Answer
$\dfrac{(e-1)^2}{e}$
Work Step by Step
Here, we have $f(x, y)=e^{x-y}$ and $g (x, y)=e^{y-x}$
This gives: $\dfrac{\partial f}{\partial x}=e^{x-y}$ and $\dfrac{\partial g}{\partial y}=e^{y-x}$
Next, we will compute the flux integral.
$\oint_C F \cdot n \ ds=\int_R (\dfrac{\partial f}{\partial x} +\dfrac{\partial g}{\partial y}) \ dA$
or, $=\int_{0}^{1} \int_0^{x} (e^{x-y}+e^{y-x}) \ dy \ dx$
or, $=\int_{0}^{1} (e^x-e^{-x}) \ dx$
or, $=\dfrac{e^2 }{e}-\dfrac{2e }{e}+\dfrac{1}{e}$
or, $=\dfrac{(e-1)^2}{e}$
