If an object moves along a straight line with position s(t), then its velocity is v(t) = s'(t), so
[tex]\int ^3_0v(t)dt=s(t_3)-s(t_0\text{)}[/tex]
is the net change of position or displacement during time period from t=3 to t=0 seconds.
Lets find the integral of our velocity equation:
[tex]\int ^3_0(-4t+4)dt=(-4\frac{t^2}{2}+4t)^3_0[/tex]
where the last part means the evaluation at t=3 and t=0. Then, we get
[tex]\int ^3_0(-4t+4)dt=-4\frac{3^2}{2}+4(3)-(-4\frac{0}{2}+4(0))=-18+12=-6[/tex]
Then, the displacement during time period from t=3 to t=0 seconds is equal to -6 meters
