Answer
$4$
Work Step by Step
Consider $I=\iint_{D} (2+x^2y^3-y^2 \sin x) \ dx \ dy ....(a)$
and $D=\left\{ (x, y) | |x|+|y| \leq 1 \right\}
$
Let us set $ x=-u$ and $ y=-v$
Therefore, $\iint_{D} f(x,y) dA=\iint_{R} (2+(-u)^2(-v)^3-(-v)^2 \sin (-u)) (-du) \ (-dv) \\=\iint_{R} (2-u^2v^3+v^2 \sin u) \ du \ dv$
or, $I= \iint_{R} (2-x^2y^3+y^2 \sin x) \ dx \ dy...(b)$
After adding equations (a) and (b), we get:
$I=2 \iint_{D} d A$
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 $2$. This implies that $ \iint_{D} d A =2$
Thus, $I=2 \iint_{D} d A=(2)(2)=4$
