In our course of differential geometry we defined the integral $\int_{U} \omega$ of a differential form $\omega=f dx_1\wedge \ldots \wedge dx_n: T^nU \rightarrow \mathbb R$ with $U\subseteq \mathbb R^n$, $U$ open and compact support $\operatorname{supp}(\omega)$ via $$\int_U \omega := \underbrace{\int_U f(x_1,\ldots,x_n) dx_1 \ldots dx_n}_\text{Riemann integral}$$
This is the same definition Manfredo do Carmo used in his book "Differential Forms and Applications" [chapter 4.1, page 55].
Unfortunately there was no motivation given for this definition in our course. For me as a student it seemed that our professor set $\int_U f dx_1\wedge \ldots \wedge dx_n$ to be $\int_U f(x_1,\ldots,x_n) dx_1 \ldots dx_n$ because both notions are similar. But the reasoning must be vice versa: Because both concepts of integration lead to the same result the notations are similar.
So why does the above definition make sense? Is there some kind of intuitive idea behind the integral of $\omega$ from which I can see that it is the usual Riemann integral? What is the concept/idea behind integrating differential forms? (For example I can think of the Riemann integral as the limit of the area under step functions)
Best Answer
There are a couple confusing things going on here.
First, basis covectors are often denoted something like $\mathrm dx^i$ which is visually similar to $dx^i$ (I'm writing them like so, differently, to emphasize these are different notions; it wouldn't surprise me to see $dx^i$ as a basis covector either though), so the relationship between a basis covector and a differential appears clear, even obvious.
And yet these things are not really even alike.
When you integrate a $k$-covector, what are you really doing? You're integrating on some (usually) $k$-dimensional manifold. If $e_1, e_2, \ldots$ are basis vectors of this manifold*, then an infinitesimal patch of this manifold is described using a $k$-vector $dV = (e_1 \wedge e_2 \wedge \ldots \wedge e_k) dx^1 dx^2 \ldots dx^k$.
What do $k$-covectors do? Well, usually we're told they eat $k$ vectors in all to give a scalar (or a scalar field, if in fact we have a $k$-covector field). Alternatively, they can be seen to eat a single $k$-vector.
So what you're really doing when you integrate a $k$-covector is this:
$$\int_U \omega \equiv \int_U \omega(e_1 \wedge e_2 \wedge \ldots \wedge e_k) \, dx^1 dx^2 \ldots dx^k$$
For some reason, people seldom even talk about the existence of $k$-vectors. Knowing they exist at all is really important. It turns the Riemann tensor into a map from $2$-vectors to $2$-vectors, for instance. Anyway, the object $\omega(e_1 \wedge \ldots \wedge e_k)$ is a scalar function and as such you clearly a classic Riemannian integral now.
*I don't denote that these basis vectors depend on the point of the manifold you take them at, but they do have this dependence.