# Prove that $\lim_{n\rightarrow \infty} \int_{[0,1]^n}\frac{|x|}{\sqrt{n}}=\frac{1}{\sqrt{3}}$

where $[0,1]^n=[0,1]\times \dots \times [0,1]$ is the unit cube in $\mathbb{R}^n$.

## Answer

**Answers can be viewed only if**

- The questioner was satisfied and accepted the answer, or
- The answer was disputed, but the judge evaluated it as 100% correct.

The answer is accepted.

Join Matchmaticians Affiliate Marketing
Program to earn up to 50% commission on every question your affiliated users ask or answer.

- answered
- 232 views
- $20.00

### Related Questions

- Describing Bayes Theorem succinctly in words
- Riemann Sums for computing $\int_0^3 x^3 dx$
- Notation question. Where does the x in the denominator come from?
- Probability maximum value of samples from different distributions
- Differentiate $f(x)=\int_{\sqrt{x}}^{\arcsin x} \ln\theta d \theta$
- Convergence in probability and uniform integrability implies convergence in Lp
- Calculus / imaginary numbers and S^2
- Let $X$ be a single observation from the density $f(x) = (2θx + 1 − θ)I[0,1](x)$ with $−1≤ θ ≤ 1$. Find the most powerful test of size $α$ and its power