Talk:Fundamental theorem of calculus

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Mathematics grading: B Class Top Importance  Field: Analysis
Article starts well, but proof section is a mass of equations with insufficient prose. More information on history needed. Tompw 18:52, 7 October 2006 (UTC)

These articles on calculus seem unable to get through a sentence without two or three inappropriate uses of the word "you". Instead of saying "two plus three equals five", the author of these articles would write "Suppose you want to know what two plus three is. You will find that it is five." -- Mike Hardy

I agree with this assessment. The use of the first person plural is an almost universal convention in mathematics writing. This is primarily because the use of the second person ("you") often tends to assume a condescending attitude toward the reader, as in "I already understand this, but you're still struggling with it." Using "we" gives the impression that the author and reader are "on the same side". (Although, anyone who has read an advanced math book knows this is little consolation when he or she has read "we find", "we see", "we observe", etc. for the 100th time and doesn't understand.

I must admit I find it grating too. I prefer to write maths with "we", but sparingly. -- Tarquin 20:46 Jan 6, 2003 (UTC) (but look at the TeX equations! the soft curves of the integral sign! the variables leaping from tree to tree, as they float down the mighty rivers of British Columbia. ! The Giant Redwood. The Larch. The Fir! The mighty Scots Pine! ... Sing! Sing! [singing] I'ma lumberjack, and I'm okay. ... (men in white coats enter and carry Tarquin offstage)

I, personally, find "one" to be the best, although this tends not be befitting in most cases of mathematics articles. "One might find that three plus two equals five." doesn't quite carry the same... power? He Who Is 22:42, 7 June 2006 (UTC)

Contents

[edit] Proof

May I suggest using a simpler example for those of us that are not engineers. Perhaps something like in short the distance traveled is equal to the speed in miles per hour divided by the time spent traveling?

I have been having trouble finding a rigorous, but still easy to understand proof of the fundamental theorem of calculus. All theorems should have proofs (the definition of a theorem is basically something that can be proved) including this one.202.165.85.99 05:38, 23 Sep 2004 (UTC)



Can we add something perhaps on Taylor expansions etc? —Anonymous?

(I trimmed the redundant proofs out. They're available on my user page, and of course the history.)

The proof I presented on the article is a little messy. I believe it needs wikifying. It also doesn't lead from the previous section quite right, but the intuition section is so well written I hardly want to touch it. There seems to be a community in the math wiki pages, but I'm having trouble tracking it down. Anyhow, I hope this meets with their approval.

—Daelin 09:19, 29 Nov 2004 (UTC)

The proof I used appears to be the Riemann Integral, however I use plainer notation than our article on that (\sum_{i=1}^n f(c_i)(x_i-x_{i-1}) vs \sum_{i=1}^{n-1} f(t_i)(x_{i+1}-x_{i}) for instance). The Riemann Integral also includes far more discussion than mathematical expression. I'm not certain it's exactly the Riemann Integral, however. Verification, anyone? —Daelin 18:31, 2 Dec 2004 (UTC)

I must add that it does state explitly in WP:MSM that proofs should be excluded unless they directly add to the content. (e.g. act as a bridge to help explain other points.) He Who Is 22:46, 7 June 2006 (UTC)

The first proof sites the 'mean value theorem for integration' whose result, although not dependant on FTC, is one that is not even a consequence of the mean value theorem and has subtle details that may not require a formal proof, but at least some lip service as to why it is the case. The 'mean value theorem for integrals' really relies on the crucial result of the intermediate value theorem rather than the mean value theorem, which is what is linked. I imagine a curious reader wanting verification of this step, and then being led down a blind alley by checking the mean value theorem. So rather than citing the 'mean value theorem for integration' maybe put a small explanation like this: because f is continuous it attains its max and min on [Δx,Δx+h], at say m and M by the intermediate value theorem. By the definition of the integral,

\int_{x_1}^{x_1 + \Delta x} f(t) dt is bounded above and below by Δx*f(m) and Δx*f(M). Again, apply the intermediate value theorem to the inequality, one arrives at the desired result. -Kevin.t.joyce 10:20, 19 Oct 2006 (UTC)

[edit] example of function which is integrable but has no antiderivative?

in the article it states "Part II of the theorem is true for any Lebesgue integrable function f which has an antiderivative F (not all integrable functions do, though)". I would like to know an example of a function which is integrable but doesn't have an antiderivative. That is, it seems to me that if the Lebesgue integral exists, then

x
x

should be an antiderivative almost everywhere, which I guess would only be true when the FTC holds? -Lethe | Talk

