43
$\begingroup$

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 $$ \sum_n n^3= \frac{n^2(n+1)^2}4+k?$$

|improve this question
$\endgroup$
  • 12
    $\begingroup$ Just explain that it is a weird notation for the antiderivative, that the class might want to change, but you won't be able to go against the mathematical establishment $\endgroup$ – vonbrand May 18 '14 at 13:28
  • 19
    $\begingroup$ Additionally, $+ C$ is not always sufficient. For example, the general antiderivative of $1/x$ is $$\begin{cases} \ln(x) + C_1 & \text{when $x > 0$,} \\ \ln(-x) + C_2 & \text{when $x < 0$.} \end{cases}$$ $\endgroup$ – François G. Dorais May 18 '14 at 14:36
  • 1
    $\begingroup$ Even if one takes the trouble to develop a self-consistent, coherent, optimized notational and conceptual system, there is no enforcement mechanism (well, ...) to make people behave sensibly in this or any other way. In particular, I guess we find ourselves needing to teach people how to cope with ambiguity or amorphousness, rather than telling them that there are absolutely-reliable universal conventions that will never change, etc. We even find ourselves forced (!) to deal with self-inconsistent, misleading, annoying conventions and "definitions". Dang. Hm. $\endgroup$ – paul garrett Apr 14 '16 at 23:23
  • 4
    $\begingroup$ Long ago, as a teaching assistant, I taught with a professor (a specialist in PDEs and harmonic analysis) who insisted that all integrals be written with limits. We were teaching a calculus course directed at students in biological science ("pre-meds"). He found that this approach, carefully enunciated, helped these students understand the ambiguities (notice the ambiguities don't vanish - rather they are absorbed into the sometimes indeterminate choice of limits). My sense was that the approach was quite successful. Other approaches can be successful too. $\endgroup$ – Dan Fox Oct 28 '18 at 17:50
  • $\begingroup$ @FrancoisG.Dorais Technically you can still write $+C$; is just that in general $C$ is only guaranteed to be locally constant, not globally constant. $\endgroup$ – J.G. Nov 9 '18 at 17:21
37
$\begingroup$

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 $x=3$.

|improve this answer
$\endgroup$
  • 7
    $\begingroup$ Also I find it helpful to do analogous problems completely graphically: sketch 2 different functions which have derivative $f$, where $f$ is graphed below. $\endgroup$ – Steven Gubkin May 18 '14 at 21:22
  • 3
    $\begingroup$ A trick I've always liked: when teaching the antiderivative, tell student's it's "C, plus the other stuff." That is, make "C +" the first thing you do, rather than something you can forget to tack onto the end. (Obviously caveats for antiderivative of 1/x and such) $\endgroup$ – bbayles May 19 '14 at 2:57
  • 3
    $\begingroup$ I like your idea "Find three different functions ...", which I don't believe I ever tried (or even had an awareness of trying). As for what I did do, see this 5 December 2008 AP-Calculus post at Math Forum. (If I get less busy later, I might LaTex the essence of it here unless someone else posts the same idea.) $\endgroup$ – Dave L Renfro May 19 '14 at 15:51
27
$\begingroup$

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.

|improve this answer
$\endgroup$
  • 3
    $\begingroup$ Your answer makes sense, but the concept of solution set is commonly a serious issue with first year students in France (at least in my experience). $\endgroup$ – Benoît Kloeckner May 18 '14 at 19:06
20
$\begingroup$

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.

|improve this answer
$\endgroup$
  • 3
    $\begingroup$ Since in first years I try to enforce the rule of distinguishing the function $f$ from its value $f(x)$ at $x$, even considering $x^2/2+c$ as a set of function bothers me. There is almost no issue at all when one masters these notions, but I do not see how to avoid deep confusions for already struggling students. $\endgroup$ – Benoît Kloeckner May 18 '14 at 20:47
  • 5
    $\begingroup$ @BenoîtKloeckner I share your frustration with getting students to distinguish $f$ and $f(x)$. But, I suspect their inability to get the concepts has much more to do with their reluctance to see math as more than problem solving. I can't fault them, they've been fed a steady diet of canned problems replete with their algorithmic solution mantras. It's a shock when they find that math is a story and you can choose your own adventure. $\endgroup$ – James S. Cook May 19 '14 at 5:01
  • 3
    $\begingroup$ Many teachers believe it's the students fault if they can't distinguish $f$ from $f(x)$. That's a good point of view, which we should apply to every problem in teaching. But consider your suggestion $\int f(x)dx =\{F(x)\mid F'(x)=f(x)\}$: the RHS is probably $\mathbb{R}$, so maybe you meant $\int f(x)dx =\{F \mid F'=f \}$? Ok, now lets drop $\{\}$ and say $=$ denotes congruence. Then we should still correctly write $\int x dx = (x\mapsto \frac{x^2}{2})$. Now try doing any concrete computation, like integr. by subst. or parts with this notation, without confusing $f$ for $f(x)$. $\endgroup$ – Michael Bächtold May 28 '18 at 8:35
  • $\begingroup$ Hey James, can we talk? $\endgroup$ – I am Back May 31 at 19:54
7
$\begingroup$

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.

