Answer
$0$
Work Step by Step
Need to compute the derivative for the parametric vector equation.
Since, $ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$
We have: $x= t \implies \dfrac{dx}{dt}=1$
and $y= 2-t \implies \dfrac{dy}{dt}=-1$
Therefore, $ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}=\sqrt {(1)^2+(-1)^2} dt=\sqrt 2 dt$
Now, the line integral is:
$\int_C 3(x-y) ds=\int_0^2 3[t-2+t)(\sqrt 2) dt= 6\sqrt 2 \int_0^2 (t-1) dt= 6 \sqrt 2 [\dfrac{t^2}{2}-t]_0^2=6 \sqrt 2(\dfrac{4}{2}-2)=0$
