Would most analysts say that $(2/3) x^{3/2}$ is an antiderivative of $x^{1/2}$ on $[0,\infty)$, or just on $(0,\infty)$?

More generally, is there a standard interpretation of the assertion "$F$ is an antiderivative of $f$ on $I$" when $I$ contains one or both of its endpoints?

It could be taken to mean that $F$ is continuous on $I$ and $F'(x)=f(x)$ for all $x$ in the interior of $I$, or it could include a condition on the one-sided derivatives at the endpoint(s), specifically that the right-sided (resp. left-sided) derivative of $F$ at the left (resp. right) endpoint of $I$ should agree with the value of $f$ there.

I suspect one can find an example (of the $x^2 \sin 1/x$ variety) that would show that the two definitions are inequivalent.

This is related to the issue of what one means when one says that a function is continuous on an interval. Does one want to say that the right-continuous Heaviside function is continuous on $[0,\infty)$ or does one want to say that it is merely continuous on $(0,\infty)$? That is, should one use a one-sided limit or a two-sided limit at the endpoint? Clearly the former choice makes more sense if the function isn't even defined on the other side of the endpoint (as in the case of $\sqrt{x}$), but if the function is defined on the other side of the endpoint, then arguments could be made for the latter choice.

My guess is that these questions occupy a fuzzy zone where different authors make different choices of terminology, and where there's no consensus. But I'd be glad to learn otherwise.

open domainswith possibly the additional assumption of "... extends continuously to the boundary". Note that in more than 1 dimensions, continuous extension to the boundary is more than just issue of blow-up on the boundary point, but also continuityonthe boundary which is now more than 0 dimensional. – Willie Wong Feb 19 '11 at 9:25