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

Section 5.4 Indefinite Integrals and the Net Change Theorem 407
where the object moves in the positive direction, but now we have proved that it is
always true.
● If we want to calculate the distance the object travels during the time interval,
we have to consider the intervals when vstd > 0 (the particle moves to the right)
and also the intervals when vstd < 0 (the particle moves to the left). In both cases
the distance is computed by integrating | vstd | , the speed. Therefore
3
y t2
t 1
| vstd | dt − total distance traveled
√
√(t)
A¡
A£
0 t¡ A t t
FIGURE 3
Figure 3 shows how both displacement and distance traveled can be interpreted
in terms of areas under a velocity curve.
displacement − y t2
distance − y t2
t 1
vstddt − A 1 2 A 2 1 A 3
t 1
| vstd | dt − A 1 1 A 2 1 A 3
● The acceleration of the object is astd − v9std, so
y t2
astd dt − vst 2 d 2 vst 1 d
t 1
is the change in velocity from time t 1 to time t 2 .
Example 6 A particle moves along a line so that its velocity at time t is
vstd − t 2 2 t 2 6 (measured in meters per second).
(a) Find the displacement of the particle during the time period 1 < t < 4.
(b) Find the distance traveled during this time period.
SOLUTION
(a) By Equation 2, the displacement is
ss4d 2 ss1d − y 4
vstd dt − y 4
st 2 2 t 2 6d dt
1
−F t 3
3 2 t 2
4
2 2 6t G1
1
− 2 9 2
This means that the particle moved 4.5 m toward the left.
(b) Note that vstd − t 2 2 t 2 6 − st 2 3dst 1 2d and so vstd < 0 on the interval f1, 3g
and vstd > 0 on f3, 4g. Thus, from Equation 3, the distance traveled is
To integrate the absolute value of vstd,
we use Property 5 of integrals from
Section 5.2 to split the integral into
two parts, one where vstd < 0 and one
where vstd > 0.
y 4
| vstd | dt − y 3
f2vstdg dt 1 y 4
vstd dt
1 1
3
− y 3
s2t 2 1 t 1 6d dt 1 y 4
st 2 2 t 2 6d dt
1
3
−F2 t 3
3 1 t 2
3
G 1F
2 1 6t t 3
3 2 t 2
4
2 2 6t 1
G3
− 61 6 < 10.17 m n
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