Fundamental Theorem of Calculus: definition of differentiable interval? I have a question on the Wiki definition:
$$
F(x) = \int_a^x f(t) dt
$$
where $f : [a,b] \rightarrow \mathbb{R}$, $F : [a,b] \rightarrow \mathbb{R}$ and $F$ is differentiable on $(a,b)$ then:
$$
(F'(x) = f(x)) \land (x \in (a,b))
$$
Questions $F'(x) = f(x)$ is defined on $(a, b)$ and earlier $f(x)$ is defined on $[a,b]$.  Does this make the antiderivative a partial function?  Also how can $F'(x) = f(x)$ if the domain is different then the earlier definition?
 A: $F(x)$ is not a partial function, since it is defined everywhere on $[a,b]: F(a) = \int_a^a f(t)dt = 0,$ and $F(b) = \int_a^b f(t)dt$.
Generally, when we are looking at a real-valued function $G$ that is only defined on an interval $[a,b]$, we think of the derivatives at $a$ and $b$ as simply being the right and left derivatives at those points (respectively). That is, when we look at the derivative $G'(a) = \lim_{x \rightarrow a} \frac{G(x)-G(a)}{x-a}$, the limit should be taken using only values of $x$ that lie inside $[a,b]$, since that is the domain we are working in. This is equivalent to taking the limit only from the right of $a$.
In the case of the FTC, these one-sided limits $F'(b) = \lim_{x \rightarrow b^-} \frac{F(x)-F(b)}{x-b}$ and $F'(a) = \lim_{x \rightarrow a^+} \frac{F(x)-F(a)}{x-a}$
do exist, and are equal to $f(b)$ and $f(a)$ respectively, as explained in my answer to this linked question.
A: $\frac{F(x+h)-F(x)}{h}=\frac{1}{h}\int_x^{x+h}f(t)dt\approx \frac{f(x)\times(x+h-x)}{h}\to f(x)$ as $h\to 0$  Approximation from definition of Riemann integral.
End points excluded since for $x=a$ and $h\lt 0$ or $x=b$ and $h\gt 0$, the integral cannot be defined.
$f(x)$ must be continuous at $x$ for $F(x)$ to be differentiable at $x$.
