confusion on determining integral limits in bivarite distribution

$f(x_1,x_2)= 8x_1x_2 , 0<x_1<x_2<1 $ $\text{and}$ $0 , \text{elsewhere}$

I want to find $E(X_1X_2^2).$

According to the book: $$E(X_1X_2^2) = \int_{0}^{1}\int_{0}^{x_2} (x_1x_2^28x_1x_2)dx_1dx_2$$

My question is about the border of $x_2$. Accoding to book , integral limits of $x_2$ is $(0,1)$ ,but i think that it is wrong ,because the inequalities are strict.For example , if $x_1 =0.13$ , then how can we assume that $x_2$ can be any value less than $0.13$ ,so i think that the limits should have been $(x_1,1)$ .Hence ,the correct one must be : $$E(X_1X_2^2) = \int_{x_1}^{1}\int_{0}^{x_2} (x_1x_2^28x_1x_2)dx_1dx_2$$

Am i correct ? If not ,can you please explain clearly , i am a beginner in this topic. Thanks in advance

Let's describe the domain, what are the values that $x_2$ can take? It takes value between $0$ and $1$.

Now, let's describe the values that $x_1$ can take in terms of $x_2$. We have $0 < x_1<x_2$.

In summary, we want to have $0<x_2 < 1$ and $0<x_1 < x_2$.