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Suppose, I have a function $\cos(x)$. Now,

$$\int{\cos(x)dx}$$

$$\sin(x)+c\\ {\text{[c is a constant]}}$$

Now, there could be an infinite number of values for $c$. For example, $c=1,2,-2,\pi,-\pi, 0, \frac{1}{3}, 7, 500, ...$

And using each value of $c$, we can formulate an infinite number of functions:

$$\sin(x)+3\tag{1}$$

$$\sin(x)+\pi\tag{2}$$

$$\sin(x)+\frac{1}{3}\tag{3}$$

$$...\tag{...}$$

Questions:

  1. According to the 1st comment to this post made by @EricTowers, "indefinite integral" and "antiderivative" are interchangeable terms:

These aren't definite integrals. They are "antiderivatives" or "indefinite integrals". And yes, an integrable function has infinitely many antiderivatives, differing only by a vertical shift of their graphs. – Eric Towers

However, according to Greg Martin's 2nd comment to his answer, "indefinite integral" and "antiderivative" aren't interchangeable terms:

"Antiderivative" and "indefinite integral" are not synonyms. An indefinite integral is literally an integral, and evaluating an indefinite integral leads to the set of all antiderivatives of the integrand (the set of all functions whose derivative equals the integrand). So: $∫\cos xdx$ is an indefinite integral; $\sin x+C$ is the set of all antiderivatives of $\cos x$; and $\sin x+3$ is one antiderivative of $\cos x$

Are they disagreeing? Am I misunderstanding them?

  1. According to the 2nd comment to this post made by @EricTowers, $\sin(x)+3$ or $(1)$ is an indefinite integral.

@EricTowers So, even if the constant is specified (for example $c=3$), $\sin(x)+3$ will still be called an indefinite integral of cos(x)? – tryingtobeastoic

Yes. A definite integral is a number, obtained by integrating over an interval. An indefinite integral is a function. – Eric Towers

However, according to the 2nd comment made by @GregMartin to his answer, $\sin(x)+3$ or $(1)$ is just an antiderivative out of the infinite number of antiderivatives found by evaluating the indefinite integral $\int{\cos(x)dx}$.

Thanks for the clarification kind sir. I had another question: $∫\cos xdx=\sin x+c$. $\sin x+c$ is an anti-derivative/indefinite integral. Now, $c$ is a constant, and it could have any of the following values $c=π,13,4,5,3,−3,...$. Now, if I specify the value of $c$ ($c=3$ for example), will $\sin x+3$ still be called an indefinite integral? - tryingtobeastoic

"Antiderivative" and "indefinite integral" are not synonyms. An indefinite integral is literally an integral, and evaluating an indefinite integral leads to the set of all antiderivatives of the integrand (the set of all functions whose derivative equals the integrand). So: $∫\cos xdx$ is an indefinite integral; $\sin x+C$ is the set of all antiderivatives of $\cos x$; and $\sin x+3$ is one antiderivative of $\cos x$

Are they disagreeing? Am I misunderstanding them?

  1. (Addendum point): Eric Towers and Greg Martin don't seem to be in agreement about what a definite integral is either. According to @EricTowers(2nd comment by him),

A definite integral is a number, obtained by integrating over an interval. An indefinite integral is a function.

So, according to him, a definite integral is a plain old arithmetic number: for that, the upper and lower limit must be constants, like so:$\int_{a}^{b}{f(x)dx}$.

On the other hand, according to @GregMartin (first comment by him),

A definite integral can have constants as endpoints, in which case it results in a numerical answer, or it can have variables as endpoints, in which case it results in an answer that is a function of those variables.

So, according to Eric, the lower and upper limits must be constants ($∫_{b}^{a}f(x)dx$). However, according to Greg, the lower and upper limits can be constants or variables.

What is the more correct view in your opinion or are they both equally correct?

