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

374 Chapter 5 Integrals
take the velocity during that time interval to be the initial velocity (25 ftys), then we
obtain the approximate distance traveled during the first five seconds:
25 ftys 3 5 s − 125 ft
Similarly, during the second time interval the velocity is approximately constant and
we take it to be the velocity when t − 5 s. So our estimate for the distance traveled
from t − 5 s to t − 10 s is
31 ftys 3 5 s − 155 ft
If we add similar estimates for the other time intervals, we obtain an estimate for the
total distance traveled:
s25 3 5d 1 s31 3 5d 1 s35 3 5d 1 s43 3 5d 1 s47 3 5d 1 s45 3 5d − 1130 ft
We could just as well have used the velocity at the end of each time period instead
of the velocity at the beginning as our assumed constant velocity. Then our estimate
becomes
s31 3 5d 1 s35 3 5d 1 s43 3 5d 1 s47 3 5d 1 s45 3 5d 1 s41 3 5d − 1210 ft
If we had wanted a more accurate estimate, we could have taken velocity readings
every two seconds, or even every second.
n
√
40
20
0
10 20
FIGURE 17
30
t
Perhaps the calculations in Example 4 remind you of the sums we used earlier to
estimate areas. The similarity is explained when we sketch a graph of the velocity function
of the car in Figure 17 and draw rectangles whose heights are the initial velocities
for each time interval. The area of the first rectangle is 25 3 5 − 125, which is also
our estimate for the dis tance traveled in the first five seconds. In fact, the area of each
rectangle can be interpreted as a distance because the height represents velocity and the
width represents time. The sum of the areas of the rectangles in Figure 17 is L 6 − 1130,
which is our initial estimate for the total distance traveled.
In general, suppose an object moves with velocity v − f std, where a < t < b and
f std > 0 (so the object always moves in the positive direction). We take velocity readings
at times t 0 s− ad, t 1 , t 2 , . . . , t n s− bd so that the velocity is approximately constant
on each subinterval. If these times are equally spaced, then the time between consecutive
readings is Dt − sb 2 adyn. During the first time interval the velocity is approximately
f st 0 d and so the distance traveled is approximately f st 0 d Dt. Similarly, the distance traveled
during the second time interval is about f st 1 d Dt and the total distance traveled during
the time inter val fa, bg is approximately
f st 0 d Dt 1 f st 1 d Dt 1 ∙ ∙ ∙ 1 f st n21 d Dt − o n
i−1
f st i21 d Dt
If we use the velocity at right endpoints instead of left endpoints, our estimate for the
total distance becomes
f st 1 d Dt 1 f st 2 d Dt 1 ∙ ∙ ∙ 1 f st n d Dt − o n
i−1
f st i d Dt
The more frequently we measure the velocity, the more accurate our estimates become,
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