|improve this answer
$\endgroup$
  • $\begingroup$ Hughes-Hallett applied (which is probably the opposite side of Apostol in terms of difficulty/rigor!) also does definite integrals before antiderivatives, and probably some other "reform" texts do too. $\endgroup$ – kcrisman Oct 24 '15 at 20:58
6
$\begingroup$

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.

|improve this answer
$\endgroup$
6
$\begingroup$

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.

|improve this answer
$\endgroup$
5
$\begingroup$

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 $+k$.

|improve this answer
$\endgroup$
  • 6
    $\begingroup$ But $\sqrt{\cdot}$ is usually defined as the operator giving the positive square root of its argument, so as to make it single-valued. This makes it quite a bad analogy doesn't it? As for $\mathrm{d}x$, I agree that it is somewhat mysterious, but it is simply a weird notation. It does not conflict too violently with other notation. $\endgroup$ – Benoît Kloeckner May 19 '14 at 18:24
  • 2
    $\begingroup$ Do you try to formalize the notion of $\int$ and $d$ as operators, e.g., specifying their ranges and domains? This seems a little thorny to me, since the notation was designed to work on variables and expressions, not functions. Say $a$ is the area of a circle. Then $da$ historically meant just an infinitesimally small change in $a$. There's no implication that $a$ is a function of anything. It could be a function of the radius of the circle, or it could be a function of time if the circle is growing. Stuff like $dv=dx\:dy\:dz$ also doesn't seem to fit the operator paradigm well. $\endgroup$ – Ben Crowell May 19 '14 at 22:03
  • $\begingroup$ @BenCrowell: Definitely not for the students. I have a way of thinking about it for myself (d maps functions to a vector space over functions with basis elements differentials). This is good enough to keep me from saying outrageously wrong things, but it's probably not entirely mathematically sound. $\endgroup$ – Aeryk May 20 '14 at 11:24
  • 1
    $\begingroup$ @Ben: One can model this as variables being scalar fields on some implicit domain and $d$ is the exterior derivative (defined only for sufficiently smooth scalars). If you work exclusively with scalars and differentials (and equations involving them), then the domain itself never actually needs to make an appearance in the formalism, although you can imagine it being behind the scenes as encoding the domain of variation and the prior relations between the variables. $\endgroup$ – Hurkyl Oct 22 '15 at 22:20
3
$\begingroup$

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 $d/dx$

|improve this answer
$\endgroup$
  • 3
    $\begingroup$ A problem with this approach is that it doesn't make $\int^x$ and $d/dx$ into inverses of each other: you don't get $\sin(t)$ out of $\frac{d}{dx}( \cos(x) + k )$. $\endgroup$ – Ilmari Karonen May 19 '14 at 18:32
  • 13
    $\begingroup$ Teaching students a nonstandard notation like this is a bad idea IMO. They need to be able to cope with how calculus is notated in the real world. $\endgroup$ – Ben Crowell May 19 '14 at 20:00
3
$\begingroup$

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

"Given a function with the mapping $x\mapsto \sin x$, its antiderivatives are exactly those functions whose mappings are $x \mapsto -\cos x + k$, for some real number $k$."

  • The above statement is not "utterly difficult", is fairly precise, and can be understood by even average high school/introductory calculus students (provided the teacher has already thoroughly and adequately schooled them in the idea of functions as mappings and not as mere formulae).

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

|improve this answer
$\endgroup$
  • 1
    $\begingroup$ I know that, as I stated "I personally know how to translate this sequence of symbols into a proper mathematical sentence"; my point is pedagogical. If you use the shorthand again and again, and do not use the long-winded statement that often, then the long-winded statement cannot be expected to stay in student's minds. $\endgroup$ – Benoît Kloeckner Oct 22 '18 at 20:09
  • 1
    $\begingroup$ Well, then your question is similar to that of how we might teach 8-year-olds to remember how to spell a long and difficult word. There are different views on this, but how I'd do this is by (spaced) repetition, examples, exercises, homework, and questions on quizzes/tests. (The practice of most mathematicians would be to simply state the above long-winded statement at most once -- probably because that's all they themselves need to remember it. But good teachers know when and how much repetition is necessary.) $\endgroup$ – dtcm840 Oct 22 '18 at 23:12
  • $\begingroup$ You stated that "for students it seems utterly difficult to give a precise meaning to this". I don't think the above long-winded statement is "utterly difficult". And if taught, I think most students at the high school/introductory university level will be able to understand it and remember it. $\endgroup$ – dtcm840 Oct 22 '18 at 23:18
  • 1
    $\begingroup$ @dtcm840 do you treat functions and mappings as different kinds of objects? If so, what do you tell your students is the difference between them? $\endgroup$ – Michael Bächtold Dec 7 '18 at 8:51
  • $\begingroup$ @MichaelBächtold: Formally, a function $f$ from $A$ to $B$ is a subset of the cartesian product $A \times B$ (with the property that if $x\in A$, then $\exists !y\in B$ such that $(x,y) \in f$). But informally and at a high school level, I teach that a function consists of three "components", namely the domain $A$, the codomain $B$, and the mapping. And so informally, the mapping is the "rule" that tells us how to assign each element in the domain $A$ to an object in the codomain $B$. To specify a function, one must specify all three "components", of which the mapping (rule) is only one. $\endgroup$ – dtcm840 Dec 8 '18 at 2:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.