Should we avoid indefinite integrals?

Question

I am very uncomfortable with indefinite integrals, as I have a hard time giving them a precise sense that matches the way they are written and the usual meaning of other symbols.

For example, when one writes $$ \int \sin(x) \,\mathrm{d}x = -\cos(x) + k$$ then the status of both $x$ and $k$ is pretty unclear (which quantifier in front of each of these variables?)

Of course, I personally know how to translate this sequence of symbols into a proper mathematical sentence, but for students it seems utterly difficult to give a precise meaning to this, in particular at the stage when we try to explain the distinction between a function and its value at a point, or when we consider functions of several variables.

In my experience, this kind of notation tend to reinforce the student's habit to see mathematical notation as a kind of voodoo formulas that can be manipulated using certain incantations: no one probably knows what the incantation mean, but using the wrong incantation is forbidden for some reason (maybe it will summon an efreet?). On the contrary, I would like to show them the meaning behind everything we teach them.

For this reason, I try to never use indefinite integrals, relying instead on moving bounds, e.g.: $$ \forall a,x \quad \int_a^x \sin(t) \,\mathrm{d}t = -\cos(x)+\cos(a).$$

Questions: what possible issues are there in avoiding completely indefinite integrals? Is there any pedagogical advantage to using them? Is there a third way to go?

Edit: let me add another issue with the notation $$ \int \sin(x) \,\mathrm{d}x = -\cos(x) + k$$ In the right-hand side, $x$ is implicitly a variable (as opposed to the parameter $k$), but on the left-hand side it is both a global variable and a local (mute) variable of integration. Given the (already somewhat weird) role we give to the integration variable in definite integrals, this is a source of confusion that bothers me a lot. Does anyone even imagine writing something like $$ \sum_n n^3= \frac{n^2(n+1)^2}4+k?$$

Answer 1

On quizzes, homeworks, and tests, I repeatedly ask questions like this:

Find three different functions that have derivative equal to $x^2 + x$.

Forcing them to do antiderivatives and deal with the quantifier on the +C without staring at the notation helps some of them separate the +C from the voodoo magic.

I do a similar thing in college algebra classes to deal with unpleasant quantifiers:

Find three different polynomials with variable x that have roots at $x=2$ and $x=3$.

Answer 2

No, it is a bad idea to avoid indefinite integrals, the reason being simply that your students will encounter them elsewhere, and therefore need to be familiar with them. Calculus is a service course. The purpose of the course is to make science and engineering majors fluent in the language of calculus as used in their fields.

Rather than always using moving bounds, why not just tell your students that when we write $$ \int \sin(x) \,\mathrm{d}x = -\cos(x) + k$$ we're really describing a set of functions on each side of the equals sign, with an implied quantifier over $k$ on the right? In the US educational system, students are introduced to the notion of a "solution set" very early on, so this should be natural to them.

Answer 3

I go a step further than Thomas (see Henry Towsner's answer). In my view, $$ \int f(x) \ dx = \{ F(x) \ | \ F'(x)=f(x) \} $$ On a connected domain, it is true that $F'(x)=G'(x)$ implies $F(x)-G(x)=c$ hence, given an integrand which is continuous (or piecewise continuous, insert your favorite weakened set of functions here) we may write: $ \int f(x) \ dx = \{ F(x)+c \ | \ c \in \mathbb{R} \}. $ Then, I tell the students that nobody wants to write this all the time so we drop the $\{ \}$ and simply summarize it with a slogan: the indefinite integral is the most general antiderivative. This really means it is the set of all functions which form antiderivatives of the integrand. Moreover, I warn them, for this reason the usual rules of equality do not apply. In fact, $\int x \, dx = x^2/2+c$ and $\int x \, dx = x^2/2+42+c$ are the same answer.

Truth is, we are working on equivalence classes of functions as we study indefinite integration as the notion of equality has properly been replaced with congruence. Moreover, if we take function space and quotient by the subspace of constant functions then for some connected domain the indefinite integral and derivative operator are inverse operations. This I do not tell first semester calculus students, however, in a good semester of linear algebra I think it makes a nice quotient space discussion.

