Find variance and expectation of $(n+1)X_{(1)}$ Consider a $X_{1} \dots X_{n}$ be a i.r.v with uniform distribution $[0,\theta)$. Now we may consider $Y = (n+1)X_{(1)} = (n+1)\min (\{ X_{i}\}^{n}_{i=1})$ 
Suppose we want to know variance and distribution of this r.v.
First of lets consider : $\mathrm{P}((n+1)^{2}X^{2}_{(1)} < x) = 1 - \mathrm{P}(\forall i | X_{i} \ge \frac{\sqrt{x}}{n+1}) = 1 -\prod(1-\frac{\sqrt{x}}{\theta(n+1)}) = 1-(1-\frac{\sqrt{x}}{\theta(n+1)})^{n}$
So we may find $f_{Y^{2}}(x) = \frac{n}{2\theta(n+1)\sqrt{x}}(1-\frac{\sqrt{x}}{\theta(n+1)})^{n-1}$. 
Then we may try to find second moment by $\mathrm{Var}(Y^{2}) = \int^{\theta}_{0} x^2 f_{Y^{2}}  \mathrm{d}x$.
Am I right? 
If yes , how can we find this integral? Or I should tend $n$ to infinity and consider limit in distribution of PDF?
UPD: 
Also it's easy to get that $f_{Y}(x) = \frac{n}{\theta(n+1)}(1-\frac{x}{\theta(n+1)})^{n-1}$
. So does we need to integrate $\int_{0}^{\theta} \frac{x}{\theta(n+1)}(1-\frac{x}{\theta(n+1)})^{n-1}$ ? 

You already know that
$$
f_{X_{(1)}}(x) = \frac n\theta \left(1 - \frac x\theta\right)^{n-1}.
$$
I'm going to ignore the scaling by $n+1$ as the hard part is the moments of $X_{(1)}$, not the scaling.
This means that
$$
E(X_{(1)}^p) = \frac n\theta\int_0^\theta x^p \left(1 - \frac x\theta\right)^{n-1}\,\text dx.
$$
Let $t = x/\theta$ so
$$
E(X_{(1)}^p) = \theta^p n\int_0^1 t^p \left(1 - t\right)^{n-1}\,\text dt = \theta^pn\,\text{B}(p+1, n)
$$
so you can work out the moments of your $Y$ in terms of the Beta function $\text B$. And for integer inputs, you can use its relation to the Gamma function to get some ratio of factorials. 
Here's how that'd go. 
$$
\text{B}(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}
$$
so
$$
n\text{B}(p+1, n) = n\frac{\Gamma(p+1)\Gamma(n)}{\Gamma(n+p+1)} = \frac{p!n!}{(n+p)!} = {n + p \choose n}^{-1}
$$
so overall
$$
E(X_{(1)}^p) = \theta^p {n + p \choose n}^{-1}
$$
when $p \in \mathbb N$.

My overall approach here is that I'd rather try to work out expectations with respect to a simpler distribution if I can, so my first attempt was to use the Law of the Unconscious Statistician to get the moments of $Y$ by using expectations with respect to $X_{(1)}$'s distribution. In this case it happened to work. I could have still used $u$-substitution to get to the same place even if I started by integrating with respect to $f_Y$ but this way was more direct and less work.
