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I teach the definite integral before the indefinite for a few reasons, one of which is that I want students to recognize that the definite integral means area (not anti-derivative). If we do indefinite integrals before the Fundamental Theorem of Calculus, the students are likely to think the FTC is not a big deal. ("Yep, just find the anti-derivative, that's what we already learned to do when we see this tall skinny S.")

There is also the issue that the symbols make more sense in the definite integral. $$\int_1^2 x^2\, dx$$

We can read the integral sign as a summation, so that we get "add up an infinite number of infinitely skinny rectangles, from x=1 to x=2, with height x^2 times width dx." Each part of the symbol makes sense.

But with indefinite integrals, the tall skinny S does not mean summation, and the dx has no real meaning. I think it's a very unfortunate symbol choice. [See Should we avoid indefinite integrals? for more on this.] But, since we're stuck with it, I'd like to know if there is any explanation one can give for the symbols. I tell my students that the S part means "find the anti-derivative", and the dx means "and x is the variable". Is there a better way to describe the meaning of the dx (and maybe even the S) here?

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  • $\begingroup$ The way things are usually done $\mathrm dx$ has no meaning. If it was me being taught that, I'd either be very confused or call you out on this. $\endgroup$ – Git Gud Nov 24 '14 at 15:22
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    $\begingroup$ If we think of it as an infinitely small change, it does have meaning. And if it had no meaning, why would it be there? I don't know the history well enough to guarantee that this is why the notation developed this way, but it makes sense. I also don't know enough about the infinitesimal approach to guarantee that it's what makes this make sense, but it seems to. Why would you be confused, Git? $\endgroup$ – Sue VanHattum Nov 24 '14 at 15:34
  • $\begingroup$ But the usual treatment is not the infinitesimal approach. I know nothing about it, but I assume it makes perfect sense there. Using $\varepsilon$-$\delta$, which is how I learned about these things, $\displaystyle \int _a^bf(x)\,\mathrm dx$ simply denotes a certain limit (which has nothing to do with the variable $x$, in fact I prefer the notation $\displaystyle \int _a^bf$). Despite the fact that the use of the variable might help communication, it serves no mathematical purpose. $\endgroup$ – Git Gud Nov 24 '14 at 15:43
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    $\begingroup$ My students aren't ready for formalizing this. They need to start with conceptualizing it. If you don't think of dx as an infinitely small change in x, how do you understand the volume integrals (volumes for rotations around an axis of y=f(x))? $\endgroup$ – Sue VanHattum Nov 24 '14 at 20:52
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    $\begingroup$ "They need to start with conceptualizing it". I don't disagree, I even said before that I get the idea of infinitesimal. But I maintain that $\mathrm dx$ doesn't have a meaning in this context ($\varepsilon$-$\delta$). I think it's possible we understand the word 'meaning' in different ways. It seems to me you understand it (or are using it here like that) as 'the intuition/conceptual idea behind the thing/notation' where as I can only attribute meaning (in my sense of the word) to objects which actually exist within the given framework. $\endgroup$ – Git Gud Nov 24 '14 at 21:23
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It sounds like what you want to do is:

1) Do definite integrals first using $\int_a^bf(x)\ dx$.

2) Then do antiderivatives using Newtonian notation: use capital letters to represent antiderivatives. So for instance, if $f(x)=2x$, then $F(x)=x^2+C$ so that $F'(x)=f(x)$.

3) Do the Fundamental Theorem of Calculus.

4) Since the two ideas are related, now introduce the indefinite $\int f(x)\ dx$ as another way of saying antiderivative (which, by the way, is Leibniz's notation).

