Talk:Fundamental theorem of calculus/Archive 3

Archive 1

Archive 2

Archive 3

Re: FTC Proof

Area is positive by definition, rendering the absolute value operation nugatory.

On the other hand, by your concern, the original proof is flawed due to lack of accountability for left-hand sums for convex functions. I propose adding a second graph illustrating a sample such sum, and editing the original definition as follows:

$A(x+h)-A(x)=f(x)h+({\text{Red Excess}})$

to

$A(x+h)-A(x)=f(x)h\pm ({\text{Red Excess}})$

Preceding unsigned comment added by OverLordGoldDragon (talk • contribs) 16:06, 17 February 2016 (UTC)

Hi @OverLordGoldDragon: Just because you've written the above, it does not mean a consensus exists for you to re-add your content. Please wait until other editors get involved and make further comments. Thank you -- samtar ^whisper 16:11, 17 February 2016 (UTC)

@Samtar Fair.

As I explained above, the proof you have been attempting to add to the article is incorrect. The flaw has nothing to do with convexity but rather with the possibility that the function f is decreasing.

It is a fair point that area is always positive. However, the rectangle excess and red excess can be negative, so they are not always areas. Perhaps the article is not as clear about this as it should be. But the solution is to make clear that these excesses are not areas, not to banish the absolute value signs. Absolute value signs (or some proxy) are necessary somewhere. Ozob (talk) 00:31, 18 February 2016 (UTC)

@Ozob: "The flaw has nothing to do with convexity but rather with the possibility that the function f is decreasing." Correct. That's what I meant. " However, the rectangle excess and red excess can be negative, so they are not always areas." The values do represent areas - and they are always positive. What you are referring to is whether we add or subtract the Excess that is produced. I addressed this in my previous reply, which you have dismissed. "But the solution is to make clear that these excesses are not areas" Then what are they? "Absolute value signs (or some proxy) are necessary somewhere." A vague suggestion. See the revision I suggest as a resolution:

___________________________________________

In fact, this estimate becomes a perfect equality if we add the red portion of the "excess" area (shown in the diagram), and subtract it when the rectangles produce overestimates (over the decreasing regions in the function, not shown). So:

$A(x+h)-A(x)=f(x)h\pm ({\text{Red Excess}})$

Rearranging terms:

$f(x)={\frac {A(x+h)-A(x)}{h}}\pm {\frac {\text{Red Excess}}{h}}$.

The nullification of the Red Excess can be proven more intuitively as follows:

$0\leq {\text{Red Excess}}\leq {\text{Rectangle Excess}}$

Thus, if

${\text{Rectangle Excess}}=0$

then

${\text{Red Excess}}=0.$

As seen from the graph,

${\text{Rectangle Excess}}=(f(x+h)-f(x))h.$

Since red excess is zero when rectangle excess is zero, we shall replace red excess in the equation above with rectangle excess. Then,

$f(x)={\frac {A(x+h)-A(x)}{h}}\pm {\frac {(f(x+h)-f(x))h}{h}}$

Taking the limit,

$f(x)=\lim _{h\to 0}\left({\frac {A(x+h)-A(x)}{h}}\pm (f(x+h)-f(x))\right),$

OverLordGoldDragon (talk) 19:36, 24 February 2016 (UTC)OverLordGoldDragon

You say that the rectangle excess and red excess are always positive and are always areas. I disagree, but I believe the problem is that we're using different definitions of the rectangle excess and the red excess. For me, the red excess is defined to be $A(x+h)-A(x)-f(x)h$, and the rectangle excess is defined to be $h(f(x+h)-f(x))$. I think we both agree that these quantities can be negative. If I understand you correctly, you want to define the red excess to be $|A(x+h)-A(x)-f(x)h|$ and the rectangle excess to be $|h(f(x+h)-f(x))|$. Both of these can be interpreted as areas. With this understanding and some minor changes, the proof presently in the article becomes the one you gave above.

I still prefer the proof presently in the article. I do not see any reason to insist on making the red excess and the rectangle excess be areas, and I think it is more natural not to. Making them areas forces you to use ± operations instead of + operations because you must undo the absolute value you just took. The proof in the article does not have to keep track of this extra sign, and for that reason, I think it's simpler. Ozob (talk) 03:07, 25 February 2016 (UTC)

Proof of the first part is circular

The proof of the first part starts with:

For a given f(t), define the function F(x) as

$F(x)=\int _{a}^{x}f(t)\,dt.$

Later, it uses the mean value theorem to show that

${\frac {F(x_{1}+\Delta x)-F(x_{1})}{\Delta x}}=f(c)$

The problem is that the mean value theorem applies if and only if $F'(x)=f(x)$. But that's what we are trying to prove, hence the proof is circular.

The proof linked at the end of the section (http://www.imomath.com/index.php?options=438) is correct, but uses a completely different approach: it starts with the mean value theorem, then shows that f(x) is linked to the integral using Riemann sums.

