Solution
= Solution
Put $q=2t-t^2$. The second coordinate is exactly $q^2$, so this is the same parabola as in part (i), with a different increasing parametrization. Since $q$ runs from zero to one and $dq=(2-2t)dt$, the <line integral> is
$$
\int_0^1 5q^3\,dq=\boxed{\frac54}.
$$
Equivalently, substitute the original parameter to get $\int_0^15(2t-t^2)^3(2-2t)dt$. This demonstrates that a <line integral> depends on the oriented path, not on its speed of traversal.