Show that the volume of the polyedron ${x=(x_1,x_2,⋯,x_n)\in \mathbb{R^{n}} ; 0 \leq x_i \leq 1, \ x_1+x_2+...+x_n\leq (n/2)}$ is $1/2$

Question

Show that the volume of the polyedron $ (x_1,x_2,⋯,x_n)\in \mathbb{R^{n}};0 \leq x_i \leq 1, \ x_1+x_2+...+x_n\leq (n/2)$ is $1/2$

I've seen a lot of ways to calculate the volume of an n-simplex, but on polyhedra of this type I can't find anything .. the article I'm studying comments on using integration and changing variable but if anyone knows any way to calculate, I'll be grateful :) ..

Answer 1

Making the change of variables $y_i=1-x_i$ your polyhedron becomes:

$$ \{y=(y_1,y_2,⋯,y_n)\in \mathbb{R^n}\ :0 \leq y_i \leq 1, \ y_1+y_2+...+y_n\geq (n/2)\}. $$

But the union of this with your set (i.e. with a congruent polyhedron) is the unit cube, hence your polyhedron is half the unit cube.