Obviously, the question remains, why on earth should we use the same symbol $\int$ for $\int f(x) \, dx$ and $\int_{a}^{b} f(x) \, dx$? These are radically different objects. The indefinite integral is a set of functions whereas the definite integral is a number. The answer is the FTC. That said, I think it important to make a point of emphasizing just how surprising it is that these two ideas have any connection at all.

Edit added 6/4/19 I see what Michael is saying in the comments about the error of me confusing a function with its value. The thing is, the notation $\int f(x) \, dx$ already indicates a variable for the functions in play. I'm not at peace with an answer which has $x$ on the LHS but not the RHS (say $\int f(x) \, dx = \{ F \ | \ F' = f \}$. If we are to go this route to talk about functions rather than their values then I'd probably adopt the notation $\int f$ for the indefinite integral of $f$ and keeping with the legalism of my current answer here I'd write: $$ \int f = \{ F \ | \ F' = f \} $$ I think I prefer my answer with its abuse, but I see why others would rather engage in the subtlety which Michael points towards.

Answer 4

Here's one problem with trying to replace indefinite integrals with definite ones.

With the indefinite integral, we can say

$$ \int \frac{\mathrm{d}x}{1+x^2} = \arctan x + C $$

where $C$ can be any constant. However, the only antiderivatives we can express as

$$ \int_a^x \frac{\mathrm{d}x}{1+x^2} = \arctan x - \arctan a $$

are those with $C \in [-\pi/2, \pi/2]$. (or $C \in (-\pi/2, \pi/2)$ if you don't want to allow $a = \pm \infty$)

Does anyone even imagine writing something like $$\sum_n n^3= \frac{n^2(n+1)^2}4+k?$$

Yes, actually; it is a very natural thing to do when what you're looking for is an anti-difference. Even if your problem actually called for a definite summation, keeping track of the boundary can be more complicated than just leaving things up to a constant which you solve for later by plugging in values.

Answer 5

You can emphasize definite integrals, by treating them first -- before the indefinite integrals and even before the derivative. Apostol's calculus text was famous for this.

After enough definite integrals, students may both understand and appreciate the indefinite integrals better. This order corresponds to the history also: the earliest texts we have which look like calculus are calculations of areas by Archimedes.

Answer 6

The textbook we use (a fairly standard one in the US, Thomas) is actually pretty careful about this: it says that an indefinite integral is a collection of functions, namely, all the antiderivatives. The resulting possibilities ("an antiderivative" versus "the indefinite integral") are a bit confusing for students just learning the topic, especially since "$\cos x+k$" could mean either "the indefinite integral, i.e. all the possible functions at once" or "a particular antiderivative with $k$ some constant", but I've found that Chris Cunningham's suggestion---emphasizing that this represents multiple functions by having them explicitly instantiate multiple cases---helps.

Answer 7

I think it's funny that you/your students have a problem with the indefinite integral but don't mind the mysterious placeholder $dx$ at the end of a definite integral.

In my Calculus course, I take a differentials approach (see here), so in that sense $\int$ is the inverse operator of $d$. However, since $d$ is a many-to-one operation, the $+k$ is required in the output of $\int$ in the same way that $\pm$ is required when undoing a square. When I first introduce indefinite integrals (which is after definite integrals which motivate the notational symbols), I'll do an easy example like $$\int 3x^2\ dx = \int d(x^3) = x^3+k$$ (which most of my students get pretty quickly, especially when compared to $\sqrt{9}=\sqrt{(\pm 3)^2} = \pm 3$). I'll also do a more advanced example: $$\int \frac{x}{x^2+3}\ dx = \int \frac{\frac{1}{2}d(x^2)}{x^2+3} = \frac{1}{2}\int\frac{d(x^2+3)}{x^2+3} = \frac{1}{2}\int d(\ln(x^2+3)) = \frac{1}{2}\ln(x^2+3)+k$$ (which all of them need to think about before understanding).

Anyway, my point is treat $\int$ as the inverse operator of $d$ and draw upon the analogy of square roots to explain the $+k$.

Answer 8

Teach your students that $$ \int \sin x \,\mathrm{d}x = -\cos x + k$$ is simply (very convenient) shorthand for this precise but long-winded statement:

Suppose the function $f$ is defined on an interval and has an antiderivative.

