Talk:Fundamental theorem of calculus

"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)

Proof of Theorem part a and b

I was looking to modify the proofs here to avoid calling on the mean value theorem and focus solely on definitions and properties of the definite integral:

For a given function f that is integrable on the interval $[a,b]$, for all $x\in [a,b]$, define the function F(x) as $$F_{a}(x)=\int _{a}^{x}f(t)\,dt.$$Note that, from the definition of a derivative and a definite integral:

${\begin{aligned}{\frac {d}{dx}}F_{a}(x)=\lim \limits _{h\rightarrow 0}{\frac {F_{a}(x+h)-F_{a}(x)}{h}}=\lim \limits _{h\rightarrow 0}{\frac {\int _{a}^{x+h}f(t)dt-\int _{a}^{x}f(t)dt}{h}}=\lim \limits _{h\rightarrow 0}{\frac {\int _{x}^{x+h}f(t)dt}{h}}\\=\lim \limits _{h\rightarrow 0}\lim \limits _{n\rightarrow \infty }{\frac {\Sigma _{i=1}^{n}f(x+{\frac {hi}{n}}){\frac {h}{n}}}{h}}=\lim \limits _{h\rightarrow 0}\lim \limits _{n\rightarrow \infty }\Sigma _{i=1}^{n}f(x+{\frac {hi}{n}})({\frac {1}{n}})=\lim \limits _{n\rightarrow \infty }\Sigma _{i=1}^{n}f(x)({\frac {1}{n}})\\=\lim \limits _{n\rightarrow \infty }nf(x)({\frac {1}{n}})=f(x)\end{aligned}}$

Thus, $F_{a}(x)$ is differentiable on $[a,b]$ and satisfies the definition of an antiderivative of $f(x)$ on that interval.

Now, for any definite integral of $f(x)$ that exists, $\int _{x_{0}}^{x_{1}}f(x)dx$, we may let $a$ vary so that $x_{0}<a<x_{1}$in order to say:

$\int _{x_{0}}^{x_{1}}f(t)dt=\int _{x_{0}}^{a}f(t)dt+\int _{a}^{x_{1}}f(t)dt=-\int _{a}^{x_{0}}f(t)dt+\int _{a}^{x_{1}}f(t)dt=F_{a}(x_{1})-F_{a}(x_{0})$.

If then $g(x)$ is some other antiderivative of $f(x)$, then $g(x)$ and $F(x)$ may only differ by a constant term, so we may write: $g(x)=F(t)+C$.

However, we then have:

$\int _{x_{0}}^{x_{1}}f(t)dt=F(x_{1})-F(x_{0})=g(x_{1})+C-(g(x_{0})+C)=g(x_{1})-g(x_{0})$.

So, for any antiderivative of $f(x)$, $F(x)$, the integral $\int _{x_{0}}^{x_{1}}f(t)dt$ is equal to $F(x_{1})-F(x_{0})$.

I am concerned that I am dodging some of the properties that we would like to consider, but the current proof does not include some of the formal statement either, so maybe this is up to the standards. I don't think I'm calling on continuity of f(x) here either, but maybe it is too late and I'm not seeing the obvious. Thank you! Cable30 (talk) 07:42, 8 November 2024 (UTC)