Talk:Constant of integration

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joke

It's a subject of one of the great Math jokes... See http://www.netfunny.com/rhf/jokes/old90/constant.html AnonMoos 14:29, 26 September 2006 (UTC)

untitled

Isn't a bit of overkill to use FTC to prove F' identically zero implies F identically constant? It follows immediately from the MVT... 207.62.177.227 (talk) 23:55, 18 November 2009 (UTC)

It is, and it does. The downside to that approach, I think, is that the mean value theorem isn't usually given much emphasis in calculus courses, so if we used it here, then I think we'd be more likely to confuse the reader. (I could be wrong, though.) It's too bad, really, because the Fundamental Theorem of Calculus is really just a souped-up mean value theorem. Ozob (talk) 00:25, 19 November 2009 (UTC)
Upon further reflection, I think the problem may be considerably worse than I originally imagined. Let's say there are the 2 basic versions of FTC we could be using: (1) the one that assumes continuity of f, or (2) the one that does not. In the former case, we're using the corollary of the fact that the accumulation function is an antiderivative, and the proof of this corollary requires using the fact that two functions with the same derivative differ by a constant. In short, the former case is a circular proof. In the latter case, if you work through the proof (say, baby Rudin proof for Reimann-integrability) for the case when f = 0, you'll see that at the pivotal point of the proof, it reduces to nothing more than the MVT (take the trivial partition). In short, the latter case is a rip-off of MVT.
So, choose your poison: either the proof given is circular, or a rip-off. In any case, the conclusion has absolutely nothing to do with integration. If that's true that MVT isn't taught in calc now (maybe this is true just in high school), it's a shame. 76.20.109.233 (talk) 10:22, 20 November 2009 (UTC)

"... where C is an arbitrary constant"

Necessity - wrong calculation?

Mathematical jibberish

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