Generalization to $N$ dimensions of distribution function evaluation over an hyper- rectangle Let's consider an absolutely continuous random vector $V \equiv (X,Y)$ and its associated joint distribution function $F(x,y)=Pr(X \le x,Y \le y) = \int_{-\infty}^{x}\int_{-\infty}^{y} f(x,y)\,dx\,dy$.
If we take four points in the $xy$ plane that are vertexes of a rectangle $R$, $A \equiv (x_1,y_1) \;, B \equiv (x_1,y_0) \;,C \equiv (x_0,y_0) \;,A \equiv (x_0,y_1)$, with $x_1 > x_0$ and $y_1>y_0$, it is well known that the  probability that the values of the random vector $V$ are within the rectangle $R$ is given by the value of the distribution function $F(x,y)$ taken at those points according to the below formula:
$Pr(x_0<X \le x_1,y_0<Y \le y_1)=F(x_1,y_1)-F(x_1,y_0)+F(x_0,y_0)-F(x_0,y_1)$
Is there an explicit formula, generalizing the one above, that applies when we move from $2$ to $N$ dimensions?
In other words, given the distribution function $F(x_1,...,x_N)=Pr(X1 \le x1,\ldots,X_N \le x_N)$, it is there a formula that allows to compute 
$Pr(a_1<X_1 \le b_1,\ldots,a_N<X_N \le b_N)$ by the values of the N-dimensional distribution function computed in the vertexes of the hyper-rectangle $[a_1,b_1]\times\ldots\times[a_N,b_N]$?
What happens if values $a_i$ or $b_j$ are allowed to be $+\infty$?
In two dimensions we find the value of the $1$-dimensional marginal distributions $F_i(x)$, what's found in the $N$-dimensional case?
From computational point of view, is this formula applicable in practice for value of $N$ equal to $10$? I suppose it involves $2^{10}$ vertexes....

Here is the formula you are looking for:
$$
\color{red}{\mathrm P(a_i\lt X_i\leqslant b_i\ \text{for all}\ i)=\sum_c(-1)^{n(c)}F(c_1,c_2,\ldots,c_n)}
$$
The sum in the RHS runs over the $2^n$ vectors $c=(c_i)_{1\leqslant i\leqslant n}$ such that $c_i\in\{a_i,b_i\}$ for every $1\leqslant i\leqslant n$, and $n(c)$ denotes the number of indices $i$ such that $c_i=a_i$.
The simplest proof might be to note that the LHS if the expectation of the random variable
$$
\prod_{i=1}^n\left(\mathbf 1_{X_i\leqslant b_i}-\mathbf 1_{X_i\leqslant a_i}\right)=\sum_c(-1)^{n(c)}\prod_{i=1}^n\mathbf 1_{X_i\leqslant c_i},
$$
and to take the expectation of the RHS, remembering that the expectation is a linear operator and that, for every $c$,
$$
\mathrm E\left(\prod_{i=1}^n\mathbf 1_{X_i\leqslant c_i}\right)=\mathrm P(X_i\leqslant c_i\ \text{for all}\ i)=F(c_1,c_2,\ldots,c_n).
$$
