VC4.4

Here,C is the straight line joining (0,0,0) and (1,1,1) whose equations are

$\frac{x-0}{1-0}=\frac{y-0}{1-0}=\frac{z-0}{1-0}=t\,$ (say)

Therefore,on C, x=t,y=t,z=t and so $r=xi+yj+zk=ti+tj+tk,dr=(i+j+k)dt\,$

Also, $F=(3t^2+6t)i-14t^2j+20t^3k\,$

The given points correspond to t=0 and t=1.

Therefore, $\int_{C} F\cdot\,dr=\int_{0}^{1} [(3t^2+6t)i-14t^2j+20t^3k]\cdot(i+j+k)dt\,$

= $\int_{0}^{1} [(3t^2+6t)-14t^2+20t^3]\,dt=[t^3+3t^2-\frac{14t^3}{3}+5t^4]_{0}^{1}=\frac{13}{3}\,$