How do I show that $Var(Y) = n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}$? My original pdf is $f(y) = \frac{n (y_{n} - \theta_{1})^{n-1}}{(\theta_{2} - \theta_{1})^{n}}$ for $\theta_{1} < y < \theta_{2}$.
After using U-substitution, I obtain $E(Y) = \frac{n \theta_{2} + \theta_{1}}{(n+1)}$. 
For variance of $Y$, I need to find $E(Y^2)$ first. 
This is what I have with U-substitution: $E(Y^2) = \frac{n\theta_{2} - n \theta_{1} + 2n \theta_{1} \theta_{2} - n\theta_{1}^{2} + \theta_{1}^{2}}{(n+1)}$.
But with this the $Var(Y)= E(Y^2) - \left [E(Y)  \right ]^{2} \neq n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}$
My mistake is definitely the $E(Y^2)$ part, so I would appreciate it if someone here can go over it with me. 
Edit:
$$\begin{eqnarray}
   \mathbb{E}\left(Y^2\right) &=& \mathbb{E}\left(\left(Y-\theta_1\right)^2\right) + 2 \theta_1 \cdot \mathbb{E}\left(\left(Y-\theta_1\right)\right) + \theta_1^2 \\ &=& \frac{n}{n+2} \left(\theta_2-\theta_1\right)^2 + 2 \theta_1 \cdot \frac{n}{n+1} \left(\theta_2-\theta_1\right) + \theta_1^2
\end{eqnarray}$$ from below and also my own work.
But now how do I simplify $$Var(Y) =  \mathbb{E}\left(Y^2\right) - \left [ \mathbb{E}\left(Y\right)  \right ]^{2}$$ to  $$Var(Y) = n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}?$$

Instead of $\mathbb{E}(Y^2)$ consider 
$$ 
  c_2 =\mathbb{E}\left(\left(Y-\theta_1\right)^2\right) = \int_{\theta_1}^{\theta_2} \frac{n}{\left(\theta_2-\theta_1\right)^n} \left(y-\theta_1\right)^{n+1} \mathrm{d} y = \frac{n}{n+2} \left(\theta_2-\theta_1\right)^2
$$
Similarly
$$
 c_1 = \mathbb{E}\left(\left(Y-\theta_1\right)\right) = \frac{n}{n+1} \left(\theta_2-\theta_1\right)
$$
Now clearly
$$\begin{eqnarray}
   \mathbb{E}\left(Y^2\right) &=& \mathbb{E}\left(\left(Y-\theta_1\right)^2\right) + 2 \theta_1 \cdot \mathbb{E}\left(\left(Y-\theta_1\right)\right) + \theta_1^2 \\ &=& \frac{n}{n+2} \left(\theta_2-\theta_1\right)^2 + 2 \theta_1 \cdot \frac{n}{n+1} \left(\theta_2-\theta_1\right) + \theta_1^2
\end{eqnarray}$$
