Answer
$$\int_C(\sqrt {1+x^{3}}dx+2xydy=3$$
Work Step by Step
Green's Theorem states that
$$\int_C Pdx+Qdy=\int\int_D(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$$
$$\int_C(\sqrt {1+x^{3}}dx+2xydy=\int\int_D2ydy$$
$$=\int_{0}^{1}\int_{0}^{3x}2ydydx$$
$$=\int_{0}^{1}9x^2dx$$
$$=3$$
Hence, the result is
$$\int_C(\sqrt {1+x^{3}}dx+2xydy=3$$
