Answer
$$1$$
Work Step by Step
Green's Theorem states that: $\int_C fdx+g dy=\iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA$
Here, we have $f=xy ; g =x^2+x$
Now, $\dfrac{\partial g}{\partial x} =2x+1$ and $\dfrac{\partial f}{\partial y}=x$
Therefore,
$ \iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA=\int_{0}^1\int_{y-1}^{1-y} (2x+1-x) \ dx \ dy \\=\int_{0}^1 [x^2+x]_{y-1}^{1-y} \ dy\\=\int_0^1 (2-2y) \ dy \\=[2y-y^2]_0^1 \\=2(1-0)-(1-0) \\=1$
