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This question already has an answer here:

For Example:

$\int x^2 dx$

This is a very simple integral yet I have trouble grasping the meaning of various notations. Does the placement of '$dx$' next to $x^2$ mean multiplying $dx$ by the object of integration which is $x^2$?

If it did, then this would be solved by taking the integral of $x^2$ (this equals $\frac{x^3}{3}$) and multiplying it by the derivative of $x^2$ , the derivative of $x^2$ is $2x$.

$\frac{x^3}{3}$ $*$ $2x$ = $\frac{2x^4}{3}$

This is not correct.

Instead when we have $\int x^2 dx$ we use the power rule and come to the result:

$\frac{x^3}{3}$

In my mind this seems to ignore 'dx', so why do we even include it in the first place? Why is the integral not written as:

$\int x^2 $

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marked as duplicate by JMoravitz, user223391, Foobaz John, S.C.B., J. M. is a poor mathematician Feb 3 '18 at 2:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Then what do you do when you see $\int x^2y$? $\endgroup$ – Doug M Feb 3 '18 at 0:23
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    $\begingroup$ Related: Leibniz Notation. $\endgroup$ – JMoravitz Feb 3 '18 at 0:26
  • $\begingroup$ It is a notation that tells you what the variable of integration is (and for a Lebesgue integral, what the measure is). It also has historical antecedents in terms of infinitesimals. $\endgroup$ – Xander Henderson Feb 3 '18 at 0:26
  • $\begingroup$ @JMoravitz , that question is similar but I believe my question (and confusion) is stemming form the appearance of $dx$ as a value which is multiplied by. The other question does not address this. $\endgroup$ – sawreals2 Feb 3 '18 at 0:31
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    $\begingroup$ Related: math.stackexchange.com/questions/1991575/… $\endgroup$ – Ethan Bolker Feb 3 '18 at 0:36
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The $dx$ is not an important part of the actual calculation, no. If you like the view that the integral notation means "integrate whatever is between $\int$ and $dx$", you can do that. However, it will make anything more advanced than simple anti-differentiation difficult to remember and do correctly.

Historically, $dx$ comes from the transformation that underlies the definition of integrals $$ \sum_i f(x_i)\Delta x \to \int f(x)dx $$ and it can be seen simply as a relic from that, if you want.

However, an actual meaning may be given to $dx$, and having one in mind that will make more advanced things like theoretical arguments and substitution a lot easier to do correctly.

In fact, there are several possible meanings, depending on your taste. I think the simplest one to explain intuitively may be the one from measure theory: $f(x)dx$ can here be seen as a density, and you integrate up density along a bit of the number line to find total mass. This would make $dx$ the standard, unit density of the number line, while $f(x)$ tells you the actual density at any point, in units of $dx$.

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In the early 1970's, (I want to say 1973, but I'm not 100% certain) this question was debated in the letter column of several issues of the American Math Monthly. There were three schools of thought, as I remember.

  1. It's just a punctuation mark that just tells you where the integrand ends.
  2. It's a dummy variable.
  3. It's an essential part of the notation; we integrate differential forms, not functions.

No conclusion was reached, and the editors eventually refused to publish any more letters on the subject.

As for me, I'm not happy with any of these answers, but I have nothing to offer in their place.

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  • $\begingroup$ Could anybody else maybe link to at least the first of these AMM letters? $\endgroup$ – J. M. is a poor mathematician Feb 3 '18 at 2:23
  • $\begingroup$ @J.M.isnotamathematician Unfortunately, I don't think that the Monthly is freely available on the Web, except through JSTOR. If you've got access to JSTOR through an institution like your university, you may be able to browse through the issues and find it. But if you only have free access as an individual, like me, you can only look at three articles every 2 weeks, so it would take a prohibitive amount of time. $\endgroup$ – saulspatz Feb 3 '18 at 2:36
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The following was originally my comment, but I want to put it in an answer because a conclusion has been reached, which is contrary to the answer most above.


