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