Evaluation of the double integral $\int_{[0,1]×[0,1]} \max\{x, y\} dxdy$ 
Evaluate:  $$\int_{[0,1]×[0,1]} \max\{x, y\} dxdy$$


I am totally stuck on it. How can I solve this?
 A: Hint: break it into two parts, one where $x > y$ and the other $x \le y$.
A: $$
\begin{align*}
\int_0^1 \int_0^1 \max(x, y) dx dy &= \int_0^1 \int_0^y y dx dy + \int_0^1 \int_y^1 x dx dy \\
&= \int_0^1 y^2 dy + \int_0^1\frac{1 - y^2}{2} dy \\
&= \frac{1}{2}\int_0^1 (y^2 + 1) dy \\
&= \frac{1}{2}\left( \frac{1}{3} + 1\right) = \frac{2}{3}
\end{align*}$$
A: In general, the integral $$I_n = \int_{[0,1]^n} \max\{x_1,x_2,\ldots,x_n\} dx_1 dx_2 \ldots dx_n$$can be written as
$$I_n = n \int_{x_1=0}^1 \int_{0\leq x_2,x_3,\ldots,x_n \leq x_1} x_1 dx_1 dx_2 \cdots dx_n = n \int_{x_1=0}^1 x_1^n dx_1 = \dfrac{n}{n+1}$$
since $$\max\{x_1,x_2,\ldots,x_n\} = \begin{cases} x_1; & x_k \leq x_1\\ x_2; & x_k \leq x_2\\ \vdots& \vdots\\  x_n; & x_k \leq x_n\end{cases}$$
and this divides $[0,1]^n$ into $n$ regions.
A: In general, the integral
$$I_n=\int_{[0,1]^n}\max\{x_1,\cdots,x_n\}\,\mathrm{d} x_1\cdots\mathrm{d}x_n$$
can be viewed as an integral of the largest order statistic, say $X^{(1)}$.
Let $x^{(1)}=\max \{x_1,\cdots,x_n\}$, the distribution of $X^{(1)}$ will be $$F_{X^{(1)}}(x)=F_{X_1}(x)\times\cdots\times  F_{X_n}(x)=x^n.$$
Hence
$$I_n=\int_{[0,1]^n}\max\{x_1,\cdots,x_n\}\,\mathrm{d} x_1\cdots\mathrm{d}x_n=\int_{[0,1]}x\,\mathrm{d}F_{X^{(1)}}(x)=\int_0^1 nx^n\,\mathrm{d}x=\frac{n}{n+1}.$$
