Solution
= Solution
The same <beta integral> with one additional power of $y$ gives
$$
\mathbb E Y_{(r)}^2=\frac{r(r+1)}{(n+1)(n+2)}.
$$
Subtracting the square of the mean yields
$$
\operatorname{Var}(Y_{(r)})=\frac{r(n+1-r)}{(n+1)^2(n+2)}.
$$
At $r=1$ and $r=n$, therefore,
$$
\boxed{\operatorname{Var}(Y_{(1)})=\operatorname{Var}(Y_{(n)})=\frac n{(n+1)^2(n+2)}.}
$$
Their equality also follows by reflecting every uniform observation via $y\mapsto1-y$.