I've dug around a bit, and it seems that Cantor function is an example of a function which doesn't obey the fundamental theorem of calculus, because it is not absolutely continuous which is apparently a requirement for the FTC? why doesn't it mention that in the article? So, does this mean that the Cantor function has no antiderivative? or that its antiderivative just doesn't obey the FTC? -Lethe | Talk
How about my faviorite, exp(-x^2)? My teacher once showed how to integrate it, but it has no anti-derivative. -- Taku 05:36, Dec 6, 2004 (UTC)
That function has an antiderivative. It's called the error function -Lethe | Talk

A step function H(x) that is 0 for x<=0 and 1 for x>0 is integrable on [-1,1], but it is not a derivative because derivatives have intermediate value property, as can be seen from applying the intermediate value theorem to the difference quotient (f(x+h)-f(x))/h. So we see that H(x) does not have an antiderivative. Any piecewise-continuous function with simple jumps will do as an example. Counterexamples in Analysis By Bernard R. Gelbaum, John M.H. Olmsted Dover June 2003, ISBN 0486428753 is a good reference.

I'm having a hard time believing your claim. Why isn't H(x) the derivative of the function f(x)=0 for x<=0, f(x)=x for x>0? -Lethe | Talk 01:00, 16 September 2005 (UTC)
Because your f is not differentiable at 0, that's why. michaelliv 15:43, 26 May 2006 (UTC)

[edit] Misprints corrected

Today I made changes to Part II, Corollary, and the proof of Part II. Part II and Corollary assumed F only to be continuous and nevertheless delt with F'. It looks like a simple interchange of f and F had happened.

All elementary (if not all) formulations of the fundamental theorem of calculus suffer from the inability to give a simple characterization of the regularity properties of F (in addition to the differentiability) that imply F(b)-F(a) = integral from a to b of F'(x) dx. The problem is complicated since one could fine tune the notion of the integral or even this of the derivative to obtain a simple and sufficiently general formulation of the fundamental theorem of calculus.

One interesting question results from this, and I would be glad to get answers, hints, or opinions about it: Consider a simple setting: a finite closed interval [a,b], a<b, and a real-valued differentiable function f defined on that interval (obvious what this means even at the end points of the interval). What do we know about the real valued function f' ????. Can we characterize these properties ('to be a derivative') in terms of this function alone without mentioning the function from which it can be obtained by differentiation?ulrich 07:03, 10 Jun 2005 (UTC)

If I understand you correctly then I see no answer to your question. What does it mean to be a derivative, other than to imply the existence of an antiderivative? That is the only meaningful interpretaion of your question that I can consider. If you accept that, then any proof would require the construction of an antiderivative, and then the mean value theorem would ensure that it is basically the usual F in the FTOC.

[edit] The connection atop the article

It's unwise to exaggerate the relationship of the derivative with the integral. The Fundamental Theorem (in all its forms) is the beginning and the end of their relationship; everything else is intuition. --VKokielov 19:29, 27 July 2005 (UTC)

I agree. Differentiation and integration are in no way inverses. Differentiation and antidifferentiation are inverses. He Who Is 23:51, 7 June 2006 (UTC)

[edit] Various Comments

The 'Intuition' section is not intuitive. The FTOC simply states that the the displacement of a moving object is equal to the net area under the velocity graph. I said that without any symbols or any mention of derivatives! This is most easily seen in the case of "distance = rate*time". Thus, the fundamental theorem is an attempt to generalise this forumla for non-constant velocity. "We've just learned how to differentiate. Now what?"

If you use the word "infinitesmal" in a calculus course you have lost your students. This is really a shame. Most modern US textbooks make some attempt at discussing differentials, but I've yet to see a single one connect this to the fundamental theorem. It's beyond me what their point is. It is differentials we integrate. What's worse is when you treat substitution, you end up saying vague things, like everything really is a differential...sigh

The FTOC is stunningly beautiful (thanks to Leibniz) when we realise that it simply says the "dx's cancel" in dF/dx dx and we end up integrating (summing) dF instead, thus arriving at the net change of F!

The reason for constructing the function F should be phrased in English: Every continous function has an antiderivative. Some sense should be given to how remarkable this is, or perhaps how succesful the calculus is in "clearing things up". The class of differentiable functions is sooo much smaller than the class of continuous functions -- the fundamental theorem requires reflection on this fact.

