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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?


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

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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