ce amaths -- integration

A ball is thrown vertically upwards from the ground with a velocity v m/s, where v=4t-2t^2, at the time t seconds after prorojection.Find
a) the distance of the ball from the ground at the time t seconds after projection.
ds/dt = v = 4t - 2t^2
s = 2t^2 - 2t^3/3 + C, C is constant
0 = 2(0)^2 - 2(0)^3/3 + C, C = 0
s = 2t^2 - 2t^3/3
b)the maximum height of the ball bfrom the ground.
v = 4t - 2t^2 = 0
(2 - t)t = 0
ball reachs maximum at t = 2
maximum height = 2(2)^2 - 2(2)^3/3 = 8/3 m
c) the distance it travels from t=2 to t= 2.5. (Give your answers correct to 2 d.p.)
8/3 - [ 2(2.5)^2 - 2(2.5)^3/3 ] = 0.58 m