How to interpret FTC1 that assumes integrability on $[a,b]$ and continuity at $c \in [a,b]$, as opposed to continuity on $[a,b]$? Consider the statement of the first fundamental theorem of calculus in Chapter 14 of Spivak's Calculus

Let $f$ be integrable on $[a,b]$, and define $F$ on $[a,b]$ by
$$F(x)=\int\limits_a^x f\tag{1}$$
If $f$ is continuous at $c$ in $[a,b]$, then $F$ is differentiable at
$c$, and
$F'(c)=f(c)$
(If $c=a$ or $b$, then $F'(c)$ is understood to mean the right- or
left-hand derivative of $F$)

In Chapter 13, entitled "Integrals", there is a theorem which says

$f$ continuous on $[a,b]$ $\implies$ $f$ integrable on $[a,b]$

Why didn't the statement of the fundamental theorem simply say

Define $F$ on $[a,b]$ by
$$F(x)=\int\limits_a^x f$$
If $f$ is continuous at $c$ in $[a,b]$, then $F$ is differentiable at
$c$, and $F'(c)=f(c)$.

Here is my attempt at explaining this

Consider the function
$$f(x)=\begin{cases} x, \text{ if } x \text{ rational } \\ 0, \text{
 if } x \text{ irrational }\end{cases}$$
and consider the point $x=0$. $f$ is continuous at $0$ (and nowhere
else).
Let $[a,b]$ be any interval containing $0$, and let $P$ be a partition
on $[a,b]$. Then
$$L(f,P)=\sum\limits_{i=1}^n m_i \Delta t_i = 0$$
$$U(f,P)=\sum\limits_{i=1}^n M_i \Delta t_i>0$$
Hence $L(f,P) \neq U(f,P)$ for all partitions, and hence $f$ is not
integrable on $[a,b]$.
Thus the right-hand side of $(1)$ isn't even defined.

So is the reason we can't use the second proposed version of the theorem because of cases such this?
It seems that most of the statements of the first fundamental theorem of calculus use continuity of $f$ on an interval $[a,b]$.
This rules out the example I gave above.
The first version above seems to be more general, because it accommodates continuity on an interval. If $f$ is continuous on $[a,b]$ then it is integrable on $[a,b]$, and so we can apply the first theorem to each point in this interval to conclude that for all $x \in [a,b]$, $F'(x)=f(x)$.
Is this analysis correct?
 A: Your second definition of $F$ is not well-posed in general. You can play on the hypothesis though and look for the minimal ones. It turns out that asking $f$ discontinuous on a null-measure set (Lebesgue measure here) is enough to assure Riemann integrability for $f$, then your $F$ is well defined and $F'(c)=f(c)$ if f is continuous there.
A: Fundamental theorem of calculas is all about existence of primitive.
First version of F.T.C  is about the existence of primitive of a continuous function $f$  on compact interval $[a, b]$ and the primitive is defined by $F(x) =\int_{a}^{x}f(t) dt$.
Second version is known as fundamental theorem for Riemann integral. $f\in \mathcal{R}([a, b]) $ then the indefinite Riemann integral $F$ defined by $F(x) =\int_{a}^{x}f(t) dt$ is differentiable and $F'=f$ on $[a, b]$
$f\in C[a, b]\implies f\in \mathcal{R}([a, b]) $.But a Riemann integrable function need not be continuous i.e a discontinuous function may have anti-derivative Or primitive and for that function the indefinite Riemann integral is still differentiable and it's derivative agree with the function.
For an example choose $f:[0, 1]\to \Bbb{R}$ defined by $f(x) =\begin{cases} x^2\sin(\frac{1}{x}) & x\neq 0 \\ 0 &\text{ otherwise} \end{cases}$
Now consider $g(x) =f'(x) =\begin{cases} 2x\sin(\frac{1}{x})-\cos(\frac{1}{x}) & x\neq 0 \\ 0 &\text{ otherwise} \end{cases}$
Then the function $g$ has a primitive $f$ but $g$ is not continuous at $0$. Hence $f(x) =\int_{0}^{x} g(x) dx$ and $f'(x) =g(x) $ on $[0, 1]$.(Note: here $g$ is integrable in Riemann sense as $g$ has only discontinuity at $0$ )
