Answer
$-\dfrac{1}{30}$
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 $f=x+y ; g =x^2-y$
Now, $\dfrac{\partial g}{\partial x} =2x$ and $\dfrac{\partial f}{\partial y}=1$
Therefore,
$ \iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA=\int_{0}^1\int_{x^2}^{\sqrt x} (2x-1) \ dy \ dx \\=\int_{0}^1(2x-1) (x^2-\sqrt x) \ dx\\=\int_0^1 -2x^3+x^2+2x^{3/2}-x^{1/2} \ dx \\=-\dfrac{1}{30}$
