# What is $dx$ in integration?

When I was at school and learning integration in maths class at A Level my teacher wrote things like this on the board.

$$\int f(x)\, dx$$

When he came to explain the meaning of the $dx$, he told us "think of it as a full stop". For whatever reason I did not raise my hand and question him about it. But I have always shook my head at such a poor explanation for putting a $dx$ at the end of integration equations such as these. To this day I do not know the purpose of the $dx$. Can someone explain this to me without resorting to grammatical metaphors?

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This question is somewhat related and may be useful to you: math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio –  Dane Sep 21 '12 at 17:51
Much of the math curriculum is designed for people who want to understand math, rather than for people whom one should be trying to seduce into understanding math. That's one of the reasons the teaching methods sometimes don't work well. –  Michael Hardy Sep 21 '12 at 18:03
Well, it's a full stop. –  artistoex Sep 22 '12 at 0:57

The motivation behind integration is to find the area under a curve. You do this, schematically, by breaking up the interval $[a, b]$ into little regions of width $\Delta x$ and adding up the areas of the resulting rectangles. Here's an illustration from Wikipedia:

Then we want to make an identification along the lines of

$$\sum_x f(x)\Delta x\approx\int_a^b f(x)\,dx,$$

where we take those rectangle widths to be vanishingly small and refer to them as $dx$.

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The symbol used for integration, $\int$, is in fact just a stylized "S" for "sum"; The classical definition of the definite integral is $\int_a^b f(x) dx = \lim_{\Delta x \to 0} \sum_{x=a}^{b} f(x)\Delta x$; the limit of the Riemann sum of f(x) between a and b as the increment of X approaches zero (and thus the number of rectangles approaches infinity). –  KeithS Sep 21 '12 at 19:14
$dx$ is the differential of the function $x$ which behaves $dy = \frac{dy}{dx}dx$ under a change of coordinate. –  Makoto Kato Sep 21 '12 at 20:46
This is a good explanation of the origin of the notation, but it doesn't quite explain why we bother to write it down. A big part of what the dx notation does is telling you the variable you're integrating over, and as a bonus parenthesizing the integrand. (The latter is probably why the teacher made the 'full stop' remark.) –  Joren Sep 21 '12 at 21:03
I agree with @Joren that one of the most important parts is that it's telling you where you're taking those deltas from, in the case of dx, with respect to x. –  ernie Sep 21 '12 at 21:27
Motivation for finite integration is area. I suggest editing the question since integration is very useful for anti-derivative which also there the dx indicates that we are integrating over x –  raam86 Sep 21 '12 at 22:51
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There are multiple ways of explaing what the $dx$ means.

• Practical explanation: It says we are integrating over variable $x$. If we were to integrate over variable $t$, we would write $dt$ instead, and so on.

• Infinitesimal explanation: We can think of an integral as the limit of a sum: The area under the graph of a (positive) function $f$ can be approximated by the sum $\sum_x f(x) \Delta x$, and in the limit, we make $\Delta x$ arbitrarily small and call it $dx$ (an "infinitesimal" quantity). Jonathan's answer explain that in detail.

• Advanced explanation: In vector analysis, $dx$ takes meaning as a differential form (roughly, something that behaves like an infinitesimaly small piece of a curve).

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Quite a good explanation –  Barranka Sep 21 '12 at 21:22
Great answer. The second point is probably the most helpful to the OP, but the first and last are IMO just as good/important: differential forms are what really gives meaning to integrals for the general case (not necessarily just over intervals), and for the simple examples the part of bringing $x$ into scope is a very important conceptual thing. –  leftaroundabout Sep 23 '12 at 17:03

Leibniz, who introduced this notation in the 17th century, thought of $dx$ as an infinitely small increment of $x$, and at least as a heuristic, that is an immensely useful idea.

However, note some other points:

