Does $f(x)\,dx$ denote multiplication of $f(x)$ by $dx$?

Question

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 $\mathrm{d}x$?

Answer 1

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 $f(x) \mathrm{d}x$.

Answer 2

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.

Answer 3

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.