Don't understand the Fundamental Theorem of Calculus If $f$ is continuous on $[a, b]$ and defining 
  $$
F(x) = \int_a^x \! f(t) \, dt
$$ 
  for $x \in [a, b]$, then $F'(x) = f(x)$ for $x \in (a,b)$.


I don't understand what function the variable $t$ is for.  What is the definition of $f(t)$?  And is $F(x)$ a set?  Because it seems like $\int_a^x$ can have varying answers depending on what $x$ is.
Remember that a function is like a machine where you for each input there is exactly one output. We usually write $y = f(x)$ when $x$ denotes the input and $y$ denotes the output. $f$ is the function (rule) that tells you how you can find $y$ given $x$.

Ok, so now let's say that you have a function $f$. Let's call the input variable $t$. So for each number $t$ you get the (output) number $f(t)$. Now, given two numbers $a$ and $b$, you can find the integral
$$
\int_a^b f(t)\; dt.
$$
This integral gives you a number. Here we use $t$ in $f(t)$ and $dt$ to say that $t$ is the (input) variable for the function $f$.

Now if you choose another $b$ then you will get another number for the integral. So, for example, you might have
$$
\int_a^5 f(t)\;dt, \quad 
\int_a^8 f(t)\;dt, \quad
\int_a^{11.5} f(t)\;dt.
$$
So, you actually have a function. Let's call the input variable for this function $x$ (we don't want to use the same variable as the one for $f$). So for each (input) number $x$ we get a number (output):
$$
\int_a^x f(t)\;dt
$$
If $x=5$ then you get $\int_a^5 f(t)\;dt$ (a numbers).

We haven't given this function a "name", so let's call it $F$ (again we don't want to call it $f$ because we already have a different function with that name).

So for each input $x$ you get the output $F(x)$ and this is defined by
$$
F(x) = \int_a^x f(t)\; dt.
$$
This is then a new function - a function who's output depends on the input $x$. What the Fundamental Theorem of Calculus tells you is that (under the given assumptions) this function is differentiable (that is you can find the derivative) with derivative
$$
F'(x) = f(x).
$$
So, for example
$$
F'(5) = f(5).
$$
Remember that the derivative of a function is just a new function. So again we have an output ($F'(x)$) for each input $x$. The way you calculate the output of the derivative of $F$ is by evaluating $f$ at the input.
A simple and concise answer is as follows:

$$F(x)=∫_a^b f(x)\,dx$$

Therefore,

$$F(x)=f(b)-f(a)$$

So $F(x)$ is the integral of $f(x)dx$ from $a$ to $b$ , right? All the fundamental theorem of Calculus is saying is that along this set (interval $a,b$) exists the inverse operation:

$$F'(x)=f(x) $$