Does $dx$ in the formula $\int f(x)dx$ represents a differential of x? While I asked a question about integrals Is $dx\,dy$ really a multiplication of $dx$ and $dy$?, I found out that many of the answers were assuming that dx is just a notation in the formula $\int f(x)dx$. I'm not convinced with that, so I decided to ask this question, because the answer of this question will imply the answer of the question I've set before.
So, does $dx$ in the formula $\int f(x)dx$ represents a differential of x or it's just part of a notation?
 A: From what I understand, historically, one worked with infinitesimal elements dx (which, in standard analysis aren't defined) and $\int$ is actually a long s denoting sum, as one thought of an integral as being the "sum" of infinitesimal elements $f(x)\text{d}x$. In modern theory of integration, one integrates with respect to a measure and in this context $\text{d}x$ ( we sometimes write $\text{d}\lambda$) denotes the fact that we are in fact integrating $f$ with respect to the Lebesgue measure. To find a link with differential forms (and therefore line integrals) the following link might be a good place to start: Integration of forms and integration on a measure space
A: Yes, you can think of $\omega = f(x)\, dx$ as a differential form and $\int f(x)\, dx$ the set of function $F$ whose differential $dF = \omega$. This however makes sense mostly for function of one variable, since in general a differential form might not have primitives. Also one could consider the integral
$$
\int_a^b f(x)\, dx
$$
as the path integral
$$
\int_\gamma \omega
$$
where $\gamma$ is the curve parameterizing the oriented segment $[a,b]$ and $\omega = f(x)\, dx$. However, again, the path integral requires for its definition the integral over the real line. Hence one is forced to define $\int_a^b f$ without speaking of differential forms (Riemann or Lebesgue integrals).
