Find multiple integrals $I_{\max}(k,n)$ and $I_{\min}(k,n)$ in various ways 
*

*$I_{\max}(k,n)=\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_k\left(\max\limits_{1\le i\le k}x_i\right)^n\,dx_1dx_2\dots dx_k$

*$I_{\min}(k,n)=\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_k\left(\min\limits_{1\le i\le k}x_i\right)^n\,dx_1dx_2\dots dx_k$


Various solutions are welcome.
P.S. I added my solution as an answer.
 A: Denote by $C$ the part of the unit cube $[0,1]^k$ where $\max_{1\leq i\leq k}x_i=x_k$. The plane $x_k=x$ intersects $C$ in the set $$C_x=\{(x_1,\ldots,x_{k-1})\,|\,0\leq x_i\leq x \ (1\leq i\leq k-1)\}$$ of volume $x^{k-1}$. By Fubini's theorem we therefore obtain
$$I_\max(k,n)=k\int_C x^n {\rm d}(x_1,\ldots, x_{k-1},x)=k \int_0^1x^n\>x^{k-1}\>dx={k\over n+k}\ .$$
Similarly, denote by  $D$ the part of $[0,1]^k$ where $\min_{1\leq i\leq k}x_i=x_k$.  Then $$D_x:=\{(x_1,\ldots,x_{k-1})\,|\,x\leq x_i\leq 1 \ (1\leq i\leq k-1)\}$$ has volume $(1-x)^{k-1}$. By Fubini's theorem we therefore obtain
$$I_\min(k,n)=k\int_D x^n {\rm d}(x_1,\ldots, x_{k-1},x)=k \int_0^1x^n\>(1-x)^{k-1}\>dx= {n+k\choose k}^{-1}\ .$$
A: To show another approach, consider that in general we can write
$$
\eqalign{
  & I(f,A) = \int\limits_{(x_{\,1} , \cdots ,x_{\,k} )\, \in \,A} {f(x_{\,1} , \cdots ,x_{\,k} )\,dx_{\,1}  \cdots dx_{\,k} }  =   \cr 
  &  = \int\limits_{{\bf x}\, \in \,A} {f({\bf x})\,dA}  = 
 \int\limits_{{\bf x}\, \in \,A} {\left( {\int\limits_{z = 0}^{f({\bf x})} {dz} } \right)\,dA} \quad \left| {\;f: \mathbb R^{\,k}  \to \mathbb R} \right. \cr} 
$$
that means that $I(f,A)$  , in the space $(\bf x, z)$, is the
volume comprised between the plane $z=0$ and the surface $z=f(\bf x)$, 
and inside the cylinder ${{\bf x}\, \in \,A}$.
When $0 \le f(\bf x)$ in all the integration domain,  we can compute the volume by integrating over the cross sections $A(z)dz$, intending by
A(z) the "area" (measure) of the domain in which  $z < f({\bf x})$, i.e.
$$
A(z) = \mu \left( {{\bf x}\, \in \,A:f({\bf x}) = z} \right)
$$
so
$$
\eqalign{
  & I(f,A) = \int\limits_{(x_{\,1} , \cdots ,x_{\,k} )\, \in \,A} {f(x_{\,1} , \cdots ,x_{\,k} )\,dx_{\,1}  \cdots dx_{\,k} }  =   \cr 
  &  = \int\limits_{z = 0}^{\max \left( {f({\bf x})} \right)} {A(z)dz}  \cr} 
$$
Coming to our case, we have 
$$
\eqalign{
  & \left\{ \matrix{
  0 \le x_{\,j}  \le 1\quad \left| {\;j \in \left\{ {1, \cdots ,k} \right\}} \right. \hfill \cr 
  \left[ {z < \min (x_{\,1} , \cdots ,x_{\,k} )^{\,n} } \right]\;\mathop  = \limits^{0\, \ne \,n} \;\left[ {z^{\,1/n}  < \min (x_{\,1} , \cdots ,x_{\,k} )} \right] =  \hfill \cr 
   = \left( {\left[ {z^{\,1/n}  < x_{\,1}  \le 1} \right] \cdots \left[ {z^{\,1/n}  < x_{\,k}  \le 1} \right]} \right)\;\quad  \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad A_{\,\min } (z) = \left[ {0 \le z \le 1} \right]\left\{ {\matrix{
   1 & {0 = n}  \cr 
   {\left( {1 - z^{\,1/n} } \right)^{\,\,k} } & {0\, \ne \,n}  \cr 
 } } \right. \cr} 
$$
and
$$
\eqalign{
  & \left\{ \matrix{
  0 \le x_{\,j}  \le 1\quad \left| {\;j \in \left\{ {1, \cdots ,k} \right\}} \right. \hfill \cr 
  \left[ {z < \max (x_{\,1} , \cdots ,x_{\,k} )^{\,n} } \right]\;\mathop  = \limits^{0\, \ne \,n} \;1 - \left[ {\max (x_{\,1} , \cdots ,x_{\,k} ) \le z^{\,1/n} } \right] =  \hfill \cr 
   = 1 - \left[ {0 \le x_{\,1}  \le z^{\,1/n} } \right] \cdots \left[ {0 \le x_{\,k}  \le z^{\,1/n} } \right] \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad A_{\,\max } (z) = \left[ {0 \le z \le 1} \right]\left\{ {\matrix{
   1 & {0 = n}  \cr 
   {\left( {1 - z^{\,\,k/n} } \right)} & {0\, \ne \,n}  \cr 
 } } \right. \cr} 
$$
where $[P]$ denotes the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
   1 & {P = TRUE}  \\
   0 & {P = FALSE}  \\
 \end{array} } \right.
$$
Therefore
$$ \bbox[lightyellow] {  
\eqalign{
  & I_{\,\min } (k,n) = \int_0^1 {A_{\,\min } (z)dz}  = \int_0^1 {\left( {1 - z^{\,1/n} } \right)^{\,\,k} dz}  =   \cr 
  &  = n\int_0^1 {\left( {1 - t} \right)^{\,\,k} t^{\,\,n - 1} dt}  = nB(k + 1,n) = {{\Gamma (k + 1)\Gamma (n + 1)} \over {\Gamma (n + k + 1)}} = {1 \over {\left( \matrix{
  n + k \cr 
  k \cr}  \right)}} \cr} 
} $$
and
$$ \bbox[lightyellow] {  
I_{\,\max } (k,n) = \int_0^1 {A_{\,\max } (z)dz}  = \int_0^1 {\left( {1 - z^{\,\,k/n} } \right)dz}  = {k \over {k + n}}
} $$
Note that the method can be  applied to  integration
bounds different from $0 \cdots 1$, and  to real 
 values of $n$ (and also of $k$ ..)
