5128_Ch07_pp378-433

Section 7.1 Integral as Net Change 383
Step 3: Evaluate the integral. Using an antiderivative, we have
Net velocity change 2]
8
8
2.4tdt 1.2t 76.8 mph.
0
0
So, how fast is the car going when the 8 seconds are up? Its initial velocity is 5 mph and
the acceleration adds another 76.8 mph for a total of 81.8 mph.
(b) There is nothing special about the upper limit 8 in the preceding calculation. Applying
the acceleration for any length of time t adds
t
2.4u du mph u is just a dummy variable here.
0
(b) to the car’s velocity, giving
vt 5 t
The distance traveled from t 0 to t 8 sec is
8
0
0
2.4u du 5 1.2t 2 mph.
vt dt 8
5 1.2t 2 dt Extension of Example 3
0
8
[
5t 0.4t 3]
0
244.8 mph seconds.
Miles-per-hour second is not a distance unit that we normally work with! To convert to
miles we multiply by hourssecond 13600, obtaining
1
244.8 0.068 mile.
36 00
m i h
sec mi
h se c
The car traveled 0.068 mi during the 8 seconds of acceleration. Now try Exercise 9.
Consumption Over Time
The integral is a natural tool to calculate net change and total accumulation of more quantities
than just distance and velocity. Integrals can be used to calculate growth, decay, and,
as in the next example, consumption. Whenever you want to find the cumulative effect of a
varying rate of change, integrate it.
EXAMPLE 5
Potato Consumption
From 1970 to 1980, the rate of potato consumption in a particular country was Ct
2.2 1.1 t millions of bushels per year, with t being years since the beginning of 1970.
How many bushels were consumed from the beginning of 1972 to the end of 1973?
SOLUTION
We seek the cumulative effect of the consumption rate for 2 t 4.
Step 1: Riemann sum. We partition 2, 4 into subintervals of length Δt and let t k be a time
in the kth subinterval. The amount consumed during this interval is approximately
Ct k Δt million bushels.
The consumption for 2 t 4 is approximately
Ct k Δt million bushels.
continued