This proof: https://math.berkeley.edu/~ogus/Math_1A/lectures/fundamental.pdf starts in a similar way to the one in the Wikipedia article, but uses the continuity of f and a few properties of integrals instead of the mean value theorem.

— Preceding unsigned comment added by 82.56.38.63 (talk) 09:45, 26 February 2016 (UTC)

MIT non-functional 146.115.82.167 (talk) 01:34, 4 September 2016 (UTC)

No. See Mean_value_theorem#First_Mean_Value_Theorem_for_Definite_Integrals for a proof of the required statement. Ozob (talk) 14:41, 4 September 2016 (UTC)

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Unfortunate link for "indefinite integral"

The current link to "indefinite integral" links to the article on Antiderivative, which would lead a reader to the conclusion (reinforced by that article) that "indefinite integral" is merely a synonym for "antiderivative". With that interpretation, the statement of the first part of the FTC as given in the current article, which is:

"The first part of the theorem, sometimes called the first fundamental theorem of calculus, is that the indefinite integral of a function is related to its antiderivative, and can be reversed by differentiation.[Note 1] This part of the theorem guarantees the existence of antiderivatives for continuous functions.[2]"

seems to claim that the content of the FTC is to say that two synonyms are related, which is inane.

To clarify the content of the FTC, there needs to be a distinction between the definition of "indefinite integral" and the definition of "antiderivative".

Tashiro~enwiki (talk) 00:48, 25 May 2017 (UTC)

I have edited the lead for fixing this. D.Lazard (talk) 08:45, 25 May 2017 (UTC)

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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. — Preceding unsigned comment added by 69.23.220.138 (talk • contribs) 09:25, 13 August 2005 (UTC)

FORMULA needed

Shouldn't the formula just be stated outright before the proofs like on the top? — Preceding unsigned comment added by 71.138.215.62 (talk • contribs) 02:46, 25 November 2006 (UTC)

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. — Preceding unsigned comment added by 58.187.53.161 (talk • contribs) 05:30, 2 March 2008 (UTC)

Apparent typo in equation

There is an apparent typo in the equation following the words " tiny black-bordered rectangle. More precisely," The end of the equation has a vertical bar indicating absolute value just before the comma at the very end of the equation, immediately before the words "where x + h 1 {\displaystyle x+h_{1}} {\displaystyle x+h_{1}} and x + h 2 {\displaystyle x+h_{2}} {\displaystyle x+h_{2}} are points". This vertical bar at the rightmost side of the equation is a closing vertical bar which is not matched by a corresponding "opening" vertical bar anywhere in the equation. It's the vertical-bar equivalent of mismatched parentheses. — Preceding unsigned comment added by 64.223.129.132 (talk • contribs) 23:41, 17 December 2018 (UTC)

I took care about the vert. bar. Purgy (talk) 07:01, 18 December 2018 (UTC)

Regarding the example of timing a car

I knew *nothing* of calculus until right now, and I'm "not good with math". The example about the car worked perfectly on me; I found it easy to follow, and after being clueless on the topic, I now do have a clue - a rudimentary one I'm sure, but clear and memorable.

To whoever conceived and wrote that example, my sincere thanks - and please write about more things I don't understand yet! :) TooManyFingers (talk) 11:47, 17 October 2021 (UTC)

Proof of the second part

There's a slight problem. The statement of the second part states that F'(x) = f(x) for almost all x in [a,b] (in the sense that F' ≠ f only on a measure-zero subset of [a,b]). However, the proof of the second part assumes that F' = f for all x in [a,b]. — Preceding unsigned comment added by 2001:569:7753:9b00:7887:719f:c887:47ef (talk • contribs) 00:54, 20 July 2017 (UTC)

There is a version of the FTC part II which does demand only that F' = f for almost all x in [a,b], as first assumed. A source for this is (Botsko, 1991: https://www.maa.org/sites/default/files/pdf/mathdl/MM/0025570x.di021172.02p0038j.pdf). Now, for most practical purposes, it is enough that f is continuous on (a,b). That secures that a piecewise continuous function can still be analytically integrated by dividing the interval exactly along the discontinuities. For example, the signum function, f(x) = x/|x| has exactly one discontinuity, in x = 0. One can integrate the function on the interval [-1,2] by adding the results from integrating on the intervals [-1,0) and (0,2]. On [-1,0) one antiderivative F(x) = -x, while on (0,2] the antiderivative G(x) = x can be used. But using the result of Botsko, one can integrate this even easier, by observing that the absolute value function H(x) = |x| is the antiderivative of f(x) almost everywhere (in fact, everywhere except for the point x = 0). Then H(x) can be used to calculate the integral on the whole interval [-1,2] without the need for any subdivision or area cancellation tricks. Vegarius (talk) 08:54, 9 February 2023 (UTC)