Exterior derivative of differential p-form $\omega$ can be defined by "(p+1)-linear part of the value of $\omega$ integrated over the boundary of infinitesimal (p+1)-parallelotope".


More specifically, $d\omega(v_1,v_2,...,v_{p+1})=\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}\omega$, where $[tv_1,tv_2,...,tv_{p+1}]$ is (p+1)-parallelotope spanned by $tv_1,tv_2,...,tv_{p+1}$.

This aspect of exterior derivative is already mentioned by Petya and MathCrawler, but there is no proof why it is equal to the standard definition of exterior derivative 
. So I'll give you.


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It suffices to prove that

$$d\big(f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\big)(v_1,v_2,...,v_{p+1})\\=
\sum_{i\in \{ 1,2,3,...,n \} }\frac{f(x_1,x_2,...,x_n)}{\partial x_i} x_i\wedge x_1\wedge x_2\wedge ...\wedge x_p (v_1,v_2,...,v_{p+1})$$

is equal to $$\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$$

Suppose $U\subset \mathbb{R^n} $ is open, $\sigma(t):[0,t]^{p+1}\to U$ is $C^{\infty}$and, $\mathbf{t} \mapsto \big(\sigma_1(\mathbf{t}),\sigma_2(\mathbf{t}),\sigma_3(\mathbf{t}),...,\sigma_n(\mathbf{t})\big)\in U$ $f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$ is differential p-form on $U$

In order to define $\partial\sigma(t)$ with induced orientation, let $d^j_-(t_1, \dots, t_{p-1}) = 
    (t_1, \dots, t_j, 0, t_{j+1}, \dots, t_{p-1}),\\d^j_+(t_1, \dots, t_{p-1}) =  (t_1, \dots, t_j, 1, t_{j+1}, \dots, t_{p-1}).$

Then$\partial \sigma(t) = \sum_{j=1}^p (-1)^j(\sigma(t) \circ d^j_- - \sigma(t) \circ d^j_+)$

$$\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$$