= Solution
A vector <line integral> along an oriented parametrized curve $r(t)$ is $\int a(r(t))\cdot r'(t)\,dt$; for a force field it is the work along that path. On the upper branch, set $y=t$, $x=t^2$, $0\le t\le1$. Then
$$
a\cdot dr=(t^2-t)(2t\,dt)+(t^2+t)dt
=(2t^3-t^2+t)dt,
$$
so
$$
\boxed{\int_Ca\cdot dr=\left[\frac{t^4}{2}-\frac{t^3}{3}+\frac{t^2}{2}\right]_0^1=\frac23.}
$$
The stated $x$ endpoints do not distinguish the two branches of $y^2=x$. If the lower branch is intended, the same parametrization runs from $t=0$ to $t=-1$, giving instead $\boxed{4/3}$. The two values correspond to different terminal points, $(1,1)$ and $(1,-1)$; choosing a branch is necessary to make the question unambiguous.
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