Fundamental theorem of calculus statement Let f be an integrable real-valued function defined on a closed interval [a, b]. Let F be the function defined, for all x in [a, b], by
F(x)=$\int _{a}^{x}\!f(t)\,dt$
Doesn't this make F(a) = $0$ ?  
 A: Yes, yes it does. In this particular formation of the FTOC, you are defining a function F(x) such that it represents the area underneath the function f(t) as t ranges from 0 to x. When x = a, you are measuring an area that is f(a) high, and of zero width, so therefore it must be zero area.
EDIT To respond to your comment:
For any integrable function f(x), there are an infinite number of antiderivatives F(x). It is true that $F(x) = \frac{x^2}{2}$ is an antiderivative of $f(x) = x$. However, $F(x) = \frac{x^2}{2} + 7$ is also an antiderivative, as is $F(x) = \frac{x^2}{2} - \pi$, and in fact so is $F(x) = \frac{x^2}{2} + C$, where $C$ is any real value. You know how when you take the indefinite integral you're supposed to include a "+C" term? Yeah, it's that one.
So, if we are interested in $F(x) = \int_1^x f(t) dt$, because this is a definite integral, we want to find the function that specifically measures the area between the vertical line $t = 1$ and the vertical line $t = x$, and as I said above that is a function that must equal $0$ at $x = 1$. So, clearly, in this particular case we are interested in the function $F(x) = \frac{x^2}{2} - 1$.
On the other hand, if we were interested in the function $F(x) = \int_2^x f(t) dt$, then we would have $F(x) = \frac{x^2}{2} - 2$, noting that $F(2) = 0$ and, in fact, $F(x) < 0 $ for $x \in [1, 2)$, which is because we are, in this case, effectively measuring the area backwards.
A: As mentioned in previous answers, yes $F(a)=0$ since as $F$ is defined
$$
F(a)=\int_a^af(t)\mathrm dt=0
$$
From your comment above, I think the source of your confusion may be notational. You are using $F$ to mean the antiderivative of $f$, or for you
$$
F=\int f
$$
which is really a class of primitive functions which differ by a constant, and not the same as the definition of $F(x)$ as a single function of $x$ in the fundamental theorem. 
Using your example in the comment, for $f(t)=t$, as defined in the fundamental theorem of calculus
$$
F(x)=\int_a^x t\mathrm dt=\frac{x^2}{2}-\frac{a^2}{2}\ne \frac{x^2}{2}
$$
in general. Taking $x=a=1$ yields
$$
F(1)=\int_1^1 t\mathrm dt=\frac{1^2}{2}-\frac{1^2}{2}=0
$$
As we claimed.
