Integral calculus

Question

The position vector in the xy plane is [t2,sin(t)]. What is the distance travelled by the particle from x=0 to x=Pie

Answer 1

The position vector R = xi + yj = t²i + sin(t)j

at x = 0 = t², then t = 0 so y = sin(0) =0 so the initial point is (0,0)

at x = pi = t², then t = (pi)^.5 so y = sin((pi)^.5 = .9797 so the final point is (pi, .9797)

But (ds/dt)² = (dx/dt)² + (dy/dt)² and s = Integral of SQRT(4t² + cos2(t)) dt from 0 to SQRT(pi)

Integrating I get distance traveled = S = 3.552