Answer
$\dfrac{31}{4}$
Work Step by Step
When $F(x,y)=pi+qj$ is a conservative field, then throughout the domain $D$, we get
$\dfrac{\partial p}{\partial y}=\dfrac{\partial q}{\partial x}$
$p$ and $q$ are the first-order partial derivatives on the domain $D$.
Here, we have $f_x(x,y)=x^3$ and $f_y(x,y)=y^3$
$f(x,y)=\dfrac{x^3}{4}+g(y)$ [g(y) : A function of y]
$f_y(x,y)=y^3+g'(y)$
Here, $g(y)=k$
Thus, $f(x,y)=\dfrac{x^4}{4}+\dfrac{y^4}{4}+k$
Now, $W=\int_C F \cdot dr =f(2,2)-f(1,0)=\dfrac{32}{4}-\dfrac{1}{4}=\dfrac{31}{4}$