• $\displaystyle\int f(x,y)\,dx$ differs from $\displaystyle\int f(x,y)\,dy$. In one case, one integrates a function of $x$, and $y$ is constant; in the other these roles are reversed and one might be integrating a very different function.
• If $f(x)$ is in meters per second and $dx$ is in seconds, then $f(x)\,dx$ is in meters, and so is the integral. These things should be dimensionally correct, and are not so without the "$dx$".
• Sometimes one has a dot-product or a cross-product or a matrix product or some other sort of product between $f(x)$ and $dx$. How would one specify that without the "$dx$" written there?
• When doing substitutions, it becomes important to distinguish between $dx$ and $du$, etc.
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I like the dimensional analysis point you make here. –  acjohnson55 Sep 22 '12 at 6:58
What do you mean by dimensionally correct? –  Sachin Kainth Oct 30 '12 at 17:19
"Dimensionally correct" is a term used in physics. It means everything's measured in the right units. For example, suppose you know that an ampere is a coulomb per second and a volt is a joule per coulomb, and a watt is a joule per second. Then $\mathrm{volt}\cdot\mathrm{amp}=\mathrm{watt}$ because the coulombs cancel. Or $\mathrm{second}\cdot\dfrac{\mathrm{meter}}{\mathrm{second}} = \mathrm{meter}$. Or $\dfrac{\$6}{\$2}= 3$ (with no "$\$$") and \dfrac{\6}{2}=\3 (with "\$$"). – Michael Hardy Oct 30 '12 at 19:42 add comment Can someone explain this to me without resorting to grammatical metaphors? It is a matter of grammar. The indefinite integral expression is a large expression organizing several pieces of information: $$\color{blue}\int \color{red}{\underline{\quad}} \color{green}d \color{purple}{\underline{\quad}}$$ The blue$\int$is a symbol expressing that this is an integral expression. The rest of the expression is the integrand. The integrand consists of three components: there is the green$d$symbol. There is the purple slot in which you place the name of the variable that you are integrating with respect to, and there is the red slot in which you place the function expression you intend to integrate (with respect to the dummy variable). There are other grammatical interpretations of integral expressions -- most importantly (IMO) the notion of a "differential form" -- but this is the one you are using in your introductory calculus class. This particular grammatical form has some symbolism. It is a useful heuristic to think of a "$dx$" as a miniature variation in a function. You can extend this heuristic by imagining the integral to be "adding" up all of these miniature variations. The symbol$\int$, I believe, originated as an elongated$S$, for "sum"; not dissimilar to the choice of sigma ($\Sigma$) for summation expressions. The notion of differential form is a very useful one you may be interested in learning more about. Unfortunately, I am not aware of any exposition that introduces it as applied to introductory calculus: it's usually only really introduced in a differential geometry course. - I don't agree that it's only grammar. Mathematicians' understanding of this has degenerated to the point where many of them think that's all it is. (See my answer posted here.) – Michael Hardy Sep 21 '12 at 17:51 @Michael: that is what it (overtly) is in elementary calculus. I really do believe differential form-like notation is a much better way to interpret it, but actually introducing the topic would be a far more ambitious answer. I do find it interesting that you object to me saying it's grammar, but your three bullet points are mostly issues of syntax... your third point especially. – Hurkyl Sep 21 '12 at 17:58 Definitely I would not introduce differential forms in first-year calculus. My first bullet point perhaps could be called "grammar". My point about dimensional correctness is not just grammar. Nor do I agree that my third point is "grammar". – Michael Hardy Sep 21 '12 at 18:01 @MichaelHardy: When I first studied calculus in university I found it incredibly confusing that people just talked about$\mathrm{d}x$as if it was some kind of object or variable you can use in normal calculations. Especially the physicists did that when using the total derivative. It really makes no sense until you define every single operation correctly and no introductory textbook that I know of does this. So, for freshmen, the grammar explanation may help to avoid frustration (it did for me). – Gregor Bruns Sep 21 '12 at 18:08 But one need not follow the textbook in all details. – Michael Hardy Sep 21 '12 at 18:15 show 1 more comment The$dx$can be given various concrete meanings, none of which one can sensibly explain to someone first learning about integrals. It is, in reality, just a notation which comes to use from the originators of calculus, motivated by the ideas behind Jonathan's answer. Today, the$dx$serves the purpose of delimiting the integrand (although the physicists, rebellious as ever, like to write$\int\mathrm d xf(x)$for what we write$\int f(x)\mathrm dx$...) and of making explicit the variable respect to which we are computing the integral (this s useful in situations like$\int f(x,y)\mathrm dx$, which is usually different from$\int f(x,y)\mathrm d y$) As for concrete mathematical meanings: the$\mathrm dx$can mean concretely all sort of things: the Lebesgue measure, a differential form, a density, and a few others. It would be impossible to explain what any of these mean to a student first encountering