Answer
$\dfrac{1}{3}$
Work Step by Step
Green's Theorem states that:
$\oint_C M \,dx+ N \,dy=\iint_{D}(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y})dA$
The integrand of the double integral becomes:
$\oint_C (y+e^{\sqrt {x}} ) dx+(2x+\cos y^2) dy=\iint_{D}(\dfrac{\partial (2x+\cos y^2)}{\partial x}-\dfrac{\partial (y+e^{\sqrt {x}} ) }{\partial y}) \ dA$
$=\int_{0}^{1} \int_{x^2}^{\sqrt x} (2-1) \ dy \ dx$
$=\int_{0}^{1} [y]_{x^2}^{\sqrt x} \ dx$
$=\int_0^1 [\sqrt x-x^2 ] \ dx$
$=\dfrac{2}{3} [ x^{(3/2)}]_0^1-\dfrac{1}{3}[x^3]_0^1$
$=\dfrac{1}{3}$