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  • $\begingroup$ possible duplicate of : math.stackexchange.com/questions/586107/… $\endgroup$ Jan 13, 2022 at 14:21
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    $\begingroup$ Surely you already know that wikipedia webpage of antiderative treats both ways as equivalent, but at least in spanish, which I am native, at university the terms used was Antiderivative (in spanish, equivalently is "Antiderivada"), but a translation of indefinite integral is not used, since "indefinite" in the transaltion aquire a more restrictive/strong meaning like if the integral were defined by its inhability of having a closed-form antiderivative (which is not the case of its definition)... so to avoid confussions it is not used (as far as I know) $\endgroup$
    – Joako
    Jan 15, 2022 at 1:25
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    $\begingroup$ I would like to clarify two issues about my last comment: (i) about the word "indefinite", when translated to Spanish (at least in Latin America), it is translated to the same word used for the translation of the word "undefined", keeping the meaning of the second one, this is why is not commonly used (but "indefinite integral" directly translated is used sometimes, but rarely), (ii) the use of "antiderivative" is of common use, but is not the preferred one, the most common name to refer to these functions is the "primitive function" (translated as "Primitiva"). $\endgroup$
    – Joako
    Jan 16, 2022 at 1:07

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Eric Towers and Greg Martin are in agreement with regards to what a "definite" integral is. It refers to a symbol of the form $$\int_a^bf(x)\,\mathrm{d}x.$$ This is what a mathematician would simply call an integral, or if more specifics are needed, a Riemann integral (see https://en.wikipedia.org/wiki/Riemann_integral for details) or a Lebesgue integral (see https://en.wikipedia.org/wiki/Lebesgue_integration), depending on the application. Both of these are what a calculus teacher would call a "definite integral."

Where Eric Towers and Greg Martin disagree is in the usage of the phrase "indefinite integral." Where is the disagreement? Well, there is a distinction between talking about a function $F$ such that $F'=f,$ and talking about a set of function $\{F:F'=f\}.$ Every element of this set can be written in the form $F+C,$ where $C$ is a constant: a real number, in the case of calculus. Eric Martin is saying that "antiderivative" refers to the first mathematical object. Greg Martin agrees. But Eric Towers is also saying that "indefinite integral" also refers to the first mathematical object, hence being synonymous with "antiderivative," while Greg Martin is saying that "indefinite integral" actually does not refer to the first object, but the second object: the set of functions with the aforementioned property.

This is a rather inconsequential disagreement that ultimately boils down to semantics. "Indefinite integral" is not even legitimate mathematical terminology. This is to say that, although it is terminology commonly used in the undegraduate calculus curriculum, you will never see it being used by scholarly resources published by mathematicians for mathematical research. You only see it used by pedagogical resources. It is not a terminology with any consensus-based meaning, and it is not standard, among mathematicians. Neither is "definite integral," as I already pointed out. Mathematicians do not speak in terms of definite integrals and indefinite integrals. They speak in terms of antiderivatives, and then they speak of Riemann integrals, Lebesgue integrals, or other type of integrals that are rigorously defined in the framework of real analysis and measure theory. This is all just to say that there is no commonly accepted definition for either of these labels, since they are not labels used outside of this very specific context of calculus education. From now on, if you want to be extra clear and avoid ambiguity, just speak of antiderivatives, and if needed, either specify one antiderivative, or talk about the set of antiderivatives otherwise, for maximal clarity. I would refrain from using such classroom-specific terminology if you are talking with people at the higher levels of mathematics.

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    $\begingroup$ "Eric Towers and Greg Martin are in agreement with regards to what a "definite" integral is" Unfortunately, they are not in agreement about that either. According to @EricTowers, "A definite integral is a number, obtained by integrating over an interval. An indefinite integral is a function." So, according to him, a definite integral is a plain old arithmetic number: for that, the upper and lower limit must be constants, like so:$\int_{a}^{b}{f(x)dx}$. (continued) $\endgroup$ Jan 14, 2022 at 5:24
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    $\begingroup$ ...On the other hand, according to @GregMartin, "A definite integral can have constants as endpoints, in which case it results in a numerical answer, or it can have variables as endpoints, in which case it results in an answer that is a function of those variables." So, according to Eric, the lower and upper limits must be constant ($\int_{a}^{b}{f(x)dx}$). However, according to Greg, the lower and upper limit can be constants or variables. $\endgroup$ Jan 14, 2022 at 5:25
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    $\begingroup$ @tryingtobeastoic That response involves a misapprehension of what a function is. A variable numerical parameter is not a function. A variable numerical parameter is still, by definition, just a number. The fact that it need not be constant is not relevant to this. A function refers to a set of ordered pairs. $\endgroup$
    – Angel
    Jan 14, 2022 at 12:49
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    $\begingroup$ Thanks for your comments and your answer sir! I had another question: according to GEdgar's answer, $\int_{a}^{x}f(t)dt$ is an indefinite integral, but according to GregMartin & EricTowers, it would've been a definite integral. I'm confused again. $\endgroup$ Jan 14, 2022 at 14:22
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    $\begingroup$ @tryingtobeastoic I would say so, yes. I can somewhat understand where Greg Martin is coming from, but I absolute have no clue as to where GEdgar got his claim from. $\endgroup$
    – Angel
    Jan 14, 2022 at 14:29
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People grew up with different mathematical literature. It is common that people use different words for the same notion or use the same word but mean different things. Which version to use mostly depends on one's taste and the context; there is really no one-fits-all agreement. Whenever in doubt with a notion, one should look at its definition.

