$
However, defining
$$F_a(x) \equiv \int_a^x f(t)dt$$
makes this definition unique since now
$$F_a(x) = \int_a^x f(t)dt = F(x) - F(a)$$
and the act of taking $F(x) \to F(x) + C$ would leave this expression invariant. In other words, specifying $a$ picks out a unique antiderivative.
A: This part of the question can also be answered:

Is its value important? 

One place the integral in the question comes up is when you want to use an integral to construct an antiderivative. You have, say, a continuous function $f(x)$ on the real line and you want a function $F$ so that $F'(x) = f(x)$. The way to get $F$ is to pick an arbitrary point $a$ and set
$$
F(x) = \int_a^x f(x)\,dx
$$
Now, you might ask, what if you pick some other point $c$ instead of $a$? Then we would have another antiderivative, $F_c$:
$$
F_c(x) = \int_c^x f(x)\,dx
$$
Now there is a simple relationship between the original $F(x)$ and the new $F_c(x)$:
$$
F_c(x) = \int_c^x f(x)\,dx = \int_c^a f(x)\,dx + \int_a^x f(x)\,dx
$$
so $F_c$ differs by a fixed amount ($\int_a^c f(x)\,dx$) from the original $F(x)$.  
So the value of $a$ is not, in general, very important. If you pick a different value, you still get an antiderivative, but maybe a different one. The antiderivative you construct will be $0$ at $a$, which can occasionally be useful, but if you just want to construct any antiderivative then you can pick the value of $a$ arbitrarily. 
The same phenomenon shows up, in a slightly different form, in multivariable calculus (e.g. Calculus 3 in the U.S.) There, when you have a conservative vector field, you can construct a "potential" (which is the antiderivative in that context) using a certain integral, but you still have to choose an arbitrary base point to do it. If you pick a different base point, you get another potential, but it will only differ from the original one by a constant, just like in single variable calculus. 
A: a could be a constant, it could also play the role of a variable.
All that a signifies is the lower boundary of your integration. In the example given above, you are integrating from a to x.
Consider an integral as the "area" under a curve. So a, and x, respectively just represent the boundaries for that area.
See: This Image
One thing to keep in mind however: the formula you have up there is incorrect. You may not use your variable of integration as your upper limit of your integral. It will make for some really wonky results.
You will want to rewrite in this way:
$$ F(x)=\int_a^xf(s)\,ds. $$
A: The "important" variable in your equation is $x$; therefore a function $F:\ x\mapsto F(x)$ is defined. But this function as an object by itself depends on the chosen $a$, where $a$ can be any point in the domain of $f$ (supposedly an interval). If you want to exhibit this dependence of $F$ on $a$ you can write your  equation in the form
$$F_a(x):=\int_a^x f(t)\ dt\ ,$$
and in the sequel you can study the properties of $x\mapsto F_a(x)$ as before.