If $f$ is defined by $f(x)=\sin x$, then its antiderivatives are exactly those functions $g$ defined by $g(x)=\cos x$.

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• The above statement is not "utterly difficult", is fairly precise, and can be understood by even average high school/introductory calculus students.

• Repeat the above statement (in speech, examples, exercises, test questions) as often as is judged necessary.

Answer 9

Have you considered the following notation: $$ \quad \int^x \sin(t) \,\mathrm{d}t = -\cos(x)+k$$ emphazing that the integral is a function of $x$ ?

Abbreviations for integration and derivation can then be written in shorthand with $\quad \int^x$ and $d/dx$

Answer 10

Indefinite integrals are easy to avoid, it only takes a will to do so.

Basic antiderivatives: $$ x^2 =\frac{d}{dx}\frac{x^3}{3},\quad \sin x =\frac{d}{dx}(-\cos x),\quad \frac{1}{x} =\frac{d}{dx}\ln|x|. $$

Antidifferentiation by change of variable: $$ e^x\sin(e^x) =\sin(e^x)\frac{d}{dx}e^x =\frac{d}{dx}\sin(e^x), $$

Antidifferentiation by parts: $$ x e^x = x \frac{d}{dx}e^x = x \frac{d}{dx}e^x + (\frac{d}{dx} x)e^x - (\frac{d}{dx} x)e^x\\ =\frac{d}{dx}(x e^x) - (\frac{d}{dx} x)e^x = \frac{d}{dx}(x e^x) - e^x\\ =\frac{d}{dx}(x e^x) -\frac{d}{dx}e^x = \frac{d}{dx}(x e^x - e^x). $$

Using differential forms, equivalent calculations can be carried out with simpler notation: $$ x^2\,dx = d\frac{x^3}{3},\quad \sin x\,dx = d(-\cos x),\quad \frac{dx}{x} =d(\ln|x|), $$ $$ e^x\sin(e^x)\,dx =\sin(e^x)d e^x =d\sin(e^x), $$ $$ x e^x\,dx = x d e^x = x d e^x + (dx)e^x - (dx)e^x\\ = d x e^x - (dx)e^x = d x e^x - e^x\,dx\\ = d x e^x - de^x = d(x e^x - e^x). $$

When I teach definite integrals (or try to teach indefinite integrals despite being fuzzy myself about their exact meaning), I prefer to carry out calculations with differential forms first and apply definite (or indefinite) integral at the end.

Answer 11

After having looked into the history of calculus and into original works by Leibniz and others, I have come to the conclusion that the schizophrenic way the "indefinite integral" is sometimes taught is a byproduct of the common schizophrenic mixture of Leibniz-time notation with Bourbaki-time foundations of calculus. I've found a satisfactory way to deal with the former, but not yet with the latter (I am working on it).

When trying to make sense of the traditional Leibniz notation in calculus, it is important to understand that even the notion of function in the 18th century was not what it is now. Now a function is something along the lines of "a set of pairs," "a functional relation," "a triple of a functional relation, a domain, and a co-domain," "a morphism in the category of sets." In the 18th century, to be a function was a property/predicate on variables: one variable could be a function of another, and also two variables could be functions of each other.

According to what I have understood, the notation of Leibniz, as used in the 18th century, heavily relied on the context. Expressions, like integrals, were not self-contained. In fact, even integration bounds were not written below and above "$\int$." Apparently, the first well-known text where the bounds of integration appeared at the integration symbol was Théorie analytique de la chaleur (1822) by Joseph Fourier -- see The history of notations of the Calculus (1923) by Florian Cajori.

I long believed that in the expression "$\int x^2dx$," the variable "$x$" is bound (is not free) because of "$dx$," but this is not at all the idea. One can have "$\int ydx$," "$\int xdy$," "$\int(xdy + ydx)$," and how this is to be integrates should be specified separately. For instance: $$ \int_{x=0}^1 x^2dx =\frac{1}{3},\qquad \int_{y=0}^1 x^2dx = 0, $$ where in the second integral the variable "$x$" represents a fixed quantity (fixed in the context of this integral), while $y$ varies from $0$ to $1$. (This is not how this would have been written in the 18th century, because bounds were not written by the integration symbol, but this is the idea.)

Now, I return to the question of the meaning of the "indefinite integral."

