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

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 9

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 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.