Talk:Fundamental theorem of calculus

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[edit] Antiderivative need being continuous throughout inteval

I found http://blog.wolfram.com/2008/01/mathematica_and_the_fundamental_theorem_of_calculus.html#more

a subtle note. Pls consider this.

[edit] ******

Isaac Barrow discovered the fundamental theorem of calculus. He's not mentioned in the article. This is dispicable. —Preceding unsigned comment added by 204.69.190.75 (talk) 23:20, 17 November 2007 (UTC)

It's hardly despicable. Chill. 24.160.240.212 (talk) 19:59, 18 March 2008 (UTC)

[edit] Comment by Michael Hardy

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)

[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] 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?"

I agree, the intuitive section was not intuitive at all, in fact it confused me. Maybe something like 'the velocity at some time multiplied by a tiny amount of time gives us a tiny displacement. If we sum up all of those tiny displacements, we get the total displacement'. It may also be helpful when saying these things to refer to what theorem you are trying to explain Taras 96 (talk) 00:50, 29 November 2007 (UTC)

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

You'll probably be able to understand this, but a 6 year old won't because there's no easy way to explain this. A derivative is the slope of a tangent line at a point. a tangent line is a line exactly touching at the one point of the curve, but can hit other points when the function is similar to a wave. A derivative is given by this formula when [f(x+deltax)-f(x)]/deltax when deltax approaches 0. You should know this type of thing. Here's the easier way to find a derivative, cnx^n-1=dy/dx of cx^n

[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] 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] "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] FORMULA needed

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

[edit] About the second theorem

For the books in Hong Kong, the second fundamental theorem, which is stated here as the process of antidifferentiation can be used to calculated definite integrals, is simply called the "fundamental theorem of calculus". Should a remark be added? --Deryck C. 09:07, 16 January 2007 (UTC)

Someday, science students will learn that the -DERIVATIVE- should have been designated as the "fundamental theorem" of calculus.

[edit] The connection atop the article: "derivation and integration are inverse operations"

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] Only an integral with variable upper limit is invertible (with some restrictions)

I don't completely agree with He Who Is. If we define antidifferentiation as the process of computing and arbitrarily selecting one of the possible antiderivatives of a function, differentiation and antidifferentiation are not inverses. They are quasi-inverses. Neither the differentiation is the inverse of the antidifferentiation, nor viceversa. For one of them to be the inverse of the other, they should be able to reverse each other (and for that, they need to be bijective, and they are not).

Four important notes:

  1. Antidifferentiation cannot be defined differently, because antidifferentiation and antiderivative are strictly related concepts (as well as differentiation and derivative, they represent, respectively, the operation and its result), and all authors agree about the definition of antiderivative.
  2. According to most authors, antidifferentiation (as defined above) is synonym of indefinite integration. The article about antidifferentiation currently espouses this approach. Other authors say that indefinite integration is a process which yields an infinite set of antiderivatives, rather than an arbitrarily selected element of that set.
  3. The definite integral with variable upper limit, used in the first part of the Fundamental Theorem, non-arbitrarily selects a specific antiderivative (the antiderivative with zero at time t = a). Thus, it is something more than an indefinite integral. Due to the fixed lower limit (a), it yields a specific antiderivative, rather than any antiderivative, and as a consequence it might be regarded as bijective and invertible, provided that its codomain is restricted to its range R (int: X → R), and of course the domain of the derivative is restricted to R (der: R → X), otherwise the integral is not surjective and as a consequence cannot reverse the derivative.
  4. Even the differentiation is not invertible! Since it loses information about the constant term of its operand, it is not injective (information preserving), and as a consequence it is neither bijective nor invertible.

