= Solution
The proposed <parametric equation> traces precisely the same oriented segment, with a different speed. Set $u=t^n$, which increases from zero to one, and $du=nt^{n-1}\,dt$. Direct substitution gives the <line integrals>
$$
\begin{aligned}
\int\mathbf F\cdot d\mathbf x&=\int_0^1(2-2t^n)nt^{n-1}\,dt=\int_0^1(2-2u)\,du=1,\\
\int\mathbf G\cdot d\mathbf x&=\int_0^1(2-4t^n+2t^{2n})nt^{n-1}\,dt=\int_0^1(2-4u+2u^2)\,du=\frac23.
\end{aligned}
$$
Thus \b[both values are unchanged]. The <derivative> of $u$ vanishes at the initial endpoint for $n>1$, but the monotone <integration by substitution> is still valid. This is invariance under <reparametrization> of one curve, which applies even to a nonconservative <vector field>; it is not evidence of general path independence.
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