= Solution
A <conservative vector field> on $\mathbb R^n$ is a <vector field> $\mathbf V=\nabla F$ for a globally defined smooth <potential of a conservative vector field> $F$. The <chain rule> gives
$$
\int_C\mathbf V\cdot d\mathbf x=F(\text{endpoint})-F(\text{startpoint}),
$$
so its <line integral> is <path independent>.
<Green theorem> states that, for a bounded planar region $D$ with piecewise smooth boundary and continuously differentiable $P,Q$ on a neighbourhood of its closure,
$$
\boxed{\oint_{\partial D}(P\,dx+Q\,dy)=\iint_D(Q_x-P_y)\,dx\,dy.}
$$
The boundary is positively oriented: the region lies on the left while it is traversed, so outer components run counterclockwise and hole boundaries clockwise.
For the proposed <potential of a conservative vector field>, the <fundamental theorem of calculus> and <differentiation under the integral sign> give
$$
F_y(x,y)=Q(x,y),\qquad F_x(x,y)=P(x,0)+\int_0^yQ_x(x,s)\,ds.
$$
Since $Q_x=P_y$, the last integral equals $P(x,y)-P(x,0)$, including negative $y$ with the usual signed-integral convention. Hence \b[the proposed potential is global and]
$$
\boxed{\nabla F=(P,Q)=\mathbf V.}
$$
The fact that the <vector field> is defined on all of $\mathbb R^2$ ensures both straight integration segments stay inside its domain; a hole in the domain could obstruct a global potential.
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