# Talk:Antiderivative

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## Simple rule

here is the simple rule

## Merging with integral?

Why isn't this just merged with Integral?--Siva 20:43, 29 Aug 2004 (UTC)

Simply put, because antiderivatives and integrals are not the same thing. I think they deserve separate articles, although certainly a discussion of how they are related belongs in each article (which there currently is). Remember that it is only by the deep result of the Fundamental theorem of calculus that the seemingly otherwise unrelated concepts of integration and antidifferentiation end up being connected. and for certain classes of functions, this connection can't be made, because some functions have integrals, but not antiderivatives (within the appropriate domain). So the connection given by the Fundamental Theorem is deep and important, but it does not mean that the two ideas are exactly equivalent. -lethe talk 01:04, Aug 30, 2004 (UTC)

When is the antiderivative not equal to the integral? What kind of function would have an integral but no antiderivative?--Siva 18:22, 30 Aug 2004 (UTC)

For example, the Cantor function. -lethe talk

Also, any constant function f(x) = k. The integral denotes the area underneath the functionbetween bounds. Integrals are inverse differentiation. f(x) = 5 differentiates to f'(x) = 0, but the area beneath f(x) is at no point zero except at x = 0 itself. Another example includes hyporbolas. Integrals crossing any hyperbola's poles will always be positive or negative infinity, as is the definition of a pole. But this is not true of it's antidervite, which is simply a function for which the original hyperbola represents the derivitive.He Who Is 20:34, 28 April 2006 (UTC)

That is incorrect. Constant functions have both integrals and antiderivatives, and the two are the same (please read the article, as well as the one on the Fundamental theorem of calculus before commenting). For any continuous function the notions of integral and antiderivative coincide; that's one reason why the two are often conflated. However, the fundamental theorem of calculus does not apply to all functions; to find counterexamples you need to look at discontinuous integrands, e.g. the Cantor function cannot be obtained by integrating its own derivative -- which exists almost everywhere (yet not all discontinuous integrands are counterexamples). FilipeS (talk) 11:32, 15 May 2012 (UTC)

## Merging with integral?

I would not be so sure about a merger. It looks to me that integral is about definite integrals, and antiderivative is the operation inverse to differentiation, meaning input is a function, and output is a function. These are two different things. They are connected through the fundamental theorem of calculus, but this does not mean that they are the same.

Also note that in more than one variable, integral and antiderivative are even more distinct. As such, I believe two articles are needed, as there are two different concepts, but of course something needs to be said about how they relate. Oleg Alexandrov (talk) 19:30, 4 November 2005 (UTC)

Of course it should not be merged. I agree with Oleg's reasons, and I stand by the reasons I stated over a year ago above -lethe talk 00:18, 5 November 2005 (UTC)
Yes, yes, I forgot to give you credit. :) You indeed wrote about it above, and I used some of that in my argument (but some of my argument was original, even though very similar to what you stated earlier :) Oleg Alexandrov (talk) 01:41, 5 November 2005 (UTC)
I think I saw you arguing somewhere that holomorphic and analytic should not be merged. This is pretty much the same thing (in my opinion). Just because there's a theorem that says some things are equal, does not mean that the things are the same. -lethe talk 09:12, 5 November 2005 (UTC)
The section I added on antiderivatives of non-continuous functions should provide a pretty strong rebuttal to the merger idea Fiedorow 04:12, 11 December 2005 (UTC)
Antidifferentiation is the "inverse" of differentiation. Integrals are entirely different objects. Tomyumgoong 04:05, 29 April 2006 (UTC)

I think that should be explained in the main article as well. I don't really know much maths and so I hit this page hoping to find out what is the difference between "antidifferentiation" and "integration", but seems the main article is not helping much. Some of the arguments here should be modified and put into the main article. (But honestly, I still don't really understand it very well after all these)

Yes, it would be very helpful if the article explain what is the difference between integral and antiderivative. — Preceding unsigned comment added by 178.42.152.68 (talk) 11:19, 30 March 2013 (UTC)

Oleg would like nicer pictures for Figures 1 and 2 illustrating Example 4. These pictures are given by parametric equations

$t\mapsto\left(\sum_{n=1}^8\frac{(t-\cos(n))^{1/3}}{2^n}, \frac{1}{\sum_{n=1}^8\frac{(t-\cos(n))^{-2/3}}{3\cdot 2^n}}\right)$
$t\mapsto\left(\sum_{n=1}^8\frac{(t-\cos(n))^{1/3}}{2^n},t\right)$

respectively, where $-1\le t\le$ and where

$\frac{1}{\sum_{n=1}^8\frac{(t-\cos(n))^{-2/3}}{3\cdot 2^n}}$

is assigned the value zero if $t=\cos(1),\cos(2),\dots\cos(8)$. Perhaps someone clever enough with something like gnuplot or Adobe Illustrator or something could produce better pictures. [Note: for the actual Example 4, the upper summation limit of 8 should be replaced by $\infty$, but that is impossible to plot.]

