Answer
$\dfrac{13}{2}$
Work Step by Step
Since, we have $ds=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
or, $ds=\sqrt{(2)^2+(1)^2+(-2)^2} dt \implies ds= \sqrt 9 dt=3 dt$
Now, the line integral is: $\int_C (xy+y+z) ds=\int_0^1 ((2t)(t)+t+2-2t) (3) dt$
or, $=3\int_0^1 (2t^2-2-t) dt$
or, $=3[(2t^3/3)+2t-(t^2/2)]_0^1$
or, $=3[\dfrac{2}{3}+2-\dfrac{1}{2}]$
or, $=\dfrac{13}{2}$
