VC2.6

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\int_2^3 [r\cdot \frac{dr}{dt}]\,dt=[\frac{r^2}{2}]=[\frac{1}{2}|r(t)|^2]\, at t=2,3

=\frac{1}{2}[|r(3)|^2-|r(2)|^2]\,

Hence,from given results r(2)=2i-j+2k,r(3)=4i-2j+3k\,

Therefore,|r(2)|^2=4+1+4=9,|r(3)|^2=16+4+9=29\,

So, \int_2^3 [r\cdot \frac{dr}{dt}]\,dt=\frac{1}{2}(29-9)=10\,

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