= Solution
Set $P=10x^4-2xy^3$ and $Q=-3x^2y^2$. Their <partial derivatives> satisfy $P_y=-6xy^2=Q_x$ throughout $\mathbb R^2$. The <vector field> is smooth on the entire <simply connected> <plane>, so zero <curl> proves <path independence>.
Choose the broken straight path $(0,0)\to(2,0)\to(2,1)$. Along its horizontal segment $y=0$ and $dy=0$, giving
$$
I_1=\int_0^2 10x^4\,dx=64.
$$
Along its vertical segment $x=2$ and $dx=0$, giving
$$
I_2=\int_0^1-12y^2\,dy=-4.
$$
Therefore the <line integral> is
$$
\boxed{I=I_1+I_2=60.}
$$
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2021-resit-path-integral.png]
{title=Original sketch of the chosen oriented path through the corner at (2,0)}
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