Let $U \subset \Bbb R^n$ be open, $1 \le i \le n$. Then
$\int_U$ $d \phi(x) \over dx_i $ $d^n x = 0$ for every $\phi \in C_c^1(U)$.
Proof:
We can assume that $U = \Bbb R^n$. Furthermore, it is enough to prove the statement for $i = 1$. So let $\phi \in C_c^1(\Bbb R^n)$. We choose $R \in \Bbb R_+$ so that $Supp(\phi) \subset [R, -R]^n$ holds. Then, for every $(x_2, ..., x_n) \in \Bbb R^{n-1}$, it holds that
$\int_{\Bbb R}$ $d\phi \over dx_1$ $(x_1, ..., x_n) dx_1 = \phi(x_1, x_2, ..., x_n)_{x_1 = R}^{x_1 = -R} = 0$.
How does he conclude the last two steps here? Do integration and differentiation simply "eliminate" each other? If so, why do we still put in the values $R$ and $-R$? And why does this whole thing equal $0$?