Also as a matter of curiosity, what is the value of the Lebesgue integral of Example 4?Fiedorow 17:46, 11 December 2005 (UTC)

To answer my own question, it seems that Fatou's lemma implies that the Lebesgue integral of Example 4 satisfies the fundamental theorem of calculus.Fiedorow 01:03, 12 December 2005 (UTC)
OK, I've now uploaded new versions of the pictures, created within Maple.Fiedorow 17:02, 13 December 2005 (UTC)
Thanks, looks good! The lesson I learn again and again is that it is more productive to bug others than do the work myself. At least, it worked in this case. :) Thanks a lot! Oleg Alexandrov (talk) 20:23, 13 December 2005 (UTC)

## Computer program to take the antiderivative

Do you know any computer program to perform these operations? Thank you. --User:Oculto 11 Feb 2006

Mathematica and Maple (the computer algebra system) can do that. Oleg Alexandrov (talk) 23:42, 11 February 2006 (UTC)
Or Maxima, if you prefer free software. See also Comparison of computer algebra systems. A web-based solution: Wolfram Integrator, based on webMathematica. Torzsmokus 22:23, 21 March 2007 (UTC)

## Rename to primitive?

Shouldn't this page really be titled "primitive" or "primitive function"? This is the standard term among mathematicians. It makes sense. You have a primitive, and its derivative. Anti-derivative is mostly a term invented by undergraduate textbooks, and seems to me to be a bit unwieldy. Grokmoo 04:25, 19 February 2006 (UTC)

Well, I guess undergrad students are a bigger audience here than PhD's. :) I learned this in high school in Europe, where it was called "privitive", but in US I have heard only about "antiderivative". :) Oleg Alexandrov (talk) 06:08, 19 February 2006 (UTC)
Yes anti-derivative is a misnomer, anti means contrary not reverse. Primitive is the reverse of derivative. It is sad that in the USA they tend to complicate logical intelligibility by coming up with this type of nomenclature. I support the proposal that the title of primitive should be used. 79.168.11.181 (talk) 17:37, 11 April 2012 (UTC)

Nonsense. The prefix "anti-" can absolutely mean "opposite" (and if you have objections to a common dictionary, might I remind you that the exponential function is also called antilogarithm?), and I don't see what distinction you're trying to make between "contrary" and "reverse", anyway. While the term "primitive" is used in other languages, in English the most common word seems to be "antiderivative", and it's perfectly justified. FilipeS (talk) 11:36, 15 May 2012 (UTC)

## what's with the revertion?

Why is Oleg Alexandrov reverting my addition of alternate names of antiderivative in the intro? Moreover, the use of 3 'or's in sucession is very ugly to read. Loom91 10:59, 14 August 2006 (UTC)

Here is the diff. One should not insert links in definitions, so integral at the beginning is inappropriate.
The connection between derivatives and integrals is in the next sentence in the intro, there is no need to put it in the definition. Oleg Alexandrov (talk) 16:07, 14 August 2006 (UTC)
The 'or's are not nice, but there is no need for integral as well as indefinite integral, and Oleg is right about the link. JPD (talk) 16:23, 14 August 2006 (UTC)
I'd agree with Oleg. Antiderivatives are not integrals. The connection is explained just a short breath later. I'd say keep "integral" and "indefinite integral" out of the first sentence.
On a different note, I think I'd take out the sentence, "Finding an expression for an antiderivative is harder than calculating a derivative, and may not always be possible." It kinda clutters the intro, and isn't quite correct -- or rather isn't quite precisely worded. You might say, "Finding a closed-form expression for an antiderivative for a composition of elementary functions is harder than calculating a derivative of such an expression, and may not always be possible," but that's getting kinda wordy.
My two cents, Lunch 17:27, 14 August 2006 (UTC)
I agree with Oleg. Keep "indefinite integral" which is the same thing, and take out "integral" which is not the same thing. If people are concerned with two "or"s (which isn't really such a big deal) use "antiderivative, primitive, or indefinite integral". VectorPosse 21:37, 14 August 2006 (UTC)

