Let the pdf of $x$ be defined by $$ f(x) = \begin{cases} 0 &\mbox{ } elsewhere\ \\ \frac{4x}{\theta^2} & \mbox{for }0 < x \le\frac{\theta}{2} \\\frac{-4x}{\theta^2}+\frac{4}{\theta} & \mbox{for }\frac{\theta}{2}<x\le\theta \end{cases} $$ with $0<\theta\le2$ Find an estimator of $\theta$ by method of moments
I started computing the expected value of the first piece $$E(X_1) = \int_{0}^\frac{\theta}{2} x\frac{4x}{\theta^2}= \frac{\theta}{6}$$ and then for the second piece I get $$E(X_2) = \int_{\frac{\theta}{2}}^\theta x(\frac{-4x}{\theta^2}+\frac{4}{\theta})= \frac{\theta}{3}$$ at the end I added the two pieces together $E(X_1)+E(X_2)$ and got $\frac{\theta}{2}$ and equating it to $\bar X$ I got $\theta_{MOM} = 2\bar X$ is this the right approach to find method of moments when there are two domain to consider? Thank you in advance
Your computations are correct and you have the correct method of moments estimator, but the explanation is somewhat imprecise. The reason is that you have not defined what you mean by $X_1$ and $X_2$. Rather, $$\operatorname{E}[X] = \int_{x=0}^{\theta} x f_X(x) \, dx = \int_{x=0}^{\theta/2} x \cdot \frac{4x}{\theta^2} \, dx + \int_{x=\theta/2}^\theta x \left(\frac{4}{\theta} - \frac{4x}{\theta^2} \right) \, dx,$$ directly. There is no need to perform the computations separately nor introduce auxiliary random variables. The problem with saying something like $$\operatorname{E}[X_1] = \int_{x=0}^{\theta/2} x \cdot \frac{4x}{\theta^2} \, dx$$ is that on the interval $x \in [0, \theta/2]$, the function $4x/\theta^2$ is not a proper density: $$\int_{x=0}^{\theta/2} \frac{4x}{\theta^2} \, dx = \frac{1}{2},$$ whereas if $X_1$ were an actual random variable, this expression would have to integrate to $1$.
Alternatively, what you can write is $$\operatorname{E}[X] = \operatorname{E}[X \mid X \le \theta/2]\Pr[X \le \theta/2] + \operatorname{E}[X \mid X > \theta/2]\Pr[X > \theta/2],$$ and now if we define $$X_1 = X \mid X \le \theta/2, \quad X_2 = X \mid X > \theta/2,$$ these have densities $$f_{X_1}(x) = \frac{8x}{\theta^2}, \quad f_{X_2}(x) = \frac{8}{\theta} - \frac{8x}{\theta^2}.$$ But this is all rather circuitous because as I already pointed out, the first equation I wrote suffices. It is the same as saying, for instance, if we have any piecewise function, say $$h(x) = \begin{cases} 0, & x < 0 \\ 1, & 0 \le x \le 1, \\ x, & 1 < x \le 2, \end{cases}$$
then $$\int_{x=0}^2 h(x) \, dx = \int_{x=0}^1 1 \, dx + \int_{x=1}^2 x \, dx.$$