Solution
= Solution
For the <parametric curve>, $\mathbf r'(t)=(2t,3t^2)^T$ and
$$
\mathbf F(\mathbf r(t))=(t^6-2t^2,\ 2t^5-t^3)^T.
$$
The <line integral> is therefore
$$
\begin{aligned}
\int_C\mathbf F\cdot d\mathbf r
&=\int_0^1\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt\\
&=\int_0^1(8t^7-3t^5-4t^3)\,dt\\
&=1-\frac12-1=\boxed{-\frac12}.
\end{aligned}
$$