Calculus intergrals

Question

A particle moves in a straight line with the given velocity 𝑣(𝑡)=3𝑡^2−18𝑡+15 (in m/s). Find the displacement and distance traveled over the time interval [0,10].

(Give your answers as whole or exact numbers.)

Answer 1

v(t)=3t^2-18t+15=3(t-1)(t-5)

displacement: take integral from 0-10. Integral (3t^2-18t+15)dt = [t^3-9t^2+15t] evaluated from 0 to 10 = 250

distance you will have to break down three parts: v(t)<0 from 1 to 5, v(t)>0 from 0 to 1 and 5 to 10, so you have to set integral into three parts, from 0 to 1, 1 to 5 and 5 to 10

distance = integral from 0 -> 1, repeat function (3t^2-18t+15)dt and get antiderivative plug in 0 and 1

+ integral from 1 -> 5, -(3t^2-18t+15) dt and get antiderivative -(t^3-9t^2+15t) plug in 1 and 5, + integral from 5 -> 10, repeat function (3t^2-18t+15)dt and get antiderivative (t^3-9t^2+15t) plug in 5 and 10 = 7 + 32 + 275 = 314