Let be $S_n(1)=$$\{(x_1, x_2, ... ,x_n)\in \mathbb R^n : x_j\geq 0, x_1 + x_2 + ... + x_n \leq 1\}$ the standard simplex. My task is to calculate the geometric centre C with Lebesgue integration. I know $C=\frac{1}{\lambda(S_n(1))}\int_{S_n(1)} x \,d\lambda$. I have already shown that $\lambda(S_n(1))=\frac{1}{n!}$. But I don´t know how to solve the integral..
Any help or hints is greatly appreciated!
When you cut through the simplex with a hyperplane at coordinate $x$, you get a cross-section that:
is a simplex of dimensionality $n-1$
has lengths $u=1-x$ times those of the standard simplex with that dimensionality.
You conclude that the integral is
$\int_0^1(1-u)u^{n-1}\lambda(S_{n-1}(1))du$
where I have substituted $u=1-x$ and you already know $\lambda(S_{n-1}(1))$ according to the problem statement. Can you continue from there?
From standard geometric considerations, you should get $1/3$ for $n=2$, $1/4$ for $n=3$, ... .