James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 4.9 Antiderivatives 355
SOLUtion The motion is vertical and we choose the positive direction to be upward.
At time t the distance above the ground is sstd and the velocity vstd is decreasing. Therefore
the acceleration must be negative and we have
Taking antiderivatives, we have
astd − dv
dt − 232
vstd − 232t 1 C
To determine C we use the given information that vs0d − 48. This gives 48 − 0 1 C, so
vstd − 232t 1 48
The maximum height is reached when vstd − 0, that is, after 1.5 seconds. Since
s9std − vstd, we antidifferentiate again and obtain
sstd − 216t 2 1 48t 1 D
Using the fact that ss0d − 432, we have 432 − 0 1 D and so
Figure 5 shows the position function of
the ball in Example 7. The graph corroborates
the con clusions we reached:
The ball reaches its maximum height
after 1.5 seconds and hits the ground
after about 6.9 seconds.
500
sstd − 216t 2 1 48t 1 432
The expression for sstd is valid until the ball hits the ground. This happens when
sstd − 0, that is, when
216t 2 1 48t 1 432 − 0
or, equivalently, t 2 2 3t 2 27 − 0
Using the quadratic formula to solve this equation, we get
0 8
FIGURE 5
t − 3 6 3s13
2
We reject the solution with the minus sign since it gives a negative value for t. Therefore
the ball hits the ground after 3(1 1 s13 )y2 < 6.9 seconds.
n
1–22 Find the most general antiderivative of the function.
(Check your answer by differentiation.)
13. f sxd − 1 5 2 2 x
14. f std − 3t 4 2 t 3 1 6t 2
t 4
1. f sxd − 4x 1 7 2. f sxd − x 2 2 3x 1 2
3. f sxd − 2x 3 2 2 3 x 2 1 5x 4. f sxd − 6x 5 2 8x 4 2 9x 2
5. f sxd − xs12x 1 8d 6. f sxd − sx 2 5d 2
7. f sxd − 7x 2y5 1 8x 24y5 8. fsxd − x 3.4 2 2x s221
9. f sxd − s2 10. f sxd − e 2
11. f sxd − 3sx 2 2s 3 x 12. f sxd − s 3 x 2 1 xsx
15. tstd − 1 1 t 1 t 2
st
16. rsd − sec tan 2 2e
17. hsd − 2 sin 2 sec 2 18. tsvd − 2 cos v 2
3
s1 2 v 2
19. f sxd − 2 x 1 4 sinh x 20. f sxd − 1 1 2 sin x 1 3ysx
21. f sxd − 2x 4 1 4x 3 2 x
x 3 , x . 0
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