## Renaming to Primitive

For the sake of pluralism I think primitive should be added to the title, as i had never heard of an antiderivative until i tried to look up primitivation... In europe the word antiderivative is used only in england... In french and german for example they say "primitive", in portuguese and spanish "primitiva" and little or no references to antiderivative can be found in textbooks, undergraduate or not :D Sebastiao 14:37, 5 December 2006 (UTC)

Since it is called an antiderivative in US and England, and probably in other countries whose main language is English, I guess it is better if it stays the way it is now. The name "primitive" is mentioned in the first article sentence though. Oleg Alexandrov (talk) 15:47, 5 December 2006 (UTC)
In India we usually use indefinite integral, though primitive is also sometimes used. I don't remember seeing antiderivative. Are you sure England primarily uses antiderivative? Loom91 16:27, 5 December 2006 (UTC)
Also, the disambiguation page for the word "primitive" refers to this page (although it's a bit buried down the page). I might comment that there are all sorts of math terms that translate into other languages in confusing ways. For example, the words for "manifold" in Spanish and French (and probably all Romance languages) are cognates of the English word "variety", but we can't refer to manifolds as varieties for the sake of people from other countries looking for an article on manifolds. A variety and a manifold are, in English, two very different things. (At least the word "primitive" also has the meaning of "antiderivative" so it's not a problem in this case.) VectorPosse 16:50, 5 December 2006 (UTC)
FYI: As far as I know, "indefinite integral" is used universally in high-school and undergrad levels in England. At least it was in the 80s and 90s when I was a student there - I had never heard of "antiderivative" before I got to this page. —The preceding unsigned comment was added by Dave w74 (talkcontribs) 03:11, 21 February 2007 (UTC).
Just a little contribution, I was at secondary school in England from '98-'04 and I can't remember ever reading the term antiderivative in the textbooks (This was for the Edexcel exam board). It was always called an indefinate integral. During my undergraduate course have I never heard the term used either, and after a quick poll of the people round the room everyone else had the same experience (some even thought it was a made up term, like for a Wikipedia joke page).
Me too. I was educated to degree level in mathematics in England (graduated '89) and I never heard the term "antiderivative". "primitive" as the correct term, "indefinite integral" as a rather more informal term. When I trained as a maths teacher about 10 years later I never came across it either. 82.68.102.190 (talk) 19:31, 8 May 2009 (UTC)

## Crucial but absolutely cryptic sentence

From the "Uses and properties" section:

"It is critical to remember that an integral is not the same, in general, as the means for evaluating it; and the function that an integral implies stands apart from that means - in the case of single-variable integrals, from antiderivatives."

This sentence is absolutely cryptic. You can understand it only if you know everything a priori (and in this case you don't need to read it)!

I removed it. If it is truly critical to remember, and you want to insert it back, then in my opinion you need

• to explain its meaning (giving at least an example)
• to explain why it is so important to remember.

With regards,

Paolo.dL 09:02, 2 August 2007 (UTC)

## Inconsistency: Indefinite integral = any antiderivative or all antiderivatives?

NOTE: This section was mildly refactored, for enhancing its readibility, by correcting the notation I used to represent the square root, as suggested by Paul August. Paolo.dL 13:05, 3 August 2007 (UTC)

### Introduction

Here's the definition of "indefinite integral" given in the introduction of the article:

"In calculus, an antiderivative, primitive or indefinite integral of a function f is a function F..."

And here's the definition of "indefinite integral" given in the section Uses and properties:

"Because of this, the set of all antiderivatives of a given function f is sometimes called the general integral or indefinite integral of f"

Can you see the inconsistency between the two definitions? I believe that the second is not correct. An indefinite integral is not "all the antiderivatives of f" (an infinite set of functions), but just "any antiderivative of f", i.e. "a single antiderivative" arbitrarily selected from among infinite possible ones. This is because c, the integration constant, is not a set, but just a single constant arbitrarily selected from among a set.

Think of the nonunique solution of a square root, another example of quasi-inverse operation:

$square \ root(a^2) = \pm \sqrt {a^2} = \pm a$.
(I couldn't find a standard symbol for this operation; sqrt(x) or sqr(x) are conventionally used to indicate $\sqrt x$, the principal square root of x)

The symbol ± means either "+ or -". It does not mean both "+ and -". Therefore, it would be incorrect to write:

$square \ root(a^2) = \{ +a, -a \} \,$

where {+a, -a} is the set containing +a and -a. In symbols:

$square \ root(a^2) \neq \{ +a, -a \} = \begin{cases} +a \\ -a \end{cases}$

Paolo.dL 23:41, 29 July 2007 (UTC)

I corrected the second definition in the article. Please give me a feedback: Do you agree? Paolo.dL 18:56, 2 August 2007 (UTC)

### References

Unfortunately there is no uniformity of terminology here. So things are not a simple as they seem. Different authors define "indefinite integral" differently.

