Answer
$4(e^3-1)$
Work Step by Step
We begin with the line integral:
$$\int_{C}ye^x\,dx+2e^x\,dy$$
Green's Theorem states that:
$$\oint_CP\,dx+Q\,dy=\iint_{R}\bigg(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\bigg)dA$$
We can work out the integrand of the double integral as follows:
$$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2e^x-e^x=e^x$$
Rewriting the double integral as an iterated integral and solving, we get:
$$\int_{0}^{4}\bigg(\int_{0}^{3}e^x\,dx\bigg)dy
\\=\int_{0}^{4}\bigg[e^x\bigg]_{0}^{3}dy
\\=\int_{0}^{4}(e^3-1)\,dy
\\=\bigg[(e^3-1)\,y\bigg]_{0}^{4}
\\=4(e^3-1)$$
