To build my understanding of the FTC I attempted to reverse the steps of the proof for part 1. I started by assuming we have some continuous function that is itself a derivative.
$f(x)=F'(x)$
Next I thought I could say any point on the curve $f(x)$ would determine two points $(x, x+\Delta x)$ on the curve such that $f(c)$ is the average between the two points, due to the Mean Value Theorem. Further, we could take the limit of $f(c)$ as $\Delta x$ approaches $0$ and due to the squeeze theorem we have that $\lim_{\Delta x\to 0} f(c) = f(x)$. In other words we can view every point on the curve not just as a singular point but as an average between two points, such that every point is the limit case of two other points approaching one another.
$\therefore \lim_{\Delta x\to 0} f( c) =f(x)= F'(x) $
Next, using the definition of the derivative on the right for $F'(x)$:
$ \lim_{\Delta x\to 0} f(c) = \lim_{\Delta x\to 0} \frac{F(x+\Delta x)-F(x)}{\Delta x} $
Canceling the limits and rearranging:
$f( c).\Delta x={F(x-\Delta x)-F(x)}$
By definition $f( c).\Delta x$ is the area under the curve between $x$ and $x+\Delta x$, as we determined earlier $f(c)$ was the average between the two points, and multiplying this by the width of the interval gives the area. Therefore ${F(x+\Delta x)-F(x)}$ must also be the this area. This is only possible if $F(x)$ is a function of the area under $f(x)$.
This tells us that given $f(x)=F'(x)$, $F(x)$ which is the antiderivative of $F'(x)$ is the area function for the area under $f(x)$.
Does this make sense? Am I allowed to "cancel the limits" as I did?