Answer
$20$
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: $x= 4t \implies \dfrac{dx}{dt}=4$
and $y= 3t \implies \dfrac{dy}{dt}=3$
Therefore,
$ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}=\sqrt {4^2+3^2} dt=5 dt$
Now, the line integral is: $\int_C xy ds=\int_0^1 (4t)(3t) (5) dt=\int_0^1 60 t^2 dt=[\dfrac{60t^3}{3}]_0^1=60(\dfrac{1}{3})=20$
