Let the arc length for a curve given in parametric form be
L = integral[a,b](ds)
where ds = sqrt(dx/dt^2 + dy/dt^2) dt
If x = 1/2t^2 and y = 1/9(6t + 9)^3/2
then
dx/dt = t and dy/dt = sqrt(6t + 9)
Plugging these into the arc length formula for parametric curves, we get
L = integral[a,b](sqrt(t^2 + 6t + 9)) dt
= integral[a,b](sqrt((t + 3)^2)) dt
= integral[a,b](t + 3) dt
= [a,b](1/2t^2 + 3t)
Going from 0 to 1 gives
L = 1/2(1)^2 + 3(1) - 0 = 7/2
so 7/2 is the length of the parametric curve given above from t = 0 to t = 1.