The First and Second Fundamental Theorems of Calculus in Spivak I'm hoping for a bit of clarification. The First FTC in Spivak says that if a function is integrable on $[a,b]$ and is also continuous on $c \in [a,b]$, then
$F(x) =\int_a^xf$ is differentiable at c and $F'(c) = f(c)$.
Then the Second FTC says if the integrable function, whether continuous or not, has an antiderivative $g' = f$, then $\int_a^bf=g(b) - g(a)$, which if I'm not mistaken implies the result of the First FTC.
Initially, I thought of the 2nd FTC as a way for the result of the 1st FTC to apply to a greater variety of functions, namely those that may not be continuous. But I can't think of any such functions. The derivative of a function, if I'm correct, can't be a jump or removable discontinuity. If the function is somewhere unbounded or undefined, it won't be integrable.
And I'm not sure but I don't think an everywhere discontinuous function is integrable, if there even is such a function who's derivative is everywhere discontinuous.
So either I haven't considered all the possibilities of an integrable, somewhere discontinuous function that has an antiderivative, or I've misunderstood Spivak. Any help would be greatly appreciated.
 A: Using the terminology in Spivak, there are three results:

*

*First FTC

*Corollary of first FTC

*Second FTC

You write

I thought of the 2nd FTC as a way for the result of the 1st FTC to apply to a greater variety of functions

To be clear, the second FTC is a generalization of the corollary of the first FTC. I guess what you're asking about is a situation when the second FTC can be applied but the corollary cannot. For this, consider the function $g:\Bbb{R}\to\Bbb{R}$
\begin{align}
g(x):=
\begin{cases}
x^2\sin\left(\frac{1}{x}\right)& \text{if $x\neq 0$}\\
0& \text{if $x=0$}
\end{cases}
\end{align}
Then, $g$ is differentiable everywhere and
\begin{align}
g'(x)&=
\begin{cases}
2x\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right) & \text{if $x\neq 0$}\\
0 & \text{if $x=0$}
\end{cases}
\end{align}
Let me define $f=g'$. Then, $f$ is continuous everywhere except at the origin, and $f$ is also bounded in a neighborhood of the origin. Thus, for any $[a,b]$, it follows that $f$ is (Riemann) integrable on $[a,b]$, and clearly all the hypotheses for the second FTC are satisfied. On the other hand, $f$ is NOT continuous, so the corollary itself cannot be applied (if $0\in [a,b]$).
