= Solution
A one-form $A(x,y)\,dx+B(x,y)\,dy$ is an <exact differential> if a single-valued differentiable <potential of a conservative vector field> $f$ satisfies $A=f_x$ and $B=f_y$ on the domain. Then it equals $df$, and its integral along a path is the endpoint difference of $f$.
Integrate $f_x=10x^4-2xy^3$ with respect to $x$:
$$
f(x,y)=2x^5-x^2y^3+g(y).
$$
Its $y$ derivative is $-3x^2y^2+g'(y)$. Matching the required second component <forces> $g'(y)=0$, so the additive constant is arbitrary and
$$
df=(10x^4-2xy^3)\,dx-3x^2y^2\,dy.
$$
The <fundamental theorem for line integrals> now gives
$$
\boxed{I=f(2,1)-f(0,0)=2(2^5)-2^2=60},
$$
independently of the convenient segmented path used before. Merely checking equality of mixed partial derivatives on an unspecified domain would not by itself construct this globally defined potential.
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