• Some authors define it to be synonomous with antiderivative. in which case, there may be infinitely many indefinite integrals for a given function.
• Some define it as the most general antiderivative of f(x); that is,
$\int f(x)\, dx. = F(x) + C$
— Purcell, Calculus with Analytic Geometry p. 378. (See also Hughes-Hallett, Calculus: Single Variable, p. 288 [1]).
• Some define it to be the set of all antiderivatives of a given function (Encyclopaedia of Mathematics (2002) [2], Calculus for Dummies p. 236: [3]).
• Some define in terms of a definite integral (Courant, Differential and integral Calculus p. 110 [4] - [5])

Paul August 22:08, 2 August 2007 (UTC)

Thanks Paul, this is interesting. However,
• I can't see the difference between points 1 and 2. In both cases, there are infinitely many indefinite integrals, aren't there? I mean, "most general" means "any possible", doesn't it? Is there any author who defines the indefinite integral as a specific antiderivative (e.g. the "simplest antiderivative", with C = 0)? (I doubt it)
• About point 4, according to this search by Amazon Courant states, in page 116 "we shall no longer make any distinction between indefinite integral and primitive function", and a similar search with keywords "primitive function" shows that he defines the primitive function as any antiderivative. Shouldn't we place the reference to his book in point 1 or 2?
Other on line references (indefinite integral = any antiderivative):
Paolo.dL 15:35, 9 August 2007 (UTC)

### Analysis and classification of the different approaches

So, there exist different approaches in the literature. However:

