Integration on $\mathbb{R}^n$ in terms of differential forms One defines integration on a smooth manifold as follows: First define $\int_M \omega$ when $\omega$ is supported on a single coordinate chart by pulling back to $\mathbb{R}^n$ an integrating there, which one needs to show well-defined with respect to the choice of chart. Then define the general case by using a partition of unity, showing again that this is well-defined.
When one pulls back to $\mathbb{R}^n$, you are computing an integral $\int_{\mathbb{R}^n} f(x_1, \dots, x_n) dx_1 \dots dx_n$.  To compute this, I  treat the $dx_1 \dots dx_n$ as a formal symbol and compute the standard Riemann integral. The question is:

Is there a definition of $\int_{\mathbb{R}^n} f(x_1, \dots, x_n) dx_1 \dots dx_n$ that makes explicit use of the fact that $dx_1 \dots dx_n$ is a differential form?

I have to imagine that the definition of the Riemann integral is using an identification that isn't present for general smooth manifolds, e.g. that tangent and cotangent bundles of $\mathbb{R}^n$ have a standard choice of basis.
 A: Well, you could parameterize that region of $\mathbb R^n$ with the identity map, and pull back again. :)
But seriously...I tend to think of the ordinary Riemann integral as the "start of the induction." The fact that $dx_1 dx_2 \cdots dx_n$ can be thought of as a differential form amounts to saying "the chain rule" (or, better, "the substitution rule") works for integrals on $\mathbb R^n$. 
As for the tangent/cotangent thing, I think that the secret sauce is that $\mathbb R^n$ is parallelizable, i.e., its tangent bundle is trivial, and the parallelization is given by the derivative of translations. So the vector $\mathbf v$ at the origin and the "same" vector at some point $P$ are related by 
$$
\mathbf v_P = dT_P(0)(\mathbf v)
$$
where $T_P(A) = A + P$ is the translation that takes the origin to $P$. 
Does that help, or was all this obvious to you? 
One reason I tend to think that parallelizability is the key is that you actually can do Riemann-like integration on $S^1$ or the torus, for instance, partly because they are parallelizable. 
