AB/BC 2. (a) Graph y = (x + ex) sin (x^2 ) in [1, 3] × [−15, 20], Note that y represents velocity v and x
represents time t.
(b) The object moves to the left when the velocity is negative, namely, on the interval p < t <
r. Use the calculator to solve (x + ex)(sin (x^2 )) = 0; then p = 1.772 and r = 2.507. The
answer is 1.772 < t < 2.507.
(c) As the object moves to the left (with v(t) negative), the speed of the object increases
when its acceleration v ′(t) is also negative, that is, when v(t) is decreasing. This is true
when, for example, t = 2.
(d) The displacement of an object from time t 1 , to time t 2 , is equal toEvaluate this integral on the calculator; to three decimal places the answer is 4.491. This
means that at t = 3 the object is 4.491 units to the right of its position at t = 1, given to be x
= 10. Hence, at t = 3 the object is at x = 10 + 4.491 = 14.491.
(Review Chapter 8)