Talk:Fundamental theorem of calculus

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)

I think it should be:

$F(x)-F(a)=\int _{a}^{x}{f(x)\,dx}\rightarrow F(x)=\int _{a}^{x}{f(x)\,dx}+F(a)$

Later on, in the corollary, the lower boundary is included.

Alternatively, one should specify that there is a constant of integration C that is yet to be defined. 2A02:2454:C00A:6400:89E5:3344:B32E:8F10 (talk) 10:43, 4 June 2024 (UTC)

The original question is about the introduction; there are no equations in the introduction. This makes the meaning of your comment rather obscure. --JBL (talk) 20:30, 4 June 2024 (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)

I agree, I didn't look too hard for an authoritative source, but the AOPS Calculus text I had next to me agrees that the first "part" or First Fundamental Theorem states that:

$\int \limits _{a}^{b}f(x)dx=F(b)-F(a)$

and that the second is:

$g(x)=\int \limits _{x_{0}}^{x}f(t)dt$

Wyrdwritere (talk) 18:52, 9 July 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)

Possible very minor typo for current Reference 11

A very minor issue, but it would seem that Reference 11, as of the time of writing, refers to theorem 7.21 instead of 8.21 of Rudin's RCA, which should be the correct number, having checked a copy of the book. But since it could be a chapter numbering difference between different versions of the book, I've opted to just open a topic here. Sorry for the bother. 2001:660:6402:408:1867:ABEE:384A:B949 (talk) 12:24, 11 July 2024 (UTC)