If we simply want to write identities like $$ \int (f(x) + g(x))dx =\int f(x)dx +\int g(x)dx $$ without worrying about the bounds, like this would have been done in the 18th century, it suffices to interpret the "indefinite integrals" that appear here as indeterminate definite integrals -- definite integrals with indeterminate bounds. The meaning of this identity then is that for all $a$ et $b$ such that both sides are defined, we have: $$ \int_{x=a}^b (f(x) + g(x))dx =\int_{x=a}^b f(x)dx +\int_{x=a}^b g(x)dx. $$ In modern terms, this amounts to an identity of functions of two variables, which take $a$ et $b$ and return the value obtained by integrations from $a$ to $b$. I believed having already seen such a definition of indefinite integral given by Paul Halmos, but I have just tried to find it, and I have no idea where I could have seen it. Perhaps I deduced it myself from the definition of indefinite Lebesgue integral given by Halmos (see below).

Here is the definition of an indefinite integral that I plan to use in my teaching from now on: $\int_x f(x)dx$, written also as "$\int f(x)dx$" when it is clear from the context that the integration is with respect to $x$, is the function that takes a pair of reals $(a, b)$ and returns the value $\int_{x=a}^b f(x)dx$. We adopt the convention to not write expressions like "$(\int_x f(x)dx)(0, 1)$," but to write instead "$\int_{x=0}^1 f(x)dx$."

In addition to this definition of an indefinite integral, I plan to use the following definition of an indefinite difference: $[f(x)]_x$, written also as "$[f(x)]$" when there should be no ambiguity, is the function that takes a pair of reals $(a, b)$ and returns the value $[f(x)]_{x=a}^b = f(b) - f(a)$. We adopt the convention to not write expressions like "$([f(x)]_x)(0, 1)$," but to write instead "$[f(x)]_{x=0}^1$."

We have the following obvious property: $$ [f(x)] = [g(x)]\quad\Leftrightarrow\quad [f(x) - g(x)] = [0]\quad\Leftrightarrow\quad f - g\ \ \text{is constant}. $$

The second fundamental theorem of calculus can now be written as: $$ \int f'(x)dx = [f(x)]. $$

This interpretation is compatible with and can be applied to "non-oriented" integrals $\int f(x)|dx|$, Stieltjes integrals $\int f(x)dg(x)$, "non-oriented" Stieltjes integrals $\int f(x)|dg(x)|$, etc.

This interpretation also agrees with my recent idea to view measure-theoretic integration as a binary operation that takes a signed measure and a measurable function and returns a signed measure. Just now I've re-discovered that this is exactly the definition of indefinite Lebesgue integral given by Paul Halmos in his Measure theory. Thus, the indefinite Lebesgue integral $\int fd\mu$ is the signed measure that takes a measurable set $X$ and returns the number $\int_X fd\mu$.

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I am adding some examples of calculations:

$$ \int_t te^t dt =\int_t tde^t = [te^t]_t -\int_t e^tdt = [te^t]_t -[e^t]_t = [te^t - e^t]_t. $$

\begin{multline*} \int_x\frac{dx}{1 - x^2} =\int_x\frac{1}{2}\left(\frac{dx}{1 - x} +\frac{dx}{1 + x}\right) =\frac{1}{2}\left(\int_x\frac{dx}{x + 1} -\int_x\frac{dx}{x - 1} \right)\\ =\frac{1}{2}([\ln |x + 1|]_x -[\ln |x - 1|]_x) =\frac{1}{2}[\ln |x + 1| -\ln |x - 1|]_x =\left[\frac{1}{2}\ln\left|\frac{x + 1}{x - 1}\right|\right]_x. \end{multline*}

In the second example, it is by default understood that the obtained identity is equivalent to $$ \int_{x=a}^b\frac{dx}{1 - x^2} =\left[\frac{1}{2}\ln\left|\frac{x + 1}{x - 1}\right|\right]_{x=a}^b $$ for all $a,b\in\mathbf{R}$ such that the integral exists, which excludes the possibility that $a$ and $b$ be separated by $1$ or $-1$.

Answer 12

when one writes $$ \int \sin(x) \,\mathrm{d}x = -\cos(x) + k$$ then the status of both $x$ and $k$ is pretty unclear (which quantifier in front of each of these variables?)