integrals! - It is quite common, too, to write simply$\int f(x)$, by the way: this is possible only when this does not introduce any ambiguities (and unless the integrand is an actual formula, it is much better to write$\int f$...) – Mariano Suárez-Alvarez Sep 21 '12 at 17:52 An advantage of the notation$\int f(x)\mathrm dx$is that it can be read out quite directly as «the integral of$f$with respect to$x$», without having to add much. – Mariano Suárez-Alvarez Sep 21 '12 at 17:53 Instead of the short version$\int f(x)$I prefer the even shorter version$\int f$, where I don't have any "dangling"$x$. – Hendrik Vogt Sep 22 '12 at 8:00 The advantage of writing$\mathrm dx$at the beginning is that for nested integrals with limits, it's more easily seen which limits belong to which variable, compare$\int_1^2\mathrm dx\int_3^4\mathrm dy\,f(x^2+g(x,y))h(x+y-3)$with$\int_1^2\int_3^4 f(x^2+g(x,y))h(x+y-3)\,\mathrm dy\,\mathrm dx$– celtschk Sep 22 '12 at 9:48 Playing loose with the use of$dx$as bracket is not exclusive to physics: who hasn't written something like$\int \frac{dx}{x}$before? – Erick Wong Sep 25 '12 at 18:35 add comment Historically, calculus was framed in terms of infinitesimally small numbers. The Leibniz notation dy/dx was originally intended to mean, literally, the division of two infinitesimals. The Leibniz notation$\int f dx$was meant to indicate a sum of infinitely many rectangles, each with infinitesimal width dx. (The integral sign$\int$is an "S" for "sum.") Note that the factor$dx$in the integral is needed in order to make the units come out right. For example, if you're calculating mechanical work as$W=\int F dx$, the units wouldn't be newton-meters if you didn't have the factor of$dx$, which has units of meters. In the 19th century, mathematicians got uneasy about infinitesimals. They were afraid that a system of mathematics based on infinitesimals could not be developed in a fully rigorous and consistent way. Therefore, they rebuilt the foundations of calculus using limits, but they kept the Leibniz notation, which is extremely useful and practical. In this approach,$W=\int F dx$stands for a limit of Riemann sums of rectangles with finite widths$\Delta x$, and the$dx$becomes an archaism. Around 1960, Abraham Robinson showed that it was possible to have calculus built on a foundation of infinitesimals, and that no inconsistency would result (unless there was an inconsistency that would also affect the real number system itself, which nobody thinks is the case). Therefore it's legitimate to think of integrals and derivatives in essentially the same way that Newton and Leibniz originally conceived of them -- in fact, scientists and engineers never actually stopped thinking about them that way. - It is legitimate to think that provided one knows how to do it properly —which most people do not! – Mariano Suárez-Alvarez Sep 21 '12 at 17:54 @MarianoSuárez-Alvarez: All professional scientists and engineers know how to do it properly. They just don't know how to justify their practices in terms of NSA, or how to systematize their practices in detail. They have a set of techniques for manipulating infinitesimals. These techniques worked for Leibniz and Euler, and have worked for everyone since then, and that practitioners never stopped using. – Ben Crowell Sep 21 '12 at 20:20 Well, I have certainly seen people make mistakes —specially people who are just stating with calculus (which is the context of the OP, as one should not forget) In any case, that Robinson managed to formalize is completely orthogonal to the fact that all those people you mention do manage to work successfully...: they could probably not care less for that formalization and would probably not be able to understand it anyways. – Mariano Suárez-Alvarez Sep 21 '12 at 21:34 @MarianoSuárez-Alvarez: "they could probably not care less for that formalization and would probably not be able to understand it anyways." There is nothing that difficult or esoteric about NSA. Keisler did a very nice freshman calc book using it: math.wisc.edu/~keisler/calc.html – Ben Crowell Sep 22 '12 at 0:08 I know it can be done. The majority of those "professional scientists and engineers" you mentioned, though, have not been exposed to NSA, nor probably want to be! – Mariano Suárez-Alvarez Sep 22 '12 at 0:14 add comment Of course for something as simple as$\int{f(x)}dx$, you dont have to write$dx$if you don't feel like it, and in many situations you are allowed to just write$\int{f}$, although I don't personally make a habit of it. These things you ask about are not merely some convenient book-keeping device to let us know where the end of the intergral is, they are called differential forms, and you can add and multiply them together. The algebra of differential forms follow naturally from the simple rule that$dx^2=0$because this rule actually implies another very important rule, namely that$dx\wedge dy=-dy\wedge dx$, or in other words, that differential forms commute anti-symmetrically, see here for more info. - add comment maybe this can help you Definition of Integral - Plain links don't help... please improve this answer – Barranka Sep 21 '12 at 21:25 The linked to page has now moved. – PeteUK May 12 at 0:26 add comment I once went at some length illustrating the point that for the purpose of evaluating integrals it is useful to look at$d\$ as a linear operator.

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thanks for -1, any hints on improvement? –  Valentin Oct 4 '12 at 18:02