We have to be careful that the presence of a timing for which v = 0 may likely mean that there is a change in direction (just as a graph passes the x-axis to change from positive to negative or vice versa).

Here, we find at there is a change in direction at t = 2.16. As such, we cannot just integrate from t = 0 to t = 3 directly since the negative portion of the integral (from t = 2.16 to t = 3) will cancel out, rather than add on to, the positive portion of the integral (from t = 0 to t = 2.16).