Paolo.dL 16:38, 8 August 2007 (UTC)

Arguing about whether they are inverses or quasi-inverses is splitting hairs, and especially, in the lead. I personally would prefer "inverses", and as well as more intuitive explanation in the lead. This is not a textbook! Proofs can be safely omitted from this article, but the meaning of the theorem, call it "intuition" if you will, has to be explained. Arcfrk 01:49, 9 August 2007 (UTC)

I agree that we should focus on the explanation of the theorem, and that the difference between quasi-inverse and inverse is not important in this context. And I so wholeheartedly agree that, even before reading your comment (which I had not noticed yet because you originally inserted it above my 8 August comment), I removed the note about inverses and quasi-inverses, simplified the introduction (see next section), and wrote the following comment. However, although this doesn't matter anymore, I don't agree about using incorrect mathematical terminology in an article about mathematics. Paolo.dL 12:47, 9 August 2007 (UTC)

I don't think that this topic should be discussed in the article, simply because the theorem does not deal with it. It just says that the indefinite integral is reversible (can be undone) by a differentiation. Reversible (i.e. injective) is not synonym to invertible (i.e. injective and surjective, i.e. bijective).

As explained above, neither the indefinite integration nor the differentiation comply with the formal definition of inverse operation or function, with respect to each other. Paolo.dL 12:06, 9 August 2007 (UTC)

[edit] Differentiation, Antidifferentiation, and definite integration

I believe that the theorem (in its two parts) actually describes the relationship between three concepts: differentiation, antidifferentiation (i.e. definite integration with variable domain), and definite integration with fixed domain:

It is important to realize that the theorem uses two extremely different kinds of integral (variable domain and fixed domain). The integral used in the first part of the theorem (definite with variable upper limit) is more similar to an indefinite integral than to a "standard" (i.e. fixed domain) definite integral. Please see note 1 in the article, and my 8 August comment in the previous section.

I rearranged the introduction, according to this rationale, but respecting as much as I could the original text. Now, I believe it appears simpler to understand, and at the same time less approximative. Please, let me know if you like it.

Paolo.dL 11:28, 9 August 2007 (UTC)

[edit] Donald Duck Image

I removed this from the article.

Image:Donald_Duck_Fundamental_Theorem_Calculus.jpg The theorem is perhaps the most well known Fundamental theorem in popular culture. To impress the judges in a contest, Donald Duck figure skates the theorem on ice. The theorem is written differently and is slightly incorrect: one of the equal signs should have been a minus sign. Donald used f'(x) instead of f'(x)dx, which is usually frowned upon, but not improper. Nothing in the notation indicates the vector analysis version either. The correct version for real functions should have been: \int_b^a f'(x) \mathrm d x = f(a)-f(b)

Here is my reasoning. The main issue is that this is a nonfree image, and nonfree images are only permitted in Wikipedia articles when they make a significant addition to the article. In this case, the image is in a section on the statement of the fundamental theorem - which the image does not illustrate, and which can be conveyed perfectly via words alone. So the image doesn't make a significant addition to that section. The image might be appropriate for a section on "fundamental theorem in popular culture" except that such sections are deprecated, for good reason in my opinion, and the gist of the cartoon can be conveyed fine by a sentence.

Basically, as I see it, this image is a humorous sidebar for the article. It isn't bad for that role, and if the image were free I wouldn't worry about it (although the caption is a little too informal for my taste). But the goal of Wikipedia is to build a freely reusable encyclopedia, and so the policy doesn't permit nonfree images to be used in this way. — Carl (CBM · talk) 14:10, 20 August 2007 (UTC)