Indefinite integrals vs antiderivatives

Let's take a look at two examples of references in calculus.

In Courant,

... Accordingly, we write $$ \int_a^x f(u)\;du=\Phi(x) $$ We call this function $\Phi(x)$ an indefinite integral of the function $f(x)$.

... A function $F(x)$ such that $F^{\prime}(x)=f(x)$ is called a primitive function of $f(x)$, or simply a primitive of $f(x) ;$ this terminology suggests that the function $f(x)$ arises from $F(x)$ by differentiation.

In Stewart,

... A function $F$ is called an antiderivative of $f$ on an interval $I$ if $F^{\prime}(x)=f(x)$ for all $x$ in $I$.

... the notation $\int f(x) d x$ is traditionally used for an antiderivative of $f$ and is called an indefinite integral. Thus $$ \int f(x) d x=F(x) \quad \text { means } \quad F^{\prime}(x)=f(x) $$

Obviously, these two authors have different definitions for "indefinite integral".

In Stewart, "indefinite integral" is defined as a synonym as the notion of "antiderivative" in his book; this is equivalent to Courant's "primitive", which is essentially a solution to a differential equation.

In Courant, the notion of "indefinite integral" means the map $ x\mapsto \int_a^x f(u)\;du\;. $

Definite integrals

There is no doubt for the meaning of $\int_a^b f(x)dx$ when it refers to a Riemann integral, i.e., "definite integral" in calculus. What you may wonder, as your third question shows, is that, should one call $$ \int_a^x f(u)du $$ a definite integral or "indefinite integral" in Courant's sense.

This is similar to asking whether one treats "$\cos(x)$" as a number or a function. To make it precise, one should say

for each value of $x$, $\int_a^x f(u)\;du$ is a "definite integral";

or

the map $x\mapsto \int_a^x f(u)\;du$ is an "indefinite integral" in Courant's sense.

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There is a disagreement between the two, so you are not misunderstanding them, and so I'm going to have to go with agreeing with Greg Martin on this one. When we're taught these terms, we probably forgot the context in which our teacher/professor was specifically using them, and so it all ends up becoming synonymous in how we think about these things, but there is a difference. For easy identification, an integral has the big $\int$ while the antiderivative is the solution to it. So for the expression $$\int{f(x)}dx=F(x)+C$$ we would say that the left hand side is an indefinite integral, and $F(x)$ is one possible antiderivative of $f(x)$. It's like the difference between the left side and right side of $a\times b = c$. We could call the left a multiplication expression and the right the product, but not the other way around. Same kind of difference exists with definite integrals, except instead of an antiderivative, you just have a numerical solution.

Why is this even important? Precision of language, I guess. But also the difference between the two things (an integral being the area under a curve and an antiderivative being... an antiderivative) is also why the Fundamental Theorem of Calculus is significant in the first place. If the two were synonyms by definition, then this Theorem doesn't say anything; but it does, because they are different.

