Solution
= Solution
{c}
For the <parametric curve> $\mathbf r(t)=(t,t^2,t^3)$, $0\leq t\leq1$, the <vector field> and the tangent vector are
$$
\mathbf F(\mathbf r(t))=(9t^2,-14t^5,20t^7),\qquad
\mathbf r'(t)=(1,2t,3t^2).
$$
Consequently the <line integral> is
$$
\boxed{\int_C\mathbf F\cdot d\mathbf r
=\int_0^1(9t^2-28t^6+60t^9)\,dt
=3-4+6=5.}
$$