= Solution
When $\alpha=2\beta$, all three values reduce to $-\beta/3$. To prove <path independence> for every piecewise smooth path, construct a global potential rather than infer it from agreement on only three paths. Integrating the coefficient of $dx$ and then matching the coefficient of $dy$ gives
$$
\Phi(x,y)=\beta\left(\frac23x^3+x^2y+\frac13y^3\right).
$$
Indeed $\Phi_x=2\beta(x^2+xy)=\alpha(x^2+xy)$ and $\Phi_y=\beta(x^2+y^2)$. The integrand is the <exact differential> $d\Phi$ throughout $\mathbb R^2$. The <fundamental theorem for line integrals> therefore gives, for every such path from $A$ to $B$,
$$
\boxed{I=\Phi(0,1)-\Phi(1,0)=\frac\beta3-\frac{2\beta}{3}=-\frac\beta3}.
$$
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