Solution
= Solution
For $\mathbf r(t)=(t,t^2,t^3)$, the endpoints correspond to $t=0,1$ and $\dot{\mathbf r}=(1,2t,3t^2)$. The <vector field> on the path is $\mathbf A=(9t^2,-14t^5,20t^7)$, so the <line integral> is
$$
\boxed{\int_C\mathbf A\cdot d\mathbf r=\int_0^1(9t^2-28t^6+60t^9)dt=3-4+6=5.}
$$
Including the derivative of each path coordinate is essential; merely substituting the path into the field does not supply the line element.