TITLE: Is it true in general that $\int \dots \int_{0 \le x_1 \le \dots \le x_n,\ 0 \le x_n\le1}dx_1\dots dx_n=\left(\frac{1}{2}\right)^n?$
QUESTION [5 upvotes]: I was looking at an example with the following integral:
$$\iiiint_{0 \le x \le y \le z \le t,\ 0 \le t \le \frac{1}{2}} 1 \,dx\,dy\,dz\,dt = \frac{1}{16}$$
 Is it true in general that 
$$\int \dots \int_{0 \le x_1 \le \dots \le x_n,\ 0 \le x_n\le1}dx_1\dots dx_n=\left(\frac{1}{2}\right)^n?$$ 
(edit: here $0 \le x_n \le 1$ but in the example $0 \le t \le \frac{1}{2}$ so this can't be true...). Intuitively it would seem so since each dimension is "cut in half" by each inequality. Are there some other results for multiple integrals with this domain that have other integrand that is not a constant? Is it true that $$\int \dots \int_{0 \le x_1 \le \dots \le x_n,\ a \le x_n\le b}f(\boldsymbol{x})\,dx_1\dots dx_n=\int_a^b \int_0^{x_{n-1}} \int_0^{x_{n-2}} \dots \int_0^{x_2}f(\boldsymbol{x})\,dx_1\dots dx_n$$

REPLY [2 votes]: The formulas in the question, as well as the accepted answer, are wrong. 
Given an $L>0$, let $C_L:=[0,L]^n$ be the the $n$-cube with side length $L$. The integral
$$\int_{C_L} 1\>{\rm d}(x_1,\ldots,x_n)$$
is of course equal to the volume of $C_L$, and therefore $=L^n$. If we take only the part $S\subset C_L$ on which the coordinates satisfy $x_1\leq x_2\leq\ldots\leq x_n$ we obtain
$$\int_S 1\>{\rm d}(x_1,\ldots,x_n)={\rm vol}(S)={{\rm vol}(C_L)\over n!}={L^n\over n!}\ ,$$
by symmetry.