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I was asked this question by a student I am tutoring and I was left a little puzzled because his textbook only defines antiderivatives on intervals (which leads me to believe its author would answer the question in the title in the affirmative).

To my understanding, finding an antiderivative of $f$ means finding a function $F$ with $F' = f$. It does not matter if the domain of $f%$ is not connected. For example, $\int \frac{dx}{x} $ denotes an antiderivative on all of $\mathbb{R} - 0$, and not just on some arbitrary interval $I \subseteq \mathbb{R} - 0 $. Am I mistaking?

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You can use whatever conventions you want; the author is free to choose hers and you are free to choose yours.

One issue with defining antiderivatives on (nice) subsets of $\mathbb{R}$ which are not connected is that they are no longer just only unique up to constants; they are only unique up to locally constant functions. For example, on $\mathbb{R} \setminus \{ 0 \}$ any function which has one constant value when $x$ is negative and another when $x$ is positive has zero derivative. This is the kind of subtlety that I suspect it would be a good idea to avoid in a calculus course.

Another issue is that you'd like to write antiderivatives down using definite integrals, but for example if a function $f$ is not defined in the interval $(-1, 1)$ then it is unclear what an integral such as $\int_{-2}^x f(x) \, dx$ would mean for $x > 1$...

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    Well, to me a constant of integration $C$ denotes a locally constant function defined on the domain of $f$... I guess I am asking: which is the most common convention?2012-07-19
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    I have absolutely no idea. What kind of course is this? That seems like a pretty sophisticated notion of constant of integration.2012-07-19
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    I think that for the "Calculus" classes the standard convention is that $C$ denotes the set of constants. In this case we need to restrict the antiderivatives on intervals.2012-07-19
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    @ Qiaochu : First-year course. I suppose then that it's best to assume implicit restriction to an interval? I don't want to introduce difficulties for no reason (nor do instructors of 1st year calculus I presume).2012-07-19
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    @user18063: yes, I think so.2012-07-19