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    $\begingroup$ With all due respect, you are confusing yourself. "Antiderivative" and "indefinite integral" are synonymous, and that is how most calculus teachers use the terminology. In fact, strictly speaking, "indefinite integral" is not legitimate mathematical terminology, and it is terminology that was most certainly invented by a calculus teacher, not a mathematician. Also, you said that if there was no difference between an antiderivative and an indefinite integral, then there would not be any significance to the Fundamental Theorem of Calculus. This simply not true, because... $\endgroup$
    – Angel
    Jan 11, 2022 at 20:04
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    $\begingroup$ ...the Fundamental Theorem of Calculus does not have anything to do with the distinction. The Fundamental Theorem of Calculus deals with integrals (definite integrals, and specifically, Riemann integrals) and antiderivatives, by relating the two. It does not relate antiderivatives and indefinite integrals in any way. Again, the two are synonymous. It does not help that most calculus students, and even many calculus teachers, misunderstand what the Fundamental theorem of calculus is actually saying, $\endgroup$
    – Angel
    Jan 11, 2022 at 20:07
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Another possible disagreement is that we may say $$ F(x) = \int_a^x f(t)\;dt $$ is an "indefinite integral" of $f$ as long as $f$ is an integrable function. That could happen even if $F'$ fails to exist at some points (of course $f$ is discontinuous at such points). If so, then this $F$ is not an "anti-derivative" of $f$.

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The following statement given by Greg Martin is correct.

"Antiderivative" and "indefinite integral" are not synonyms.

They are not synonyms. If we have a set containing the functions which on differentiating gives $\cos x$, then the elements of that set will be of type $(\sin x+3),(\sin x+5)\cdots$ etc. This complete set will be called indefinite integral of $\cos x$ and each of it's element will be called an antiderivative of $\cos x$, clearly this set will have infinite elements i.e. infinite antiderivatives of $\cos x$ and all will be called under a single name as indefinite integral of $\cos x$, Therefore indefinite integral denotes a set and antiderivatives are elements of this set. All antiderivatives( or elements of the set) differ by a constant and their family is called indefinite integral.

Therefore now we can easily answer your following question-

@EricTowers So, even if the constant is specified (for example c=3), sin(x)+3 will still be called an indefinite integral of cos(x)? – tryingtobeastoic

No $\sin x +3$ will not be called indefinite integral of $\cos x$ rather it will be called an antiderivative of $\cos x$. Similarly all functions of type $\sin x+C$ will be called antiderivative of $\cos x$ and their family/set is called indefinite integral of $\cos x$

In other words indefinite integral will give a family of functions while antiderivative will be the members of that family all differing just by a constant.

If you want to find an antiderivative $F_0$ of a function $f$, which is continuous, then you can use fundamental theorem of calculus.

$$F_0=\int_0^xf(a)da$$

Now keep on varying lower limit and you will keep on getting different antiderivatives. Together you can call them indefinite integral of function $f$

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  • $\begingroup$ "They are not synonyms." On what basis do you claim this? There is no established definition for the phrase "indefinite integral" in the mathematical literature. It is a colloquial phrase, one which, when used by most calculus instructors, is used as synonymous with "antiderivative." "This complete set will be called indefinite integral of $\cos(x)$ and each of its elements will be called an antiderivative of $\cos(x)$,..." Where do you get this from? No instructor or author that I know of makes this terminological distinction, mainly because it is a useless one. $\endgroup$
    – Angel
    Jan 13, 2022 at 14:07
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    $\begingroup$ @Angel I have given the link to wikipedia page $\endgroup$ Jan 13, 2022 at 14:09
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    $\begingroup$ @Angel If the terms would have been useless , no one would have continued to use them further $\endgroup$ Jan 13, 2022 at 14:10
  • $\begingroup$ "The Wikipedia page clearly mentions this" No, it does not. I read it, and this distinction you mention is nowhere mentioned in the article. In fact, it is quite the opposite: the page explicitly states "each of the infinitely many antiderivatives of a given function f may be called the "indefinite integral" of f," which implicates that "antiderivative" and "indefinite integral" are being treated as synonymous terminology. What the page does not do is make the distinction between antiderivatives as individual functions and indefinite integral as a set thereof. $\endgroup$
    – Angel
    Jan 13, 2022 at 14:11
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    $\begingroup$ @Angel I would not like to argue more, I have mentioned what I have learnt, there is no fixed consensus between mathematicians so I think neither yours nor mine is right or wrong. $\endgroup$ Jan 13, 2022 at 14:33

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