Answer
$4$
Work Step by Step
Let us set $ x=-m$ and $ y=-n$
Consider $I=\iint_{D} (2+x^2y^3-y^2 \sin x) \ dx \ dy ....(1)$
and $D=\left\{ (x, y) | |x|+|y| \leq 1 \right\}
$
Now, $\iint_{D} f(x,y) dA=\iint_{R} (2+(-m)^2(-n)^3-(-m)^2 \sin (-n)) (-dm) \ (-dn) \\=\iint_{R} (2-m^2n^3+m^2 \sin n) dm \space dn$
or, $I= \iint_{R} (2-x^2y^3+y^2 \sin x) \ dx \ dy...(2)$
Now, we will add equations (1) and (2):
So, $I=2 \iint_{D} d A$
and $\iint_{D} d A$ is the area of region $D$.
$D$ is the region inside a square of side length $\sqrt 2$, so the area is equal to $2$. This implies that $ \iint_{D} d A =2$
and $I=2 \times \iint_{D} d A=2 \times 2 =4$