In the wikipedia page for differentiation they say that a function is differentiable on an interval if it is differentiable at every point of the interval. Since there is no mention of left or right hand derivatives, it is impossible for any function defined on a closed interval to be differentiable at the endpoints -- the two sided limit defining the derivative vacuously does not exist. Thus F'(x) = f(x) only for x in (a,b) and not for all x in [a,b].

The intermediate value theorem is very subtle and totally irrelevant to the fundamental theorem, it's a shame that lots of authors drag it into the proof. All is needed for a proof is continuity of the integrand and positivity of the integral (that is the fact that the integral of a positive function is positive). Can you see how to clean up the proof? Take it as an exercise!

I agree 100% with your last remark, the numerous authors of the numerous calculus books have been copying from each other for many decades without giving enough thought to the subject. As a result, calculus is in a very bad need of renovation today. You can visit my web page at http://www.mathfoolery.org/calculus.html to see some ideas on how to proceed. michaelliv 16:46, 26 May 2006 (UTC)

[edit] F'(x) = f(x) = G(x)

Is there a legitimate, compelling reason why f(x) is used for F'(x), instead of (e.g.) G(x)—or even just use F'(x), itself? I find F(b) - F(a) = \int_{a}^{b} f(x)\,dx can be quite confusing to an unsure reader, especially since above it is given f'(c) = \frac{f(b) - f(a)}{b - a}. It should be either F(b) - F(a) = \int_{a}^{b} G(x)\,dx or, preferably, F(b) - F(a) = \int_{a}^{b} F'(x)\,dx. ~Kaimbridge~ 23:10, 25 November 2005 (UTC)

I find the current notation very acceptable. It suggests that f and F are somehow related, in this case F'=f. This is even more useful when you have two functions, f and g. Calling their antiderivatives F and G is more helpful than calling them H and L. No? :) Oleg Alexandrov (talk) 23:25, 25 November 2005 (UTC)

I'm not saying it should be some other random letter—but Cos(x) = cos(x) = sin'(x) = Sin'(x), so someone just first attempting to understand all this may think F'=f means F'=F, especially since f did equal F earlier in the article (f'(c) = \frac{f(b) - f(a)}{b - a}). I just think there should be a uniform F/F' assignment thoughout the whole article—if you really want to identify an integrand as a different function (and personally I think the integrand should stay identified as a derivative), then let F'=G, not f (or, at the very least, change it to G'=g, to at least eliminate confusion with the earlier assignment of f to F and f' to F'). ~Kaimbridge~ 01:25, 26 November 2005 (UTC)

It is false that Cos(x)=cos(x)! (Just kidding. :) In math nobody uses Cos with big C, only mathematica does. I guess you need to get used to math notation. :) Oleg Alexandrov (talk) 02:26, 26 November 2005 (UTC)

Sure they do, whenever it is presented at the beginning of a sentence! P=) ~Kaimbridge~ 13:48, 26 November 2005 (UTC)