Go here $\longrightarrow$ https://m.youtube.com/watch?v=U_q7R5JJvb4 for a meaning of $\text{d}x$ in a general integral. In a Riemann Integral, the definition of $\text{d}x$ is a little bit different as it is derived from the Riemann Sum. For that, I suggest going here $\longrightarrow$ https://m.youtube.com/watch?v=Stbc1E5t5E4.

There have been definitions like treat $\text{d}x$ as a full stop but this is only to help notating an integral in symbols. This is not the definition of $\text{d}x$.


I also found another link to a YouTube video that is part of a series called Essence of Calculus. In this video, it specifically talks about integrals $\longrightarrow$ https://m.youtube.com/watch?v=rfG8ce4nNh0 however, this video is quite advanced and perhaps difficult to understand at first for there may be much to keep up with.

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The $dx$ term intuitively stands for a really small (infinitesimally small) change in the $x$ variable.

Remember how we define integration in the first place. $\int \limits_{a}^{b} f(x) \,dx$ is defined by first taking Riemann sums that look like $\sum f(x^{*})(x_{i+1} - x_{i})$, where $x^{*}$ is a point in $[x_{i}, x_{i+1}]$. If you remember from Calculus, this sum represents the sum of the areas of rectangles that all together give us an approximation to the area under the curve $f(x)$. In what sense are these Riemann sums an approximation of the area? In the sense that, as the base of the rectangles gets smaller and smaller (so that the rectangles get thinner and thinner), you get a better and better approximation.

When you take the limit as the base of the rectangles goes to $0$, then intuitively $(x_{i+1} - x_{i})$ goes to an infinitesimally small value, so that $\sum f(x^{*})(x_{i+1}-x_{i})$ becomes, intuitively, $\int f(x)\,dx$, where the $\int$ symbol means we are "adding" up the now infinitely many rectangles, and $f(x)\,dx$ is intuitively the area of each rectangle. So, to me, from this understanding of the limit, I always interpret the integral my mind as meaning $\underbrace{\int}_{\text{adding}} \underbrace{f(x)\,dx}_{\substack{\text{area of rectangle} \\ \text{w/ height }f(x)\text{ and base } dx}}$. Of course, this is just intuition, because $\int f(x)\,dx$ is just notation for the limit. $dx$ doesn't mean anything on its own, at least not at the level of Calculus, though as someone else pointed out, there is a way to make this intuitive interpretation rigorous. It involves defining infinitesimals (search the topic of Nonstandard Analysis if you are interested in this).

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There are several possible depths of understanding here.

At the core, d$x$ is not a notation, it is really a "$1$-differential form" on a real manifold that is the real line. You can read more about this in books about differential geometry or calculus.

Concretely, and at a more basic level, it is telling you with respect to which variable you are integrating. $\int x\operatorname{d}\!x$ is $x^2/2$ whereas $\int x\operatorname{d}\!y$ is $xy$.

In some cases, to simplify the writing in some situations where a change of variables is welcome, one might be lead to write something else than just a variable after the d. For instance $\operatorname{d}\!\log(x)$ is used for $\operatorname{d}\!x/x$ (assuming natural logarithm). More generally $\operatorname{d}\!f(x)$ or simply $\operatorname{d}\!f$ is used for $f'(x)\operatorname{d}\!x$.

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In the integral's definition $I=\lim_{n\to\infty}\sum_{i=0}^n f(x)\Delta x$ where $||P||\to 0$ when $n \to \infty$ and $P=\{a=x_0<x_1<...<x_n=b\}$, the number $\Delta x$ approaches 0 because $||P||\to 0$, then we denote that infinitesimal as $dx$ then you multiply dx times the value of f(x) and sum all those little products along the interval [a, b] . Now, we also define the diferential of f, so that if $y=f(x)$ then $dy=f'(x)dx$ is the change of 'y', notice that $\frac{dy}{dx}=f'(x)$ so you can analyse what dx means in whatever context.

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