Question: Compute volume of the set A: $A = \{(x_1,x_2,...,x_n) \text{ | } 0 ≤ x_i ≤ 1, \sum_ix_i = 1\}$
Try: We know volume is basically integrating 1 over the set. I suppose fubini’s theorem is to be used but I can’t set up. How do I use $\sum_ix_i = 1$?
Def of Volume: Volume vol(C) of a set C is the integral of the constant function 1 over C.
On
Take the simplex whose vertices are the origin and the standard basis vectors.
The volume of the simplex is $1/n!$.
Now regard it as a pyramid whose summit is the origin and base is the area you want.
The centre of the base is $(1/n,1/n,...,1/n)$ so the height of the pyramid is $1/\sqrt n$.
The volume is base × height / n = 1/n!, so $$base =\frac{n\sqrt n}{n!}$$
Start by focusing on one variable (e.g. $x_n$) and letting that "run free", then the next one can only be from $(0, 1 - x_n)$, the one after that from $(0, 1 - x_n - x_{n-1})$ and so on. So the total integral is $$ \int_0^1 \int_0^{1-x_n} \int_0^{1-x_{n-1}-x_n}...\int_0^{1-{\sum_{i=j}^n x_i}} ... \int_0^{1-{\sum_{i=3}^n x_i}} \int_0^{1-{\sum_{i=2}^n x_i}} dx_1 dx_2 ... dx_j ... dx_{n-2} dx_{n-1} dx_n. $$ You can either brute-force calculate this or try to approach it a bit more geometrically by focusing on some low dimensional cases. Some examples for specific values of $n$: $$ n=1: \quad \int_0^1 dx_1 = 1 \\ n=2: \quad \int_0^1 \int_0^{1-x_2} dx_1 dx_2 = \int_0^1 (1-x_2) dx_2 = \frac{1}{2} \\ n=3: \quad \int_0^1 \int_0^{1-x_3} \int_0^{1-x_3-x_2} dx_1 dx_2 dx_3 = \int_0^1 \int_0^{1-x_3} (1-x_3 - x_2) dx_2 dx_3 = \int_0^1 \frac{(1-x_3)^2}{2} dx_3 = \frac{1}{6} $$