Extension: Can FTC Part 2 Prove Part 1?

Previously, we saw how the first part of the Fundamental Theorem of Calculus can prove Part 2. Now, can the second part prove the first?

To see whether it can, let's review the conditions for the Fundamental Theorem of Calculus.

To recap, the main distinction between Part 1 and Part 2 is that Part 1 required $f(x)$ to be continuous, while Part 2 required $f(x)$ to be integrable. Now, let's think about how these two concepts- continuity and integrability- are related: primarily, can continuity prove integrability, or vice versa?

As we continue studying integrals, we will see that all functions that are continuous must be integral, but not vice versa. For instance, consider the function $$f(x) = \left\{ \begin{array}{lr} 1 & : x = 0\\ 0 & : x \ne 0 \end{array} \right.$$

Clearly, it is not continuous, since it "jumps" to $1$ at $x=0$. However, $f(x)$ is, in fact, integrable! (And intuitively we can see why, for instance, $$\int_{-1}^1 f(x)=0,$$ since $f(x)$ is only above the $x$-axis at $x=0$.) Examples like these show that continuity implies integrability, but not the other way around! Hence, we can say that continuity is "stronger" than integrability, and so Part 1 of the Fundamental Theorem assumes a "stronger" hypothesis than Part 2. In other words, Part 1 assumes more properties about the function in question than Part 2 does, which is why we can use Part 1 to prove Part 2, but not vice versa.