• Unwanted ambiguity. Wikipedia was the only page where the integral was defined in two different ways. In a single article, you must either use a single definition, or explicitly state that different definitions exist in the literature.
• Ambiguity of the word "arbitrary". This statement will prove to be useful later. The word "arbitrary" in the expression "arbitrary constant of integration" is somewhat ambiguous. It may be used to mean:
1. "arbitrarily selectable" from among infinitely many values (hence still indeterminate, until selection is accomplished), or
2. "arbitrarily selected" (hence determined through a process of "arbitrary selection of one value C from among all possible values $\mathbb{R}$ or $\mathbb{C}$").
• Does the integral symbol always mean integral? As far as I know, everybody writes indefinite integral examples using C (or c) as constant term, and they always call C the "arbitrary constant of integration". This is a widely accepted standard terminology, which I have seen repeated everywhere. Is there anybody who calls C the "infinite set of integration constants"?
If the indefinite ingegral were defined as a set, and the symbol $\int \,$ meant "indefinite integral" (rather than a single antiderivative), then the definition formula would contain a set in the second member. For instance, (I am not sure if this notation is correct):
$\int f(x)\, dx = \{ F(x) + c: c \in \mathbb{R} \}$
or, simplifying (I don't know if this notation is allowed):
$\int f(x)\, dx = F(x) + \mathbb{R}$
where $F(x) + \mathbb{R}$ is used as if it were the sum of a scalar plus a matrix.
Since I have never seen an indefinite integral written in a similar way, I hypothesize that those who maintain that an indefinite integral is a set might use the symbol $\int \,$ to indicate an antiderivative, rather than an indefinite integral. In this case the indefinite integral would remain an operation without a symbol, as the square root above. Is my hypothesis correct?...
• Quasi-inverses (nonunique inverses). The indefinite integral may be regarded as indeterminate. Partially determinate, but all the same indeterminate. It determines a set of possible results, i.e. a nonunique result. This is also true for the square root. Both the indefinite integral and the square root can be reagarded as nonunique inverses or "quasi-inverses". Now, what do we do with nonunique results? We can either arbitrarily select one of them or just take them all, as a set. And I am not certain about what is the best option. But even if we arbitrarily select one of the possible results, they are still nonunique and the integral is still indeterminate.
• Pseudo-inverses. I give to the term "pseudo-inverse" a different meaning. In my opinion, pseudo-inverse is an operation which gives a unique result, such as the Moore-Penrose pseudo-inverse, or the principal square root, or "the simplest" antiderivative (when defined), or a definite integral with variable upper limit. Therefore, a pseudo-inverse is pseudo because it uses as input some additional information which is not typically available to an inverse function (e.g. the assumption of orthogonality, or an arbitrarily selected sign for a square root, or an arbitrarily selected value for the constant of integration or for the lower limit of integration).
• 2nd theorem of calculus. When the integral is defined as an arbitrarily selected antiderivative, the second fundamental theorem of calculus is written and explained as follows:
$\int_a^b f(x)\,dx = F(b) - F(a),$ where F = indefinite integral = single antiderivative
The statement F = indefinite integral wouldn't work if the indefinite integral were a set! F(b) - F(a) would be the subtraction of two sets, and (if we perform an elementwise subtraction) would yield an infinite set of identical numbers. There's a possible approach to solve this problem, which appears not very satisfactory to me:
$\int_a^b f(x)\,dx = F(b) - F(a),$ where F = single antiderivative ≠ indefinite integral (indefinite integral = set of antiderivatives)
Here you define the indefinite integral as a set, but then you declare that F is just an element of that set. In this case the theorem would not relate the definite integral with the indefinite integral, but just with one element of the indefinite integral set. The indefinite integral would not explicitly appear in the theorem. That's a strong limitation of this solution, in my opinion. Why should we define the indefinite integral as an object which does not appear in the theorem?...
• Right-invertible. Correct me if I am wrong: if the result of an indefinite integral were a set, then the derivative of that set would be a set of infinitely many identical functions. This would make no sense. The indefinite integral (as well as the definite integral with variable upper limit) should be reversible by a derivative:
• An indefinite integral cannot fully reverse a derivative (i.e. it is not left-invertible), but we are not masochistic:
• why shouldn't it at least be reversible by a derivative? (i.e. right-invertible).
(see the two conditions required to be a complete inverse, in the article inverse function, that I just finished to edit yesterday; see also the first part of the fundamental theorem of calculus).
Again, the example about square root can help. A square root is reversible by a square, isn't it? If a2 = b:
$square \ root(a^2) = \begin{cases} +a \\ -a \end{cases}$ (not left-invertible; nonunique solution)
$square \ root(b)^2 = b$ (right-invertible; unique solution)
Am I right, or even in this case different definitions of the square root can be found in the literature?
• Tentative conclusion. Based on what little I know, I conclude that defining the indefinite integral as a set is possible, but it entails an appearently unreasonable entanglement. If there is a good reason to choose this approach, I cannot see it. If we want to describe this approach on the article, the rationale should be clearly explained.
• Reliable sources. By the way, are we sure that the Encyclopaedia of Mathematics (2002) [6], and Calculus for Dummies p. 236: [7] or Wikipedia or MathWorld or PlanetMath are reliable sources? Shouldn't we trust only university textbooks dealing with the specific topic? Encyclopedias and divulgative books are not primary sources.

Paolo.dL 10:53, 5 August 2007 (UTC)

Summarizing table

Approach 1 Approach 2 Approach 3 Approach 4
Indefinite integral definition Any arbitrarily selected
(a.k.a. "the most generic")
antiderivative
The set
of all antiderivatives
The specific
antiderivative
with zero at x = a
"The simplest"
antiderivative
Indefinite integral symbol $\int \,$ none $\int = \int_a^x \,$ $\int = \int_0^x \,$
Generic antiderivative symbol $\int \,$ $\int \,$ none none
Specific antiderivative symbol
(C = 0)
$? = \int_0^x \,$ $? = \int_0^x \,$ $\int \,$ $\int \,$

This table is an attempt to summarize the different approaches. Approach 3 and 4 are just possible approaches; I hypothesize that some author might have adopted them, but personally I have never seen them described in the literature. Is the table complete? Is it correct? Is it correct to state that the definition of the square root is similar to "Approach 1"? Please let me know.

I believe it is important to produce in this talk page a classification of the different approaches. Please give your contribution (if you have references please insert them above, in the appropriate section). We should eventually decide whether the article should:

1. espouse the first approach ignoring the others (as done in current version), or
2. report about all of them, without espousing one, or
3. espouse the first but in a separate section compare it with the other(s).

Paolo.dL 11:55, 6 August 2007 (UTC)

## Vandalism

There has been vandalism here lately. Even if the person thinks that his/her point is true, he/she needs to realize that going about it that way is unencyclopedic and can result in his/her IP address being blocked. All controversial claims need to be properly sourced. Here are the 2 vandalistic revisions:

http://en.wikipedia.org/w/index.php?title=Antiderivative&oldid=324283950

http://en.wikipedia.org/w/index.php?title=Antiderivative&oldid=324270368