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    $\begingroup$ I'd rather do #3 before #2. But that's not my question. My question is, when doing #4, what does dx mean? $\endgroup$ – Sue VanHattum Nov 21 '14 at 17:02
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    $\begingroup$ Asking that question probably opens a rabbit-hole few are willing to go down. Leibniz worked with differentials, infinitely small quantities. To him, roughly, $df/dx$ really was $df$ divided by $dx$. So, for example, $d(x^2)=2xdx$. Then $\int$ was the thing that undid a $d$. To Leibniz, the FTC is trivial because by definition $\int df = f + C$ (the $+C$ coming from the fact that $d$ is not a 1-to-1 function, so the inverse takes on multiple values, like $\pm$ in roots). So, $\int \frac{df}{dx}dx = \int df = f+C$. Of course, he chose the symbol $\int$ because he understood FTC as a sum, too. $\endgroup$ – Aeryk Nov 21 '14 at 18:10
  • $\begingroup$ You might want to check out: matheducators.stackexchange.com/questions/12/… and matheducators.stackexchange.com/questions/2246/… $\endgroup$ – Aeryk Nov 21 '14 at 18:11
  • $\begingroup$ Why not start with Riemann sums and build to definite integrals that way? That will ground the whole thing in the idea of "area" $\endgroup$ – shadowtalker Nov 22 '14 at 11:08
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    $\begingroup$ In addition to what Aeryk said in this answer, another thing to realize about the dx is that without it, the units are wrong. For instance, suppose we integrate velocity to get position, $x=\int v dt$. If you think of the indefinite integral as a kind of sum (not just as an antiderivative), then the units of the sum should be the same as the units of each term. If you multiply $v$ by $dt$, you get units of distance, which makes sense. Of course, thinking of the indefinite integral as a sum will only make sense to your students once they understand the fundamental theorem. $\endgroup$ – Ben Crowell Nov 26 '14 at 0:55
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If I were teaching the indefinite integral to students who already understand the definite integral and are comfortable with the usual notation for it, then I would introduce the indefinite integral as a definite integral with a variable upper limit, i.e., $$ F(t)=\int_a^tf(x)\,dx, $$ where $a$ is a constant. (Different choices for $a$ give different "constants of integration".) So far, the concept and the notation should look pretty reasonable. But then questions arise about the usual notation, which looks like $$ \int f(x)\,dx=F(x)+C. $$ Why do we omit the limits of integration ($a$ and $t$ above), allowing the role of $a$ to be taken by the $C$ on the right side of the equation, and replacing $t$ as the argument of $F$ with the variable of integration $x$? The use of $C$ can probably be explained thanks to the fundamental theorem of calculus. The integrals with various $a$'s as lower limits are some of the antiderivatives of $f$; for some purposes we'll want all the antiderivatives of $f$, and the formulation with $C$ provides these. As for the use of $x$ in place of $t$, I won't try to defend a convention that notationally identifies a bound variable (in the integral) with a free variable (in $F(x)$). In such situations, I usually tell the class something like: "I'm not responsible for this notation. It's so ancient that it was fixed even before I was born. So I never had a chance to correct it."

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The $dx$ means "with respect to $x$".

Suppose water flows over a waterfall at a rate of $(3+\sin(t))x$, where $x$ is the distance from the riverbank and $t$ is the time of year.

Do you want to know how much is flowing over the falls at once? Or how much flows over one rock over a period of time? You would set up different integrals:

$$\int (3+\sin(t))\ x\ dx = (3+\sin(t))\ x^2/\,2 + C$$ $$\int (3+\sin(t))\ x\ dt = (3t-\cos(t))\ x + C\ \ \ $$

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    $\begingroup$ This doesn't address the question. The OP is giving her students a specific interpretation of the notation for a definite integral, and is asking about how to connect that to the notation for the indefinite integral. $\endgroup$ – Ben Crowell Nov 26 '14 at 0:41
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I think that a first introduction to integration should demonstrate what it can be used for in the real world, motivating it by asking how we can find the distance traveled by a train whose velocity changes over time. First we of course have to define velocity, which is the rate of change of position at the point in time, measured by taking smaller and smaller intervals of time around the point. One natural way of expressing that would end up being the derivative of position with respect to time.

Then we employ the intuition that if a train has higher velocity than another at every point in time, it would travel at least as far in the positive direction. So we can approximate the real displacement by comparing with trains that have constant velocity over each short time interval. For each of those intervals we can see that the displacement is clearly the velocity multiplied by the time elapsed, almost by definition of velocity. Subsequently we can see that if the train's velocity is sufficiently nice we can obtain good upper and lower approximations by choosing sufficiently thin intervals. Hence we get the Riemann sum.

