Thomas' Calculus 13th Edition

$\dfrac{13}{2}$
As we know that $ds=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$ Here, $ds=\sqrt{(2)^2+(1)^2+(-2)^2} dt$ This implies that $ds= \sqrt 9 dt=3 dt$ Line integral: $l=\int_C (xy+y+z) ds=\int_0^1 ((2t)(t)+t+2-2t) (3) dt$ or, $3\int_0^1 (2t^2-2-t) dt=3[(\dfrac{2t^3}{3})+2t-(\dfrac{t^2}{2})]_0^1$ and $3(\dfrac{2}{3}+2-\dfrac{1}{2})=\dfrac{13}{2}$