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) = $0$ ?  

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.