Using this intuition, all the symbols make sense, "$\int$" being an elongated "s" for sum, "$dt$" being the size of the thin time intervals, and the bounds of the definite integral being the endpoints of the time period under discussion. Also, it is worth pointing out that using these rectangular approximations is equivalent to finding the area under the graph.

The next step is to explicitly calculate some simple examples to get a feel for what the result is like. One should then compute the displacement in terms of the time period. It is also easy to observe that like the intuition says, the displacement is additive in the time period, and so this as well as the explicitly computed examples should suggest that we should compute just the integral with a fixed lower bound, since we can obtain the general case just by a difference. This general technique of reducing the parameters needed is very useful in both mathematics and computer science.

So far we have built everything without mentioning anti-derivatives, and indeed we should not just yet. First note that the integral with a fixed lower bound can be used in all cases where the integrand is integrable to compute the integral with arbitrary bounds, even if the integrand is non-continuous. In mechanics, even though everything is quite continuous until down to the quantum level, in practice it makes sense to simplify the model of say a train that leaves a station with velocity being non-continuous. (And in other cases, acceleration. Or jerk. Or...)

Note also that the actual value of the fixed lower bound does not matter in obtaining the definite integral because the difference eliminates its effect, much as in the real world the point of reference that is used to measure displacement does not matter. Plotting the graphs obtained by using different integral lower bounds should make it clear to students that they are all the same except for a constant difference across all time, and that the constant difference corresponds to the displacement between the two chosen lower bounds.

Then the symbols "$\int f(t)\ dt$" can be said to represent simply the class of functions "$x \mapsto \int_c^x f(t)\ dt$" for all possible $c$. I find it misleading to write "$\int f(t)\ dt = \cdots + C$" because it is actually a class, not one single function. In any case, this is what the symbols should be explained to mean, and not the anti-derivative, since it may not exist. This class of functions is called the "indefinite integral", precisely because it can be used to compute arbitrary definite integrals.

But then now it is very natural to ask if there is a way to efficiently compute the indefinite integral directly without going through the approximation process. That is of course easy for certain special integrands, namely those which are continuous derivatives of some other function. The reason is that one can notice that the indefinite integral grows at a rate that is proportional to the value of the integrand at that point quite intuitively, since the thin rectangle at that point has exactly that height. A formal analysis would give the fundamental theorem of calculus, as well as the desired result that we can use anti-derivatives in the case the integrand is the continuous derivative of some other function. An integral may need to be split into separate regions before anti-derivatives can be used, and certain badly behaved cases should not be overlooked (the derivative of Volterra's function). These issues will be avoided if the indefinite integral is not confused with the anti-derivative.

Of course all these can and should be made rigorous via careful proofs at some point in time, but the above should suffice for an intuition explanation of all the basic notions. The very same ideas generalize naturally to Lebesgue integration using simple functions to approximate non-negative measurable functions, or using Lebesgue measure in place of Jordan area.