I think everywhere in the article f was a function, and F its antiderivative. No? Oleg Alexandrov (talk) 02:29, 26 November 2005 (UTC)
Ah, okay, but as I understand it F(x)'s derivative is F'(x) or G(x) (no, I'm not fixated on G, just that it is the letter after F. P=), and the antiderivative of F'(x) is \int G(x)\,dx=\int F'(x)\,dx=F(x)+C or, if you want to use f(x) as the function, \int f(x)\,dx=\int e'(x)\,dx=e(x)+C. I'm not saying it is wrong as given in the article (in fact, most articles/papers do present it this way), it's just that, IMHO, it creates a lot of unnecessary ambiguity (no, not to someone who already understands it—in which case it would be just preaching to the choir—but to someone who is first attempting to understand it....I'm saying that from previous, personal experience! P=) ~Kaimbridge~ 13:48, 26 November 2005 (UTC)
I don't understand you. What is that G and e and all? I find this article perfectly clear notatiionwise, and I think your suggestions are going to make it less so. Maybe you should get more familiar with usual math notation. :) Oleg Alexandrov (talk) 23:41, 26 November 2005 (UTC)
All I'm suggesting is that the article provide a consistent, generalized notation flow: F'(x) = G(x), F''(x) = G'(x) = H(x), etc.; and f'(x) = g(x), f''(x) = g'(x) = h(x), but there is nothing that says f has to be related to F, g to G, or h to H, so why introduce that ambiguity in this type of elementary, concept explaining article—I find it particularly ambiguitous and potentially confusing that f starts out as F' (in the Formal statements), continues on down through Part I of Proof, where F'(x_1) = \lim_{c \to x_1} f(c) is introduced, then—at the beginning of Part II—F(b) - F(a)\,\! is introduced, followed down a little further by the statements f'(c) = \frac{f(b) - f(a)}{b - a} and f'(c)(b - a) = f(b) - f(a) \,\!:
So F(b) - F(a)\,\! and f(b) - f(a)\,\! are commingled together. If it is to be consistent, then it should be f''(c)=F'(c)=\frac{F(b)-F(a)}{b-a}\,\!, though f''(c) is misleading since it suggests \frac{f'(b)-f'(a)}{b-a}\,\!, when it is actually \frac{f'(c_b)-f'(c_a)}{\frac{1}{2}(b-a)}\,\! (\mbox{where }m=\frac{a+b}{2}, \ \mathbf{F(b)-F(m)=f'(c_b)[b-m]}\,\!
 \mbox{ and }\mathbf{F(m)-F(a)=f'(c_a)[m-a]})\,\! (I think I have that set up right! P=), which isn't very helpful as an introduction.
It should just be F'(c) = \frac{F(b) - F(a)}{b - a}
I do understand what is meant, I'm just playing devil's advocate, approaching it from the view of someone who doesn't and is attempting it from scratch—though, don't worry, I'm not looking to mess with the article (at least now), as I have other projects in progress. P=) ~Kaimbridge~ 15:16, 27 November 2005 (UTC)
I fixed the occurence you mentioned. Is there any inconsistency anywhere else? Thanks. Oleg Alexandrov (talk) 21:10, 27 November 2005 (UTC)
A definite improvement! P=) ~Kaimbridge~ 15:26, 28 November 2005 (UTC)

Using F'(x)=f(x) is simply commonly accepted notation amonst the mathmatical comunity akin to arcsin(x) being defined on [-π/2 , π/2]--Tiberious726 01:04, 20 January 2006 (UTC)

[edit] What is "t"

the first equation of the "formal statements" section, the variable "t" appears out of nowhere; it is neither defined nor used in any further line. To me it appears that the variable should be "x" and that the equation works that way... either way it is very unclear.

Sure it is—"t" is defined as "time" in the previous "intuition" section and is used in the next subsection ("Corollary"), and is the variable between a and x. However, in the "intuition" section, shouldn't "v(t)" be "x(t)"?  ~Kaimbridge~14:51, 15 December 2005 (UTC)
There is both v(t) and x(t) there from what I saw. And the derivative of position, x(t), is the velocity, v(t). So everything looks right to me, I hope. Oleg Alexandrov (talk) 19:41, 15 December 2005 (UTC)

I'm with the initial question. I took calculus 7 years ago and never used it. Now I'm in economics and find many equations referring to derivatives. I have looked all over to try to find something to re-explain the basics of how to find a derivative, and everything I have found assumes I already know. I found it very easy to learn originally but time (and a head injury during that semester) has erased it. I desperately need a refresher. I really just need a very basic explanation - explained like I'm 6. Any help - please? Diane

[edit] History

This page states that the fundamental theorem was first proved by James Gregory. As I recall, he only proved a special case of the second fundamental theorem, and doesn't mention the first at all. While Barrow certainly influenced Newton's thinking on this matter, I think we should certainly mention Newton and Leibniz in this article. Grokmoo 15:59, 20 December 2005 (UTC)

I don't know the history, but I've always seen this theorem called the Leibniz-Newton theorem. I think there should be a mention of them, but I don't know enough to put it in myself AdamSmithee 22:59, 18 February 2006 (UTC)
I agree. In my business calculus class, we are learning that Issac Newton and Gottfried Wilhelm Leibniz delveloped the theorem independently.

[edit] "detrimental theorem of calculus" ?

I've never heard of anything about people arguing to call something a "more-apt" "detrimental theorem of calculus"... Is this some kind of vandalism or other such thing? A google search for this text results in only the Wikipedia match, so it sounds extremely fishy to me. 24.70.68.199 03:25, 7 February 2006 (UTC)

Weired sentence indeed. I cut it off from the article. Oleg Alexandrov (talk) 04:07, 7 February 2006 (UTC)

[edit] James Stewart

An authoritative author? Are you sure?

One should cite the relevant work by Barrow, and not that of Stewart. But I couldn't find it. Does Stewart give a reference? I don't have access to his book.

[edit] FORMULA needed

Shouldn't the formula just be stated outright before the proofs like on the top?

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