Expectation and variance of range (x(n)-x(1)) uniform I am working on calculating the expectation and then variance of the range from a Uniform(-theta, theta) distribution, but have gotten stuck. 
Basically the first page I show how I get the pdf and cdf for minimum and maximum. The second page I plug in to the joint distribution for continuous statistics using the cdfs and PDFs of minimum and maximum. 


Any help with that last line and beyond would be a great help in calculating the expectation and variance.
 A: The picture of your working is a bit hard to read, so I am going to take the liberty of ignoring this and just answering your title question myself.  I will leave it to you to compare your own working to mine.  Let $X_1, ..., X_n \sim \text{IID U}(-\theta, \theta)$ as specified in your problem.  For the trivial case where $n=1$ the range is a constant equal to zero.  I will proceed for the non-trivial case where $n \geqslant 2$.  Applying Theorem 1.3 here, the joint distribution of the minimum $X_{(1)}$ and maximum $X_{(n)}$ is:
$$\begin{equation} \begin{aligned}
f_{X_{(1)}, X_{(n)}}(a, b) 
&= \frac{n!}{(n-2)!} (F(b)-F(a))^{n-2} f(a) f(b) \\[6pt]
&= n(n-1) \Big( \frac{b-a}{2 \theta} \Big)^{n-2} \frac{1}{2 \theta} \frac{1}{2 \theta} \\[6pt]
&= \frac{n(n-1)}{(2 \theta)^n} \cdot (b-a)^{n-2} \quad \quad \quad \text{for all } -\theta \leqslant a \leqslant b \leqslant \theta. \\[6pt]
\end{aligned} \end{equation}$$
Hence, letting $R_n = X_{(n)}-X_{(1)}$ be the range, for all arguments $0 \leqslant r \leqslant 2\theta$ we have:
$$\begin{equation} \begin{aligned}
f_{R_n}(r) 
&= \int \limits_{-\theta}^{\theta-r} f_{X_{(1)}, X_{(n)}}(a, a+r) da \\[6pt]
&= \int \limits_{-\theta}^{\theta-r} \frac{n(n-1)}{(2 \theta)^n} \cdot r^{n-2} da \\[6pt]
&= \frac{n(n-1)}{(2 \theta)^n} \cdot r^{n-2} (2 \theta - r) \\[6pt]
&= \frac{n(n-1)}{2 \theta} \cdot \Big( \frac{r}{2 \theta} \Big)^{n-2} \Big( 1 - \frac{r}{2 \theta} \Big) \\[6pt]
&= \frac{1}{2 \theta} \cdot \text{Beta} \Big( \frac{r}{2 \theta} \Big| n-1, 2  \Big) . \\[6pt]
\end{aligned} \end{equation}$$
From this expression, we get the range distribution $R_n \sim 2 \theta \cdot \text{Beta}(n-1,2)$.  From here it is easy to obtain the moments and other aspects of the distribution.  The mean and variance are:
$$\mathbb{E}(R_n) = 2 \theta \cdot \frac{n-1}{n+1} \quad \quad \quad \mathbb{V}(R_n) = (2 \theta)^2 \cdot \frac{2(n-1)}{(n+1)^2 (n+2)}.$$
As $n \rightarrow \infty$ we have $R_n \rightarrow 2 \theta$ in probability, which is in accordance with intuition.
