VC2.4

$r\cdot s=(ti-t^2j+(t-1)k)\cdot (2t^2i+6tk)\,$

= $2t^3-t^2+6t(t-1)\,$

$\int_0^2 r\cdot s \,dt=\int_0^2 {2t^3-t^2+6t^2-6t}\,dt\,$

$[\frac{t^4}{2}-\frac{t^3}{3}+2t^3-3t^2]\,$ At [0,2]

$8+16-12=12\,$

$r\times s=\begin{vmatrix} i & j & k \\ t & -t^2 & t-1 \\ 2t^2 & 0 & 6t \end{vmatrix}\,$

= $-6t^3i+[6t^2-2t^2(t-1)]j+2t^4k\,$

= $-6t^3i-(8t^2-2t^3)j+2t^4k\,$

Therefore,

$\int_0^2 r\times s\, dt=\int_0^2 [-6t^3i-(8t^2-2t^3)j+2t^4k]\,dt\,$

= $-6i\int_0^2 t^3\,dt-j\int_0^2 (8t^2-2t^3)\,dt+2k\int_0^2 t^4\,dt\,$

= $-6i[\frac{t^4}{4}]-j[\frac{8t^3}{3}-\frac{2t^4}{4}]+2k[\frac{t^5}{5}]\,$ At [0,2]

= $-24i-\frac{40}{3}j+\frac{64}{5}k\,$