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.
 A: Well, the other answers are very satisfactory and more than sufficient for dealing with the problem, but I'd like to expose another point of view...
I hate this notation.
You don't integrate $f(t)$ on $[a,b]$. You integrate a function $f:[a,b] \rightarrow \mathbb{R}$ on $[a,b]$. This is illustrated by the presence of the so-called "dummy variable" being not only useless and immaterial, but also confusing. We can re-write it simply by
$$\int_a^b f(t)dt=\int_a^b f$$
And (with Lebesgue's Theory in mind), we get the much more illuminating notation:
$$\int_{[a,b]}fd\mu$$
$[a,b]$ is the set we are integrating over, and $\mu$ is the measure we are considering, which we can simply omit from notation in most cases when dealing with calculus, since we take the lebesgue measure.... arriving at:
$$\int_{[a,b]}f$$
Now, we define, for $x \in [a,b]$, a function $F:[a,b] \rightarrow \mathbb{R}$ by:
$$F(x)=\int_{[a,x]}f$$
Note that we are changing the set we are integrating over when we change $x$. The fundamental theorem of calculus is simply stating that if $f$ is continuous at $x$, then:
$$F'(x)=f(x)$$
Note: An "argument" for the notation is that it makes the theorem of change of variables intuitive. I disagree. It makes the theorem of change of variables (arguably) more easily manageable and prone to good memorizing and calculation, but it is as far as it goes. The intuition is independent of notation. Even with this "advantage", I find the theorem much less confusing theoretically when stated like this:
$$ \int_{[\phi(a), \phi(b)]} f=\int_{[a,b]} f\circ \phi . \phi '$$
A: Inside the integral, the variable $t$ is a "dummy$ variable in that it is only a label.  
One can use any "lablel" for the dummy variable.  To wit, 
$$\int_a^x f(t) dt = \int_a^x f(u) du = \int_a^x f(v) dv $$
etc.  
The result of the integration is a function of $x$, $F(x)$. 
NOTE:  $F$ is also a function of $a$, but $a$ is typically viewed as a fixed parameter.
A: 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: $t$ is a dummy variable so in a sense it's just there for notation. Think of $\sum_{t=1}^nt = n(n+1)/2$. It doesn't really matter what $t$ is as it doesn't appear in the simplified form, but it does matter what $n$ is. $F(x)$ is a function on $[a,b]$ which is sometimes called the 'antiderivative' of $f(x)$. The definition of $f(t)$ may be anything as long as it's continuous.
A: The FTC is actually a very intuitive concept that is not really anything special to just calculus. Here is an explaination with absolutely no math. The FTC basically says that the sum of small changes within a bound sums up so that the resultant affect (Net effect) only occurs at the bounds. Now here's a little math $dF =f(t)dt $ the sum of $dF$ (infintesmaly differences in $F $), which is the integral in the theorem is equivalent to just the difference of $F $ across the bounds $F (b)-F (a) $. Now why is this intuitive...here is a trivial but helpful example let's setup the sum $x_1-x_2+x_2-x_3+x_3-x_4$ what does this cancel to? $x_1-x_4$. $x_1$ and $x_4$ are the bounds. That is basically the FTC at work in a very trivial way. This allows us to also realize that the multivariable integral theorems ,Stokes, divergence and the Line FTC are all restatements  of the same 1D FTC just boosted up a dimension. Hope this helps! I think intuition is best.
A: 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:
$$F'(x)=f(x) $$
