The diagram below shows a large square with two smaller squares within it.
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Denote by 
the maximum order statistic, with 
, and similarly denote by 
the minimum order statistic. Then the CDF for is
In order for there to be some
that exceeds the value of , it must be true that exceeds the value of all the 
, so the above is equivalent to the joint probability
and since the are i.i.d., we have
where
. We have
and so
![\implies\mathbb E[X_{(n)}]=\displaystyle\int_0^1xnx^{n-1}\,\mathrm dx=n\int_0^1x^n\,\mathrm dx=\frac n{n+1}](https://tex.z-dn.net/?f=%5Cimplies%5Cmathbb%20E%5BX_%7B%28n%29%7D%5D%3D%5Cdisplaystyle%5Cint_0%5E1xnx%5E%7Bn-1%7D%5C%2C%5Cmathrm%20dx%3Dn%5Cint_0%5E1x%5En%5C%2C%5Cmathrm%20dx%3D%5Cfrac%20n%7Bn%2B1%7D)
Using similar reasoning, we can find the CDF for. We have
![\implies\mathbb E[X_{(1)}]=\displaystyle\int_0^1xn(1-x)^{n-1}\,\mathrm dx=\frac1{n+1}](https://tex.z-dn.net/?f=%5Cimplies%5Cmathbb%20E%5BX_%7B%281%29%7D%5D%3D%5Cdisplaystyle%5Cint_0%5E1xn%281-x%29%5E%7Bn-1%7D%5C%2C%5Cmathrm%20dx%3D%5Cfrac1%7Bn%2B1%7D)
In order for there to be some
and since the
where
and so
Using similar reasoning, we can find the CDF for