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

Section 5.1 Areas and Distances 373
y
y
1 y=e–®
1 y=e–®
0
1
0
1
FIGURE 15
2
2
x
x
(b) With n − 4 the subintervals of equal width Dx − 0.5 are f0, 0.5g, f0.5, 1g, f1, 1.5g,
and f1.5, 2g. The midpoints of these subintervals are x 1
* − 0.25, x 2
* − 0.75, x 3
* − 1.25,
and x 4
* − 1.75, and the sum of the areas of the four approximating rectangles (see Figure
15) is
M 4 − o 4
f sx*d i Dx
i−1
− f s0.25d Dx 1 f s0.75d Dx 1 f s1.25d Dx 1 f s1.75d Dx
− e 20.25 s0.5d 1 e 20.75 s0.5d 1 e 21.25 s0.5d 1 e 21.75 s0.5d
− 1 2 se20.25 1 e 20.75 1 e 21.25 1 e 21.75 d < 0.8557
y
y
1
1
y=e–®
y=e–®
So an estimate for the area is
A < 0.8557
With n − 10 the subintervals are f0, 0.2g, f0.2, 0.4g, . . . , f1.8, 2g and the midpoints are
x 1
* − 0.1, x 2
* − 0.3, x 3
* − 0.5, . . . , x 10 * − 1.9. Thus
0
0
1
1
2
2
x
x
A < M 10 − f s0.1d Dx 1 f s0.3d Dx 1 f s0.5d Dx 1 ∙ ∙ ∙ 1 f s1.9d Dx
− 0.2se 20.1 1 e 20.3 1 e 20.5 1 ∙ ∙ ∙ 1 e 21.9 d < 0.8632
FIGURE 16
From Figure 16 it appears that this estimate is better than the estimate with n − 4.
n
The Distance Problem
Now let’s consider the distance problem: Find the distance traveled by an object during a
certain time period if the velocity of the object is known at all times. (In a sense this is the
inverse problem of the velocity problem that we discussed in Section 2.1.) If the velocity
remains constant, then the distance problem is easy to solve by means of the formula
distance − velocity 3 time
But if the velocity varies, it’s not so easy to find the distance traveled. We investigate the
problem in the following example.
Example 4 Suppose the odometer on our car is broken and we want to estimate the
distance driven over a 30-second time interval. We take speedometer readings every
five seconds and record them in the following table:
Time (s) 0 5 10 15 20 25 30
Velocity (miyh) 17 21 24 29 32 31 28
In order to have the time and the velocity in consistent units, let’s convert the velocity
readings to feet per second (1 miyh − 5280y3600 ftys):
Time (s) 0 5 10 15 20 25 30
Velocity (ftys) 25 31 35 43 47 45 41
During the first five seconds the velocity doesn’t change very much, so we can estimate
the distance traveled during that time by assuming that the velocity is constant. If we
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