1. Yes, when an indefinite integral is specified like in the above (that is, as a general representation or as a function of the form $-\cos(x)+k,$ where $k$ is arbitrary), it doesn't really make sense to quantify $k,$ unless initial/boundary conditions are given, in which case we can assert that $$\exists k{\in}\mathbb R\;\forall x{\in}\mathbb R \int \sin(x) \,\mathrm{d}x = -\cos(x) + k.$$

2. On the other hand, when the above indefinite integral is undersood as a literal set of antiderivatives, then:

   for each real $x,$ the indefinite integral of $\sin x$ is the set of functions such that each one, for some real $k$, equals $-\cos(x)+k.$

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Appendix (excerpt from How to interpret the indefinite integral)

An indefinite integral is more a useful notational shorthand than a mathematically important object. It is conventionally called a family of antiderivatives; as such, if its integrand is of the form $g'$ and has an interval domain, then, literally, $$\int g'\,\mathrm dx=\{g(x)+C\mid C\in\mathbb R\},$$ where $C$ represents uncountably many values. How to understand the "indefinite integral" notation in calculus? contains several justifications for manipulating and adding indefinite integrals as actual sets by reading $\text‘=\text’$ as an equivalence relation such that writing \begin{align}x^2+C=x^2+3+C\quad&\implies0=3\\x^2+C=x^2+3+D\quad&\implies E=C-D=3\end{align} make sense.

More simply, we can frame an indefinite integral as the general representation of its integrand's antiderivatives, its specification containing one independent parameter (arbitrary constant) $C_i$ per maximal interval of its integrand's domain. Here, each instantiation of $$\int g'\,\mathrm dx=g(x)+C$$ has an unimportant value of $C.$ Even though this object is not a particular antiderivative, we manipulate it as if $C$ is merely undetermined.

In the context of solving differential equations with given conditions, $C$ becomes an unknown whose value is to be determined.

Answer 13

Besides the issue about the constant, there is one very particular reason why indefinite integrals should not be removed from undergraduate education:

Let's be honest: from the first year in primary school to the last day of secondary school before indefinite integrals, every mathematics class is the same: the teacher teaches a method, and when the teacher gives exercises the students must apply that method in order to solve the exercise, which means that there is a one-to-one relationship between the theory and the exercises.

This is not longer true when dealing with indefinite integrals: in order to solve those, you need to know all about basic polynomials (first and second year), trigonometric and rational functions (third and fourth year and further), you must have a perfect understanding of derivatives (fifth year), ... next to the techniques, proper to integral calculation (like partial integration).

This is nothing else than "thinking outside-the-box", which is so extremely important in our current complex economy.

Let me just give you an example from my workplace (software engineering):
A computer program is a piece of text, written by and readable for humans, but complete nonsense for computers: computers don't understand variables, functions, ..., they only care about data, being moved around at memory addresses.
In order to transform a human-readable computer program into something a computer understands (called machine language), a so-called "compilation" must be done.
This leads to the general idea that all "compilation-like" results are not human-readable.
In my workplace, there was a distinction between functionality and the recipe: the functionality was compiled (translated into machine language) and next to that a textfile was generated, containing the recipe in which those compiled files were to be used.
One day I was working at quality assurance, and I started investigating that textfile, and I was discovering lots of bugs, heavily improving the quality of the program. Hereby the reaction of my boss:

Boss: "What are you doing?"
Me: "I'm reading the recipe."
Boss: "WHAT are you doing??? It's generated by the compiler, and everything, generated by the compiler, is not human-readable."
Me: "That is true for the pure compilation, but this is the recipe, which is put in a flat textfile, and textfiles are readable, so I'm reading that."
Boss: (talking in an ironic tone) "... and what are you thinking of retrieving from that textfile?"
Me: "Here's the list of bugs I've found already."
Boss: "??? Please continue, you're doing great!!!"

For people who take this for granted: that firm existed already for more than 10 years and there were more than 120 people working there, and yet I was the first thinking about reading that file for quality reasons.

I am convinced that my "thinking-outside-the-box" mentality, for which I've known, is very related to my ability and eagerness to solve indefinite integrals.