Determine the value of $\theta$. Given two random variables $X_1$ and $X_2$ with pdf
$$
f\left(x_i\right)= \begin{cases}\frac{1}{2 \theta}, & -\theta<x_i<\theta \\ 0, & x_i \text { otherwise }\end{cases}
$$
If it is known that $X_1$ and $X_2$ are independent and $\operatorname{Var}\left(X_1 X_2\right)=\frac{64}{9}$, determine the value of $\theta$.
My working:
If the random variable X comes from continuous data with pdf
$$\operatorname{Var}\left(X_1 X_2\right)=\frac{64}{9}$$
$$\operatorname{Var}\left(X_1 X_2\right)=\int(x-\mu)^2f(x) dx$$
for $\mu$ is the expected value. For example
$$\mu=\int xf(x) dx$$
if subtitute $f(x)=\frac{1}{2 \theta}$, thus
$$\mu=\int x\frac{1}{2 \theta} dx$$
$$\mu=\frac{x^2}{4 \theta}+C$$
So
$$\operatorname{Var}\left(X_1 X_2\right)=\int(x-\mu)^2f(x) dx$$
$$\operatorname{Var}\left(X_1 X_2\right)=\int(x-\frac{x^2}{4 \theta})^2\frac{1}{2 \theta} dx$$
$$\operatorname{Var}\left(X_1 X_2\right)=\int(x^2-\frac{x^3}{2 \theta}+\frac{x^4}{16 \theta})\frac{1}{2 \theta} dx$$
$$\frac{64}{9}=(\frac{x^3}{3}-\frac{2x^4}{\theta}+\frac{5x^5}{16 \theta})\frac{1}{2 \theta}$$
$$\frac{64}{9}=(\frac{x^3}{6 \theta}-\frac{x^4}{\theta^2}+\frac{5x^5}{32 \theta})$$
How the next step and please correct if i'm wrong for my work. Thank u

Note that $EX_1 = EX_2 = 0$, hence
$$
\begin{align}
Var (X_1 X_2) &= E(X_1X_2)^2 - \left(E\left(X_1X_2\right)\right)^2 \\
&= EX_1^2 EX_2^2 - (EX_1 EX_2)^2 \\
&=(EX_1^2)^2 \\
&= \left(\int_{-\theta}^\theta\frac{x^2}{2\theta}  dx\right)^2 \\
&= \left(\frac{\theta^2}{3}\right)^2 \\
\end{align}
$$
Solve equatioin $\left(\frac{\theta^2}{3}\right)^2 = \frac{64}9$,
derive $\theta = 2\sqrt2$
