Answer
$\frac{4}{e}$
Work Step by Step
When the vector field $F(x,y)=Pi+Qj$ is a conservative field, then throughout the domain $D$, we have
$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$
$P$ and $Q$ are the first-order partial derivatives on the domain $D$.
Here, we have $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}=-2xe^{-y}$
This implies that the vector field $F$ is conservative.
Now, $f(x,y)=x^2e^{-y}+g(y)$
$f_y(x,y)=-x^2e^{-y}+g'(y)$
Here, $g(y)=y^2+C$
where $C$ is a constant.
Thus, $f(x,y)=x^2e^{-y}+y^2+C$
Now, $\int_C F \cdot dr =f(2,1)-f(1,0)=(1+4e^{-1}+C)-(1+C)=\dfrac{4}{e}$
