Talk:Fundamental Theorem of Calculus

Does the lower bound of the first part have to be $a$ of the interval, or can it be any constant$\in [a..b]$? Or does such a distinction not matter, because we can always pick a subset of the interval? --GFauxPas 07:53, 17 January 2012 (EST)
 * The constant you can always add to an indefinite integral originates at least partly from this free choice of $a$. However, in this particular case, $a$ is convenient, as it means that swapping the bounds with a minus sign is not necessary. --Lord_Farin 07:59, 17 January 2012 (EST)

I’m confused by the fact this is marked as a candidate for Proof of the Week when the page has no proof on it? — Timwi (talk) 12:37, 3 December 2012 (UTC)
 * Click on the names "first part" and "second part" to get to the actual proofs. However, since the POTW mark has been added, each part has been given an additional proof, so I'm not sure it's okay to have a POTW candidate on the theorem rather than on one of the proofs --GFauxPas (talk) 12:41, 3 December 2012 (UTC)
 * I found out about clicking on the headings by now. Personally, I find that to be highly unintuitive... as regards to UI/UX design, perhaps there should be a separate link labeled something containing the word proof at the end of each part? — Timwi (talk) 12:44, 3 December 2012 (UTC)
 * Oh dear. That was me again. Sorry. --prime mover (talk) 14:44, 3 December 2012 (UTC)

Hi, this version of FTOC is extremely weak. Instead it should go something like:

Assume $f$ is bounded on $[a,b]$. Then $F_{\pm}: [a,b] \rightarrow \mathbb{R}$ by $F_{\pm}: x \mapsto I_{\pm}(f,a,x)$ are continuous, where $I_{\pm}(f,a,x)$ are the upper/lower Riemann integrals, respectively. Further, $F_{\pm}$ are differentiable at any $c$ where $f$ is continuous, and at such a $c$, $F_{\pm}'(c) = f(c)$. Finally, suppose $f$ is Riemann integrable and $F$ is any piecewise-antiderivative of $f$. Then $\int_a^b f = F(b) - F(a)$. Somebody should fix the proof to account for these generalizations. Assuming the continuity of $f$ makes this all a huge waste of time. Xenakis (talk) 17 March 2013, 6:04PM CST


 * We may do, we may not. If you consider this place a waste of time then I would encourage you not to waste any more of it here. --prime mover (talk) 23:08, 20 March 2013 (UTC)


 * Sorry, I didn't mean to personally insult you; I have incredible difficulty in determining how you came up with this interpretation, given that (one presumes) you are not an undergraduate sophomore, the highest level of education one could possibly have attained while still thinking that one's proof of the FTOC is anything beyond mere rote; as such, you surely have no attachment to it, much less such a weakened version as that present here. Xenakis (talk) 17 March 2013, 6:18PM CST


 * @PM, he said this proof is a waste of time, not this wiki. Moreover he's taken the time to create an account to suggest a generalization (which indeed should be added; though I think it's more suitably phrased in the context of the Lebesgue integral). Just because he hasn't chosen the most diplomatic way of saying it, is there really any need to see how close to "f**k off" you can get without actually typing it? Wikis are supposed to be community efforts. --Linus44 (talk) 23:19, 20 March 2013 (UTC)


 * Lebesgue integral? Are you sure about that? FTC is historically tied to the Riemann integral, and I believe there are reasons for that. --Dfeuer (talk) 23:43, 20 March 2013 (UTC)


 * Quite sure -- a bounded function is Riemann integrable if and only if it is continuous amost everywhere, and every such function is Lebesgue integrable, so the Lebesgue's theory extends that of Riemann. We can remove the hypothesis "$f$ bounded" and have "$f$ is Lebesgue integrable" (and continuous at $c$). Moreover the statement $\int_a^b f = F(b) - F(a)$ also holds for "Lebesgue integrable" in place of "Riemann integrable"; a much wider class of functions. --Linus44 (talk) 00:00, 21 March 2013 (UTC)

Just because the statement of this result is not the most general it can be does not make it a waste of time. For students who are encountering calculus for the first time it is, not to put too fine a point on it, a useful and enlightening result. I accept that it can be made more general, and in due course that is what would happen. This will happen as soon as someone who has studied to that level, whose attitude is conducive towards constructively enhancing this website rather than using it as an opportunity to brag to everyone else how clever he is, is prepared to do the hard work to write the extended page. --prime mover (talk) 06:21, 21 March 2013 (UTC)


 * I didn't mean I agree that it's a waste of time or that it should be replaced; instead added to. I was saying "we don't do things like that" might be a preferable sentiment to "get lost". Though admittedly going by his/her second comment perhaps the latter is appropriate. --Linus44 (talk) 12:05, 21 March 2013 (UTC)


 * Perhaps a redirect to a nice and clear explanation in the FAQs of how the site is supposed to be accessible.


 * Also, this page 0.999...=1 is extremely weak. It doesn't take into account all the different numbers and bases to which it applies. Now hurry up and do it all for me otherwise it's a waste of time :D (I couldn't resist). --Jshflynn (talk) 12:10, 21 March 2013 (UTC)


 * Or, to put it another way: "Everybody on the internet is a troll until they prove otherwise." --prime mover (talk) 12:50, 21 March 2013 (UTC)

I couldn't find a proof on the page for the Fundamental Theorem of Calculus. --1is0? (talk) 12:58, 3 May 2013


 * It is an intelligence test. If you can't find the proof, then regrettably you may not be quite smart enough for this site. Hint: the answer can be found somewhere on this page. --prime mover (talk) 22:53, 4 May 2013 (UTC)

So, I don't want to re-open a bitter conversation, but I do think the statement of the theorem should be changed slightly. In particular, I believe the second version should say that if f is Riemann integrable and has a primitive F on [a,b], then int_a^bf = F(b) - F(a). In other words, remove the 'continuity' requirement. My reason is that, when f is continuous, the proof is much easier and doesn't require any upper or lower sums at all: If F is a primitive, by part 1 we know that G(x) = int_a^x f(x) is another primitive, and by the mean value theorem F(x) - G(x) is a constant function on the interval.

Awzorn (talk) 23:01, 18 February 2015 (UTC)