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This question already has an answer here:

From the definition of indefinite integral I might say: Since the derivative of a constant is zero, thus the indefinite integral of zero is a constant. Therefore: $$ \frac{dc}{dx} = 0 \quad\iff\quad \int 0dx = c, \quad\forall c\in\mathbb{R} $$

However... we know that $0\in\mathbb{R}$, and since zero is a constant, I can pull it out the integral: $$ \int 0dx = 0\cdot\int 1dx = 0\cdot(x+c) = 0 $$

And then we end up that integral of zero is zero, not an arbitrary constant. Where is wrong here?

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marked as duplicate by alexjo, Steven Stadnicki, Hakim, mfl, N3buchadnezzar Dec 15 '14 at 20:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ This is great fun! $\endgroup$ – String Dec 15 '14 at 20:17
  • $\begingroup$ I think treating zero as a constant isn't right, since it's a determined value. $\endgroup$ – Artem Dec 15 '14 at 20:26
  • $\begingroup$ @alexjo The title is somehow coincidentally identical to this question, true. But, I'm actually asking a different thing. $\endgroup$ – Physicist137 Dec 15 '14 at 20:28
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    $\begingroup$ @Physicist137 Did you read the most up-voted answer? It deals with your question. $\endgroup$ – Eff Dec 15 '14 at 20:30
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    $\begingroup$ @Physicist137 If you haven't read it, Sam DeHority answered your question as well though. $\endgroup$ – Jason Knapp Dec 15 '14 at 20:30