= Solution
An <exact differential> is an expression $P\,dx+Q\,dy$ that equals $df=f_x\,dx+f_y\,dy$ for a single-valued differentiable <function> $f$ on the domain. Its <line integral> is an endpoint difference by the <fundamental theorem for line integrals>.
Integrating $f_x=P$ with respect to $x$ gives $f=2x^5-x^2y^3+g(y)$. Matching $f_y=-3x^2y^2+g'(y)$ to $Q$ requires $g'(y)=0$. Hence
$$
df=d(2x^5-x^2y^3),\qquad
\boxed{I=f(2,1)-f(0,0)=64-4=60.}
$$
The arbitrary additive constant cancels. This constructs the <vector-field potential> globally and provides a second proof of <path independence>.
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