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In the integral form $\int \! f(x) \, \mathrm{d}x$ does $f(x)\,\mathrm{d}x$ can be seen as a multiplication of $f(x)$ and $\mathrm{d}x$?

Michael Hardy
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Emo
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    It is "analogous" to multiplication, but the reality of it is somewhat more complicated, just as $\frac{dy}{dx}$ is notation that looks like division. – Thomas Andrews May 24 '14 at 17:44
  • I'm asking for "the reality of it"! – Emo May 24 '14 at 17:46
  • In applications, "seeing" that a given quantity is represented by a certain integral is natural if one views it as a multiplication. – André Nicolas May 24 '14 at 17:50
  • This is closely related to your previous question Is $dxdy$ really a multiplication of $dx$ and $dy$? – complexist May 24 '14 at 17:57
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    Ultimately at the most basic level $\int f(x) dx$ is just a notation, where the purpose of the $dx$ is little more than to distinguish the variable of integration. For intuition, you might think of the $\int$ as analogous to a summation symbol and the $dx$ as analogous to an infinitesimally small increment, which is what I think Leibnez had in mind when he invented this notation. – Dustan Levenstein May 24 '14 at 17:59
  • I think this differs a bit from the question "Wahat is $dx$ in integration?", because there the question only asks the nature of $dx$, and here we are searching for the relation (or operation) between a function and $dx$. – Emo May 24 '14 at 18:05
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    It's a different question, but you can't approach your question without answering the other, and visa versa, so they are essentially equivalent. @Emin – Thomas Andrews May 24 '14 at 18:18

3 Answers3

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Yes. $\mathrm{d}x$ is a differential form, and the space of differential forms has scalar multiplication by continuous functions, and scalar multiplying $\mathrm{d}x$ by $f(x)$ gives $f(x) \mathrm{d}x$.

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It is not a literal product of two numbers.

There are several different "realities of it" from the point of view of advanced mathematics. Until you get to higher mathematics, it is better to think of $dx$ as purely a notational thing. For example what would it mean to try to compute:$$\int_a^b (c^2-x^2)?$$ It would be unclear that $c$ is the constant, and it is $x$ that varies. In this sense, $dx$ is just saying "with $x$ varying" and serves the same purpose as the $i=$ in the notation:

$$\sum_{i=1}^n (c^2-i^2)$$

which is obviously different from:

$$\sum_{c=1}^n (c^2-i^2)$$

A more advanced notion is one of a differential form. In that sense, $f(x)dx$ is the product of $f(x)$, a $0$-form and $dx$, a $1$-form. This is a fairly advanced view, however, and it often just feels like raw notation when first learning about them. Differential forms are more useful when you are dealing with multiple variables.

There is also the notion of measure theory, where $dx$ is often generalized to $d\mu$, where $\mu$ is a "measure." In that sense, the $x$ represents the most obvious "measure" on the real line.

Thomas Andrews
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  • It is a product of one thing that is literally a number and another that might be considered something other than a number. – Michael Hardy May 24 '14 at 20:57
  • No, it is not a product of a number. $f(x)$ is a function, not a number. @MichaelHardy – Thomas Andrews May 24 '14 at 21:55
  • For each value of $x$, $f(x)$ is a number, and that is what gets multiplied by $dx$. ${}\qquad{}$ $f$ is a function; $f(x)$ is a number. ${}\qquad{}$ – Michael Hardy May 24 '14 at 22:10
  • Sorry, but $f(x)dx$ is not really a combination of a lot of $f(x_i)dx$ values. That doesn't make any sense, unless you say what $dx$ is. There is a sense in which $dx$ is related to localization, but it certainly isn't 'literally' a single number times some thing called $dx$. It is certainly like $\sum f(x_i)\delta x$ but it isn't literally that. @MichaelHardy – Thomas Andrews May 24 '14 at 22:26
  • In particular, it is not the same thing independent of the structure of $\mathbb R$. For example, define $C$ to be the Cantor set, and define $f(x)=\chi_C(x)$, define $g(x)=\chi_{[0,1]}(x)$. The the mutli-set of values $f(x_1)dx$ and the multi-set of values $g(x_1)dx$ are the same set of values, but $f(x)dx$ and $g(x)dx$ are not the same. It is useless to try to localize at $x_1$ and then forget $x_1$. @MichaelHardy – Thomas Andrews May 24 '14 at 22:31
  • Yeah it is a product of two numbers like when we want to calculate the area or do some Physics Proofs/Calculations! Further detail see : http://math.stackexchange.com/a/200400/134694 – NeilRoy Jun 08 '15 at 05:36
  • No, it's not, @NeilRoy. If $dx$ is a number, then what number is it? It's a notation related to computation of area, but $f(x)dx$ is not a product of two numbers. Also, when commenting on an answer more than a year old, be more specific about what you mean by "it." – Thomas Andrews Jun 08 '15 at 11:21
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Yes. See my answer here.

In particular, if $f(x)$ is in meters per second and $dx$ is in seconds, then $f(x)\,dx$ is in meters, and things like that matter all the time in physics.

Community
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Michael Hardy
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