For an object whose velocity in ft/sec is given by v(t) = −2t^2 + 4, what is its distance travelled, in feet, on the interval t = 0 to t = 2 secs?

Question

I think the answer is 2.667 but I am not one hundred percent sure.

Answer 1

the distance is

s = ∫[0,2] -2t^2 + 4 dt
= -2/3 t^3 + 4t [0,2]
= -16/3 + 8
= 8/3

you are correct