Having added the picture originally, my views are hardly neutral, and I think it would be a shame to exclude it. If it might be appropriate for a section on "fundamental theorem in popular culture" it ought to be appropriate here as well.
The picture illustrates two important things 1) that the theorem is known in popular culture even though 2) it isn't written as it should. There is a knowledge of existence, but not a knowledge of exact content.
Simply mentioning that the theorem exists in popular culture is a poor substitute, it wouldn't illustrate it properly. Should the picture of Bart Simpson be removed the article and with the text "yellow cartoon boy"?Aastrup 03:42, 23 August 2007 (UTC)
Putting a picture of Bart Simpson on the chalkboard article would have the same problem as this. — Carl (CBM · talk) —The preceding signed but undated comment was added at 03:46, August 23, 2007 (UTC).
And you think it would be appropriate in the "chalkboards in popular culture" article?
I don't agree with the comparrison. What's written on a board (e.g Bart's chalkboard jokes)seldom has anything to do with the board itself, but Donald Duck skating a particular theorem, and skating it incorrectly, provides a useful illustration of how well the theorem is known, and the Fundamental_theorem_of_calculus must be one of the only mathematical theorems in the world that ever appeared in an mainstream comic. Being a single panel low resolution picture, it can't harm the sale of the original comic. Aastrup 04:27, 23 August 2007 (UTC)
Lots of math shows up in comics, in my memory. The fact that he states it wrong is not exactly evidence that the theorem is well known, and in any case we would need an actual published source to make that inference. I appreciate that this is a cute image, but our goal is to produce an encyclopedia with as little nonfree content as possible, and this image is not needed to understand the fundamental theorem or its notability. — Carl (CBM · talk) 05:12, 23 August 2007 (UTC)
A pity, I really liked that picture. But liking it isn't a valid reason for why it should stay. I'd love to hear some comments from other about this, am I the only one that thinks the picture should stay? —The preceding unsigned comment was added by Aastrup (talk • contribs) 05:47, August 23, 2007 (UTC).
That's a good idea. You could ask people who know math at Wikipedia talk:WikiProject Mathematics or people who are familiar about our image policies at this talk page. I don't know how much overlap there is between the two groups. — Carl (CBM · talk) 13:25, 23 August 2007 (UTC)

[edit] Corollary

Isn't the last corollary exactly the same as the first theorem, except written slightly differently?

The first theorem says

'if we take the integral from a to x of f, then differentiate it with respect to x, then we get the original function f(x)'

This is exactly what the last corollary listed states, except it does it one line as opposed to two Taras 96 (talk) 00:54, 29 November 2007 (UTC)

[edit] Numbering of the theorems

Most pages I've seen have numbered the theorems opposite to how they're numbered in this Wiki article, most notably Wolfram. Is there an accepted labelling scheme?Taras 96 (talk) 00:56, 29 November 2007 (UTC)

[edit] Discoverer of FTC

It is agreed by historians that Barrow discovered the FTC, but elementary mathematics books are not written by historians and so the history is simplified. Instead of acknowledging the less glorified figures, the main developers of calculus (Newton and Leibniz) are given credit for everything, including the fundamental theorem. This is a historical fib, which can be easily corrected on Wikipedia, since an article is not confined to the domain of expertise of any one person. It is perfectly possible for a person who knows the history to edit the math pages for historical accuracy, just as for a person who knows mathematics to edit a history page for mathematical accuracy.

The fact that Barrow discovered the FTC does not diminish Newton or Leibniz's claim to the discovery of the calculus. They were the ones that extended Barrows limited results to a full system capable of arbitrary generalization. That is why the bulk of the credit goes justly to them. But there is no need to lie about it, even if that lie is in many mathematics books (but in no serious work on the history of mathematics).Likebox (talk) 19:49, 16 February 2008 (UTC)

[edit] Error?

First Fundamental Theorem of Calculus.

Let f be a continuous real-valued function defined on a closed interval [a, b]. Let F be the function defined, for all x in [a, b], by

F(x) = \int_a^x f(t)\, \mathrm dt

Then, F is differentiable on [a, b], and for every x in [a, b],

F'(x) = f(x)\,.

Here you work with closed intervals, but mathworld [1] works with open intervals. I think there is a mistake. Randomblue (talk) 04:35, 17 April 2008 (UTC)

I don't think there is any mistake. The theorem is true both with open and with closed intervals.
I think the version with closed intervals is a bit more general, since any open interval can be written as a union of an increasing sequence of closed intervals, and by applying this theorem to any closed interval you end up with the antiderivative on the entire open interval. Oleg Alexandrov (talk) 05:03, 17 April 2008 (UTC)

[edit] FTC Pt 1 in words?

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In words, the value of the definite integral \int_a^x f(t)\, dt, viewed as a function of x, is an antiderivative of f.

Is it really a definite integral? It also leaves out the conditions of the start of the theorem. Does it really make the statement above it any clearer? Cheers, Ben (talk) 04:06, 12 June 2008 (UTC).