whether a function with a closed interval domain has an antiderivative Does a function f whose domain is a closed interval have an antiderivative (a function F (whose domain may not be the same as f) that is only differentiable on the domain of f, and the derivative is equal to f on the domain of f)?
For a function to be the antiderivative (with the definition above) of the above function, its domain must include values outside and near the endpoints of the closed interval in order to differentiable, and it also has to be not differentiable for those values.
 A: There appear to be two issues in question: The distinction between open and closed intervals, and differing domains of $f$ and $F$.
If $a < b$ are real, one says a real-valued function $f$ is differentiable on $[a, b]$ if $f$ extends to a differentiable function on a larger open interval $(\alpha, \beta) \supset [a, b]$.
If it's of interest, we say $f$ is continuously-differentiable if $f$ extends to a continuously-differentiable function on a larger open interval. This is equivalent to saying $f$ is continuous, differentiable in $(a, b)$, and the one-sided limits of the derivative exist at the endpoints.
Particularly, if $f$ is defined on an interval $I$, a function $F$ on $I$ is an antiderivative of $f$ if $F$ is differentiable in the preceding sense and $F' = f$ at each point of $I$. For example, if $f$ is identically zero on some interval $I$, then every constant function $F$ on $I$ is an antiderivative of $f$.
Finally, if it matters, "any function whose domain is one or more open interval has an antiderivative" isn't true, but is true if we assume $f$ is continuous (for example).
