Greenwood press

38
INTEGRATION
The area between the graph 
of
f (t) = 6.37e
−0.04t
and the t axis determined by

2
0
6.37e
−0.04t
dt.

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graph is much lower in this interval (the dark solid region), than from 0 to 2 hours
of work (the light shaded region).
This information can help managers determine when employees should take
breaks so that they can optimize their performance, because they would likely
feel more productive when they returned to work.
A definite integral can help heating and cooling companies estimate the
amount of costs needed to send power or gas to each household. On any given
day, the temperature can be modeled with a sinusoidal function, because tem-
perature increases during the day, decreases at night, and then repeats the cycle
throughout the year. For example, suppose the temperature reached a low of 50°
Fahrenheit at 2 
AM
and a high of 90° at 2 
PM
. If 
x represents the number of hours
that have passed during the day, then the temperature in degrees Fahrenheit, 
T ,
can be represented by the equation 
T = 20 cos

2π(x−14)
24
+ 70. Suppose that 
the thermostat in the house is set to 80° so that the air conditioning will turn on
once the temperature is greater than or equal to that setting. The amount of
energy used for the air conditioner is proportional to the temperature outside.
That means that the air conditioner will use more energy to keep the house cool
when it is closer to 90° than when it is near 80°. The price to cool the house
might be five cents per hour for every degree above 80°. If the temperature were
83° for the entire hour, then the cost to run the air conditioner would be fifteen
cents. However, since temperature varies according to a sinusoidal function, the
cost per minute would actually change. Therefore, a definite integral bounded by
the time when the temperature is above 80° will help predict the cooling costs.
The temperature should be 80° at x = 10 (10 
AM
) and x = 18 (6 
PM
), so the cool-
ing costs per day for days like this would be approximately $2.62 based on an
evaluation of the expression 
$0.05

18
10
(20 cos

2π(x−14)
24
+ 70 − 80)dx = $2.62.
Notice that the answer is also represented by 0.05 times the area of the curve 
between
T = 20 cos

2π(x−14)
24
+ 70 and T = 80, as shown in the following
diagram.