Answer
$4(e^3-1)$
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:
$\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}=2e^x-e^x=e^x$
Now, $\oint_CP\,dx+Q\,dy =\iint_{D}(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})dA=\int_{0}^{4}(\int_{0}^{3}e^x\,dx) \ dy$
or, $=\int_{0}^{4}[e^x\bigg]_{0}^{3}\ dy$
or, $=\int_{0}^{4}(e^3-1) \ dy$
or, $=\bigg[(e^3-1)\,y\bigg]_{0}^{4}$
or, $=4(e^3-1)$
