Why does the statement of First Fundamental Theorem of Calculus include the condition that f be continuous? By FTC, $f$ has to be continuous in order for $F$ to be defined.  I do not understand why.  What if $f$ is a step-wise function defined at every point?  If it is then we can take the "height" (ie the value of $f(x)$) of the function at every point. Then we can proceed across a given interval accumulating infinitely narrow little slices of area and in so doing we can calculate, $F(x) = A$, the area under the curve of $f$. 
I would think that since $F$ must be differentiable (since $F'=f$) then it would be more important that it be continuous.

There are formulations of the fundamental theorem of calculus, which do not assume that $f$ is continuous, but then the integrated function $F$ will not be differentiable (and hence not an antiderivative in the classical sense. To give an example, let's consider $f \colon [0,1]\to \mathbf R$ given by
$$ f(x) = \begin{cases} 0 & x \le \frac 12 \\ 1 & x > \frac 12 \end{cases} 
$$
Then the integral function is 
$$ F(x) = \int_0^x f(t)\, dt = \begin{cases} 0 & x \le \frac 12 \\ x -\frac 12 & x \ge \frac 12 \end{cases} $$
Note, that $F$ is continuous, but not differentiable (the point $\frac 12 $ is the problem here). Note that at the points, where $f$ is continuous, $F$ is differentiable with $F' = f$. 
To give a more general form:

Theorem. Let $f \colon[a,b]\to \mathbf R$ be integrable. Then the function $F \colon [a,b]\to \mathbf R$ given by 
  $$ F(x) = \int_a^x f(t) \, dt $$
  is absolutely continuous and differentiable almost everywhere with $F'=f$ almost everywhere.

