$\newcommand\R{\mathbb R}$Let $f\colon\R^p\to\R$ be a continuous function. For $u=(u_1,\dots,u_p)$ and $v=(v_1,\dots,v_p)$ in $\R^p$, let 
$[u,v]:=\prod_{r=1}^p[u_r,v_r]$; 
$u\wedge v:=\big(\min(u_1,v_1),\dots,\min(u_p,v_p)\big)$; $u\vee v:=\big(\max(u_1,v_1),\dots,\max(u_p,v_p)\big)$; 
$$\int_u^v dx\, f(x):=
(-1)^{\sum_{r=1}^p\,1(u_r>v_r) }\int_{[u\wedge v,u\vee v]}dx\,f(x).$$ 
Let $F\colon\R^p\to\R$ be any antiderivative of $f$, in the sense that 
$$D_1\cdots D_p F=f,$$
where $D_j$ is the operator of the partial differentiation with respect to the $j$th argument; it is assumed that the result of this repeated partial differentiation does not depend on the order of the arguments with respect to which the partial derivatives are taken. Let $[p]:=\{1,\dots,p\}$. For each set $J\subseteq[p]$, let $|J|$ denote the cardinality of $J$. 

Then it is not hard to establish the following multidimensional generalization of the fundamental theorem of calculus ([Lemma 5.1][1]): 
\begin{equation}
	\int_u^v dx\, f(x)=\sum_{J\subseteq[p]}(-1)^{p-|J|}F(v_J), 
\end{equation}
where $v_J:=\big(v_1\,1(1\in J)+u_1\,1(1\notin J),\dots,v_p\,1(p\in J)+u_p\,1(p\notin J)\big)$.  

Has anyone seen this or similar statement elsewhere? (I am only asking about references, not proofs.)


  [1]: https://arxiv.org/abs/1705.09159