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    $\begingroup$ @Sue VanHattum: By the way, I certainly do not agree on teaching students that "$\int$" means find the anti-derivative. Not only is it factually wrong, it is also simply incorrect as I stated in my answer. For example $\displaystyle \int \left\lfloor \frac{1}{\sqrt{|x|}} \right\rfloor\ dx$ is a perfectly well-defined class of functions that are not anti-derivatives in any meaningful sense. $\endgroup$ – user21820 Feb 5 '15 at 13:15
  • $\begingroup$ I like the way you think, but mathworld.wolfram.com/IndefiniteIntegral.html and everything else I see online disagrees on this point. $\endgroup$ – Sue VanHattum Feb 6 '15 at 13:57
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    $\begingroup$ @SueVanHattum: That's sad. This way naturally arose to me and I'm surprised people don't seem to think so. In that case my comment was also not in line with the standard interpretation of the indefinite integral. I hope that whatever I've said is nevertheless useful in motivating everything about the integral and anti-derivatives to a student. =) $\endgroup$ – user21820 Feb 6 '15 at 16:46
  • $\begingroup$ Until the indefinite integrals, I do a lot of what you describe. $\endgroup$ – Sue VanHattum Feb 6 '15 at 18:29
  • $\begingroup$ Reading your answer again almost two years later, your interpretation explains something that seemed crazy in the textbook I use (Anton, Bivens, Davis, Early Transcendentals, 10th ed). On page 367, they claim "all continuous functions have anti-derivatives." This is not true if anti-derivative means a finite combination of elementary functions. I now see that they must mean something like an area function, but indefinite. (They are not defining things the way you are, but your explanation helped me see how they were thinking.) $\endgroup$ – Sue VanHattum Jan 2 '17 at 22:54
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In a Riemann sum, we add up a bunch of skinny rectangles to get an area. For convenience, we split up the domain into equal parts of length $\Delta x$. The height of each rectangle is determined by evaluating the function $f$ at $x_i$. Rectangle $i$ therefore contributes area $f(x_i)\Delta x_i$, which gives total area $$ \sum_{i = 1}^n f(x_i)\Delta x. $$ The units of $f(x_i)$ and $\Delta x$ are both (say) meters, so the units of $f(x_i) \Delta x$ are square meters (which makes sense, considering it is supposed to be an area).

When we want to pass from adding up finitely-many rectangles to adding up infinitely-many rectangles, we use a special notation $$ \int_a^b f(x) dx. $$ The summation ceases to be about a finite-number of rectangles, but rather instructs us to visit every value of $x$ in the interval $[a,b]$ and ask "What is the value of $f(x)$?" (that is, the height of the "rectangle"). The units of $f(x)$ are still meters. The $dx$ is the width of this impossibly small rectangle, which has no particular numerical value, but crucially keeps its units: meters. Therefore, the units of $f(x)dx$ are square meters, so we're still talking about area.

I find talking about units helpful with single integrals but indispensable in motivating multiple integrals. For example, let $\rho(x,y,z)$ be the density of some solid $E$ at coordinates $(x,y,z)$. The units of $\rho(x,y,z)$ are (say) grams per cubic meter. The integral sign in $\iiint_E \rho(x,y,z) dV$ instructs us to visit every point of the solid $E$ and ask "What is the density here?". Since $dV = dxdydz$, its units are cubic meters, so the units of $\rho(x,y,z)dV$ are grams. Therefore, we must be computing the mass of this solid $E$.

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For definite integrals I also explain that the $d$ in $dx$ does not have a meaning since it does not add any information to the expression–but the $x$ in $dx$ has one: It indicates over which variable the integration should be performed.

Compare the notation for summation $$\sum_{i=n}^m a_i$$ with the notation for integrals $$\int_a^b f(x) dx$$.

Going from sum to integral, you

  • replace the sequence $a:\mathbb{N}\to\mathbb{R}$ by a function $f:\mathbb{R}\to\mathbb{R}$,
  • the bounds $n$ and $m$ for the sum by the bounds $a$ and $b$ for the integral,
  • the symbol $\sum$ by $\int$ and finally,
  • the symbol $i=$ which indicates over which variable the sum is computed by $dx$ which does basically the same thing for the integral.

So, one could, in principle, say that the $d$ in the integral serves the same purpose than the $=$ below the sum. This would suggest to write the definite integral as $$\int_{x=a}^b f(x)$$ and the indefinite integral as $$\int_x f(x)$$ but I doubt that this would work out. However, if one would also speak about indefinite sums $\sum_i a_i$ in the same way, one could make sense of this.

Indeed, it is pretty custom to write $\int f$ for the indefinite integral without mentioning the "free variable" at all when there is just one function involved. The $dx$ (i.e. the naming of the integration variable) is only needed for expressions like $\int_a^x (x-t)^n dt$ and such.

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  • $\begingroup$ D Knuth et. al. "Concrete Mathematics" introduces "indefinite sums" $\endgroup$ – kjetil b halvorsen Feb 5 '15 at 11:49
  • $\begingroup$ Good to know that! $\endgroup$ – Dirk Feb 5 '15 at 15:31

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