Why did we divide the region R into three regions?

Question

Here I cannot understand why we divided the region R into three while computing the expected value of |x-y|

Answer 1

When $1<x<2$, $y$ is necessarily less than $x$. That's one region. when $0<x<1$, there are two sub-regions: $y<x$ (that's the second integral) and $y>x$ (that's the third).

Answer 2

The entire region of integration includes some points where $x - y$ is positive and some where $x - y$ is negative. This makes it very awkward to integrate in just one integral; do you see why?

So we are going to have at least two regions of integration. One region is the triangle with vertices $(0,0),$ $(0,1),$ and $(1,1),$ and the other is the trapezoid with vertices $(0,0),$ $(1,1),$ $(2,1),$ and $(2,0).$

If you decide the outer integral should be the integral over $dx,$ and the inner integral will integrate over $dy,$ then it is hard to integrate over the trapezoid using just one region: for $0 < x < 1,$ the upper bound on $y$ is given by $y = x,$ but for $1 < x < 2$ it is given by $y = 1.$

But actually it is not necessary to use three regions (one for the triangle and two for the trapezoid); the writer of this solution merely found it convenient.

It turns out that two regions actually are sufficient. You can integrate over the trapezoid this way: $$ \frac35 \int_0^1 dy \int_y^2 dx \, (x - y)x(y+y^2). $$

You can substitute this integral into the solution instead of the sum of integrals $$\frac35 \int_1^2 dx \int_0^1 dy \, (x - y)x(y+y^2)+ \frac35 \int_0^1 dx \int_0^x dy \, (x - y)x(y+y^2).$$