Answer
$\dfrac{1}{3}$
Work Step by Step
Green's Theorem states that:
$\oint_CP\,dx+Q\,dy=\iint_{D}(\dfrac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$
We set up the line integral and find out the integrand of the double integral as follows:
$\oint_C (y+e^{\sqrt {x}} ) dx+(2x+\cos y^2) dy=\iint_{D}(\dfrac{\partial (2x+\cos y^2)}{\partial x}-\dfrac{\partial (y+e^{\sqrt {x}} ) }{\partial y})dA$
or, $=\int_{0}^{1}(\int_{x^2}^{\sqrt x} (2-1)) dy dx$
or, $=\int_{0}^{1} [y]_{x^2}^{\sqrt x} dx$
or, $=\int_0^1 [\sqrt x-x^2 ]dx$
or, $=[\dfrac{2 x^{3/2}}{3}-\dfrac{x^3}{3}]_0^1$
or, $=\dfrac{1}{3}$
