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) = $0$ ?  

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.
