How to integrate over the standard $n$-simplex directly in $\mathbb{R}^{n+1}$? Definition of standard $n$-simplex

The standard $n$-simplex contains all points $\vec{x} \in \mathbb{R}^{n + 1}$ such that $0 \le x_i \le 1$ and $\vec{x} \cdot \vec{1} = 1$.
My definition of the area of $n$-simplex
$$A_{n} =  \frac{\sqrt{n+1}}{n!}.$$
The standard $2$-simplex is an equilateral triangle with side length $\sqrt{2}$ and vertices at $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$. The area is $A_{2} = \dfrac{\sqrt{3}}2$.
The standard $1$-simplex is a line with vertices at $(1, 0)$ and $(0, 1)$. The length is $A_{1} = \sqrt{2}$
I found the area of the standard $n$-simplex by first moving the simplex from $\mathbb{R}^{n+1}$ to $\mathbb{R}^{n}$.
Area of standard simplex
Question
How to directly integrate over the standard $n$-simplex with the coordinates of $\mathbb{R}^{n+1}$?
I can't use the change of coordinate formula because the Jacobian of the transformation from $\mathbb{R}^{n}$ to $\mathbb{R}^{n+1}$ is rectangular.
Thanks.
 A: Let $\Delta^n\subseteq \mathbb{R}^{n+1}$ be the standard $n$-simplex: $$\Delta^n:=\big\{(x_0,x_1,x_2,\ldots,x_n)\in\mathbb{R}_{\geq 0}^{n+1}\,\big|\,x_0+x_1+x_2+\ldots+x_n=1\big\}.$$  An arbitrary function $f:\Delta^n\to \mathbb{R}$ can be considered as a function in $n$ free variables.  That is, there exists a unique function $F:\Sigma^n\to \mathbb{R}$ such that
$$f(x_0,x_1,x_2,\ldots,x_n)=F(x_1,x_2,\ldots,x_n)$$ for all $(x_0,x_1,x_2,\ldots,x_n)\in\Delta^n$, where $$\Sigma^n:=\big\{(x_1,x_2,\ldots,x_n)\subseteq\mathbb{R}_{\geq 0}^n\,\big|\,x_1+x_2+\ldots+x_n\leq1\big\}\subseteq \mathbb{R}^n\,.$$  In other words, $F$ is given by $$F(x_1,x_2,\ldots,x_n):=f\left(1-x_1-x_2-\ldots-x_n,x_1,x_2,\ldots,x_n\right)\,,$$ for all $(x_0,x_1,x_2,\ldots,x_n)\in\Delta^n$.  That is, if $\nu_n$ is the volume measure on $\Delta^n$ and $\lambda_n$ is the Lebesgue measure on $\mathbb{R}^n$, then
$$\int_{\Delta^n}\,f\,\text{d}\nu_n=\sqrt{n+1}\,\int_{\Sigma^n}\,F\,\text{d}\lambda_n\,.$$
This is because $$\text{d}\nu_n(x_0,x_1,x_2,\ldots,x_n)=\sqrt{n+1}\,\text{d}\lambda_n(x_1,x_2,\ldots,x_n)$$
for all $(x_0,x_1,x_2,\ldots,x_n)\in\Delta^n$ (this can be proven using the Jacobian determinant).  In particular, if $f\equiv 1$ so that $F\equiv 1$, then we get
$$\text{vol}_n\left(\Delta^n\right)=\sqrt{n+1}\,\text{vol}_n\left(\Sigma^n\right)=\frac{\sqrt{n+1}}{n!}\,,$$
where $\text{vol}_n$ is the $n$-dimensional volume.
To clarify some points, first note that there exists an isometry from the affine hyperplane $$H^n:=\big\{(x_0,x_1,x_2,\ldots,x_n)\in\mathbb{R}^{n+1}\,\big|\,x_0+x_1+x_2+\ldots+x_n=1\big\}$$
to $\mathbb{R}^n$.  We can take the isometry to be the unique affine map $\varphi:H^n\to\mathbb{R}^{n}$ which sends $e_0,e_1,e_2,\ldots,e_n\in\mathbb{R}^{n+1}$ to $$0,E_1+\alpha_n\, E,E_2+\alpha_n\, E,\ldots,E_n+\alpha_n\, E\in\mathbb{R}^n\,,$$ respectively, where $e_0,e_1,e_2,\ldots,e_n$ are standard basis vectors of $\mathbb{R}^{n+1}$, $E_1,E_2,\ldots,E_n$ are standard basis vectors of $\mathbb{R}^n$, $E:=E_1+E_2+\ldots+E_n$, and
$$\alpha_n:=\frac{\sqrt{n+1}-1}{n}\,.$$
Write $E_0:=-\alpha_n\,E$.  Let $T:\mathbb{R}^n\to\mathbb{R}^n$ be the unique linear transformation that sends  $$E_1,E_2,\ldots,E_n\text{ to }E_1-E_0,E_2-E_0,\ldots,E_n-E_0\,,$$ respectively.  Prove that $$\det(T)=1+n\,\alpha_n=\sqrt{n+1}\,.$$
The volume measure $\nu_n$ on $\Delta^n$ is inherited from the volume measure on $H^n$, which is the pullback $\varphi^*\lambda_n$ of $\lambda_n$. However, since $T$ maps the extreme points $0,E_1,E_2,\ldots,E_n$ of $\Sigma^n$ to the extreme points $0,E_1-E_0,E_2-E_0,\ldots,E_n-E_0$  of $\varphi(H^n)$, it is simpler to take the integral on $\Sigma^n$, using the Change-of-Variables Theorem.  That is,
$$\begin{align}\text{d}\nu_n(x_0,x_1,\ldots,x_n)&=\text{d}(\varphi^*\lambda_n)\left(x_0,x_1,\ldots,x_n\right)\\&=\text{d}\lambda_n\big(\varphi\left(x_0,x_1,\ldots,x_n\right)\big)\\
&=\text{d}\lambda_n\left(x_1+\alpha_n\,\sum_{i=1}^n\,x_i,x_2+\alpha_n\,\sum_{i=1}^n\,x_i,\ldots,x_n+\alpha_n\,\sum_{i=1}^n\,x_i\right)
\\&=\text{d}\lambda_n\big(T(x_1,x_2,\ldots,x_n)\big)\\&=\text{d}(T^*\lambda_n)(x_1,x_2,\ldots,x_n)\\&=\det(T)\,\text{d}\lambda_n(x_1,x_2,\ldots,x_n)\\&=\sqrt{n+1}\,\text{d}\lambda_n(x_1,x_2,\ldots,x_n)\end{align}$$
for all $(x_0,x_1,\ldots,x_n)\in\Delta^n$.
