Wedge products, dual basis and exterior algebra in my real analysis book I'm reading a book on Real Analysis and there's a chapter about surface integrals. It starts by defining the wedge product, and then it starts defining differential forms of some degree. It starts like this:

Let's use the notation $\{dx_1, \ldots, dx_m\}$ for the canonical
  basis of $(\mathbb{R}^m)^*$, dual of the basis $\{e_1, \ldots,
e_m\}\subset\mathbb{R}^m$, where $e_1 = (1,0,\ldots, 0)$ etc. For each
  set $I = \{i_1< \cdots < i_r\}\subset \{1,2,\ldots, m\}$, let's write
$$dx_I = dx_{i_1}\wedge \cdots \wedge dx_{i_r}$$
the $r$-linear alternating forms $d_{x_I}$ make the canonical basis of
  the vector space $A_r(\mathbb{R}^m)$ (space of the $r$ linear maps
  from $\mathbb{R}^m$ to $\mathbb{R}$). 
Given a list of $r$ vectors $v_1,\ldots, v_r\in\mathbb{R}^m$, we
  obtain a matrix $a = (a_{ij})$, with $m$ lines and $r$ columns, at
  which the $j$-th column is the vector $v_j = (a_{1j}, \ldots,
 a_{mk})$. In this case:
$$d_{x_I}(v_1, \ldots, v_r) = \det(a_I)$$
where $a_I$ is the matrix $r\times r$ obtained selecting the lines
  such that its indexes belong to the set $I$

I know that here, $dx_1, \ldots$ are just names for the dual basis elements, but it has something to do with $dx$ of integrals. What is this relation? Why are we even using dual basis vectors, for what they're useful?
I've been googling things and found that this is related to something called exterior algebra, which uses a lot of wedge products, but I'm completely lost here, I don't get anything about the usefulness of wedge products and its relation to dual vectors. Could somebody help me?
 A: As a brief lowbrow sketch....
Differential forms are things you integrate. The relation to $\mathrm{d}x$ in integrals is that you were being taught integration of differential forms, but introductory materials didn't want to actually introduce the idea of a differential form.
Derivatives are also most naturally differential-form valued. For example:


*

*$\mathrm{d}f$ is a far more natural object than "the vector in which $f$ increases the fastest with length equal to the rate of increase".

*$\mathrm{d}x_1$ can be viewed as applying the $\mathrm{d}$ operator to the scalar field $x_1$. $\frac{\partial}{\partial x_1}$ has no such interpretation — in fact, $\frac{\partial}{\partial x_1}$ implicitly depends on the entire choice of coordinates $\{ x_1, x_2, \ldots, x_n \}$, despite the fact only $x_1$ appears in the notation.



One reason to see why the product is alternating is to consider the two different ways to reduce integration over the unit square to integrals over paths:
$$ \int_0^1 \left( \int_0^1 f(x,y) \, \mathrm{d} y \right) \mathrm{d} x 
= \int_{\square} f(x,y) \, (\mathrm{d}x \wedge \mathrm{d}y) =  \int_0^1 \left( \int_0^1 f(x,y) \, \mathrm{d} x \right) \mathrm{d} y  $$
(normally you write $\mathrm{d}x \mathrm{d}y$ in the middle, but I want to emphasize that the meaning is the wedge product)
I'm not sure how to describe it, but once you figure out the notion of how geometric paths and surfaces and such are oriented, you'll see that the integral on the right has the opposite orientation from the integral on the left; correspondingly, $\mathrm{d}y \mathrm{d}x = -\mathrm{d}x\mathrm{d}y$ to compensate.

The reason you're learning this stuff is that the algebraic aspect of differential geometry is far more naturally expressed in terms of differentials than it is in terms of tangent vectors (i.e. parital derivative operators).
Furthermore, the generalized stokes theorem — the higher dimensional analog of the fundamental theorem of calculus — has a very natural expression when integrands are taken to be differential forms and regions of integration are taking to be chains:
$$ \int_R \mathrm{d} \omega = \int_{\partial R} \omega $$
This is related to homological algebra. In fact, de Rham cohomology is one of the big methods of the subject, where the cochains are differential forms.
For example, if $\omega$ is a $0$-form (i.e. a function) and $R$ is the path from $a$ to $b$, then the above reduces to
$$ \int_a^b \mathrm{d} f = \int_{b} f - \int_a f = f(b) - f(a) $$
(since the integral of a $0$-form over a $0$-dimensional region is simply evaluating the function at that point, and the boundary of the given path consists of the point $b$ positively oriented and the point $a$ negatively oriented)
