Talk:Fundamental theorem of calculus

Proof of the first part

The statement of the first part says that f is uniformly continuous. But the proof does not prove that f is uniformly continuous. The proof doesn't prove that f is continuous at all. — Preceding unsigned comment added by 2601:449:8400:242f:add9:abd3:3ad7:ca75 (talk • contribs) 13:50, 2 August 2021 (UTC)

"f is uniformly continuous" is a statement and thus does not need to be proved. It is a supposition for the proof that follows. If a function is NOT continuous, then the proof does not apply. 149.32.192.38 (talk) 21:55, 6 March 2023 (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)

y is a function of t, not x...

Thanks to the original author(s) of this page. There is one detail that confused me when I was learning the topic. Only later did I realize the subtlety which I consider an error.

In the "Geometric Meaning" section, it starts with "For a continuous function y = f(x) whose graph is plotted as a curve, each value of x has a corresponding area function A(x), representing the area beneath the curve between 0 and x."

This is not quite correct. The area function, A, is indeed a function of x. (x being the upper value of the integral, aka area under the curve.) However, the continuous function y, is NOT a function of x. The independent axis in the graph should be something other than x, traditionally t. The way it is presented, x is a value on the independent axis, and creates a point of the graph at [x, f(x)].

If we wanted to use x for the independent axis, then we would need to assign points on that axis to be something like x1 and x2. You can't say the independent axis is x and then say there is a value x at a certain point on it.

I know this is a subtelty, but it is worth understanding. It confused me for a long time until the light bulb went on! 149.32.192.38 (talk) 22:19, 6 March 2023 (UTC)

The function is f, and y is the value of f at x. See Function (mathematics) for the definition of this standard notation. D.Lazard (talk) 10:48, 7 March 2023 (UTC)

I agree with you 100%, but you do not refute my point. This is why I consider this a subtlety.

The point here is that f cannot be a function of x, when you pick a point on the axis and call it "x." You can call the point on the axis x1, x2, x3 or anything other than x. But it cannot be x.

I challenge you to find any other proof of the FTC that starts with "For a continuous function y as a function of x...." The area function, A, is indeed a function of x. However, the original function f (i.e. the curve that bounds the area) must be a function of a variable other than x. Again, traditionally t in other proofs of the FTC.

It may seem like I'm being pedantic, but I was hung up on this for a long time while trying to develop a deeper understanding of the FTC. 149.32.192.38 (talk) 14:46, 7 March 2023 (UTC)

This may help explain my point:

https://ximera.osu.edu/mooculus/calculus1/firstFundamentalTheoremOfCalculus/digInFirstFundamentalTheoremOfCalculus

Specifically notice the comments about the accumulation function. 149.32.192.38 (talk) 15:04, 7 March 2023 (UTC)

One last note.

In the "Proof of the first part" section of this article, it properly states:

"For a given f(t)..."

This is correct, and directly contradicts the first sentence in the "Geometric Meaning" section:

"For a continuous function y = f(x)..."

I know this seems pedantic, but it is important to know there is a differnce betweem f(t) and f(x) while proving the FTC. 149.32.192.38 (talk) 15:13, 7 March 2023 (UTC)

I have changed "For a given f(t)...” into "For a given function f ...” because the former formulation in incorrect, since this is not the value f(x) of the function at some point that is given, but the function in its entirety. I have also fixed similarly the beginning of the paragraph § Geometric meaning. You must be aware that in “the function $y=f(x)$" and "$x\mapsto A(x)$", the symbols x and y are placeholders that have no value. They could be replaced with, say, $\cdot$ and $*$ without changing the function. So, there is no harm of using x and y for specific values. D.Lazard (talk) 16:31, 7 March 2023 (UTC)

"The first part of the theorem" in the introduction

(I apologize if this is wrong - if so, please delete this)

The second paragraph on the introduction contains the statement:

The first part of the theorem, the first fundamental theorem of calculus, states that for a function f , an antiderivative or indefinite integral F may be obtained as the integral of f over an interval with a variable upper bound.

Shouldn't this be how F is defined, not the theorem itself? I.e. shouldn't it be something like:

The first part of the theorem, the first fundamental theorem of calculus, states that for a function f , if an antiderivative or indefinite integral F is defined as as the integral of f over an interval with a variable upper bound x, then

$$F'(x)=f(x).$$

BouleyBay (talk) 13:08, 31 May 2023 (UTC)

I don't understand your proposed alternative. The first part of the theorem says that, given f, if we define a new function (usually denoted F) as a certain integral of f with a variable upper bound, then F is an antiderivative of f (equiv: then F' = f). That's what the original sentence says. Your version is either circular or redundant; if the words "an antiderivative or indefinite integral" were replaced by "another function" then it would be fine (and equivalent to what is already written). I am not deeply wedded to the current wording, quite possibly it could be clearer. --JBL (talk) 17:34, 31 May 2023 (UTC)

Order of theorems

I'm wondering whether the article has the order of the Fundamental Theorems of Calculus reversed... several sources (https://math.stackexchange.com/questions/3635636/confused-about-the-fundamental-theorem-of-calculus, or Calculus, 11th edition by Larson and Edwards, to name a few) state that the First Fundamental Theorem of Calculus gives int:a-->b (f(x))=F(b)-F(a), while the Second Fundamental Theorem of Calculus states that d/dx (int:a-->x (f(t))dt) = f(x). This seems to go against the naming used in this article. Apologies for the lack of proper mathematical notation in this post. PeterRet (talk) 06:57, 13 February 2024 (UTC)

Further Explanation Required

In the section "Proof of the first part", there is a line saying "the latter equality resulting from the basic properties of integrals and the additivity of areas.". I am not aware of any Wikipedia page which lists basic properties of integrals. Also, the properties which are being used should be mentioned. 183.83.216.129 (talk) 11:53, 26 April 2024 (UTC)