Download Accumulating Change: Limits of Sums and the Definite Integral

Calculus Concepts 2/e
LaTorre, Kenelly, Fetta, Harris, and Carpenter
Chapter 6
Accumulating Change:
Limits of Sums and the Definite Integral
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Chapter 6 Key Concepts
•
•
•
•
•
•
•
Results of Change
Approximating Area with Rectangles
Limits of Sums of Rectangle Areas
Introduction to Accumulation Functions
Fundamental Theorem and Antiderivatives
Definite Integrals
Improper Integrals
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Results of Change
• The accumulated change in a quantity is
represented as the area or signed area of a
region between the rate-of-change function
for that quantity and the horizontal axis
provided the function does not cross the
horizontal axis.
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Results of Change: Example
A water tank drains at a rate of r(t) = -2t gallons
per minute t minutes after the water began
draining. Determine the change in the volume of
water in the tank during the first 4 minutes the
tank was draining.
1/2 (4 minutes) (-8 gallons per minute) = -16 gallons
The tank lost 16 gallons of water in the first 4 minutes.
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Results of Change: Exercise 6.1 #15
The rate of change of the population of North
Dakota from 1970 through 1990 is modeled on the
graph below. Find the area between the graph of p
and the horizontal axis from 0 to 15 and interpret.
(3870 people per
year)(15 years) =
58,050 people
From 1970 to 1985, the
population of North
Dakota increased by
58,050 people.
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Approximating Area with Rectangles
• The sum of the areas of rectangles may be
used to estimate the area beneath a curve.
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Approximating Area: Example
The absorption rate of a drug given in 20 mg doses
may be modeled by r(x) = 1.708(0.845)x g/mL/day
when 0  x  20. Estimate the change in the drug
concentration over the first twenty days.
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Approximating Area: Example
Right endpoint
of rectangle
x
Height
of rectangle
r(x)
Area of rectangle =
height • width
2
1.220
2.439
4
0.871
1.742
6
0.622
1.244
8
0.444
0.888
10
0.317
0.634
12
0.226
0.453
14
0.162
0.323
16
0.115
0.231
18
0.082
0.165
20
0.059
0.118
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Approximating Area: Example
The sum of the areas of the rectangles is 8.235 g/mL.
The drug concentration over the 20-day period
increased by approximately 8.24 g/mL. From the
graph, we can see that this is an underestimate.
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Approximating Area: Exercise 6.1 #13
An cyclist is clocked by her coach at 10-minute
intervals during an hour long ride. Estimate the
distance traveled by using left and right rectangles.
Time
(minutes)
Speed
(mph)
0
22
10
18
20
20
30
23
40
15
50
17
60
12
left rectangle sum = (115
mph)(1/6 hour) = 19.17 miles
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Approximating Area: Exercise 6.1 #13
An cyclist is clocked by her coach at 10-minute
intervals during an hour long ride. Estimate the
distance traveled by using left and right rectangles.
Time
(minutes)
Speed
(mph)
0
22
10
18
20
20
30
23
40
15
50
17
60
12
right rectangle sum = (105
mph)(1/6 hour) = 17.5 miles
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Limits of Sums of Rectangle Areas
• Increasing the number of rectangles used to
estimate the area between a curve and the
horizontal axis increases the accuracy of the
estimate
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Limits of Sums of Rectangle Areas
• Let f be a continuous or piecewise continuous
non-negative function from a to b. The area
of the region between the graph of f and the
x-axis from a to b is given by the limit
Area   f ( x )dx  limf ( x1 )  f ( x 2 )  ...  f ( xn ) Δx
b
a
n 
where x1, x2, . . . , xn are the midpoints of n
subintervals of length x = (b-a)/n between a
b
and b. a f ( x )dx is called the definite integral of f
from a to b.
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Limits of Sums: Example
The flow rates of the Carson River can be modeled
by f(x) = 18,225x2 -135,334.3x + 2,881,542.9 ft3/hr
where x is the number of hours after 11:45 a.m.
Wednesday. Use the idea of limit of sums to
estimate to the nearest ten thousand cubic feet the
amount of water that flowed through the river from
11:45 a.m. on Wednesday to 7:45 a.m. on
Thursday.
We need to find the area between the curve and the
x-axis from x = 0 to x = 20. We’ll use midpoint
rectangles.
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Limits of Sums: Example
Number of Approximate
Rectangles
Area
10
79,042,498
20
79,133,623
40
79,156,404
80
79,162,100
160
79,163,523
320
79,163,879
640
79,163,968
Approximately 79,160
thousand gallons of water
flowed through the river in the
20-hour period. This estimate
is accurate to the nearest
10,000 cubic feet.
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Limits of Sums: Exercise 6.2 #7
The rate of change of the weight of a mouse can be
modeled by w(t) = 13.785t-1 grams per week where
t is the age of the mouse in weeks and 1  x  15.
11
Use the limit of sums to estimate  w( t )dt.
3
Number of Approximate Between the third week
Rectangles
Area
and the eleventh week, the
8
18.02742
16
17.94007
32
17.91799
64
17.91246
mouse gained about 17.9
grams. (Table values
determined with
technology.)
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Introduction to Accumulation Functions
• The accumulation function of a function f
gives the accumulation of the area between
the horizontal axis and the graph of f from a to
x. The constant a is the starting value of the
accumulation. The accumulation function is
x
denoted by A( x)  a f (t )dt.
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Accumulation Functions: Example
f(t) is the speed of a vehicle during a 10-second
drive. The shaded area is the distance traveled
over the first x seconds with x = 1, 2, 3, 4, and 5.
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Accumulation: Exercise 6.3 #19
Find a formula for the accumulation functions
below and shade the area represented by it.
x
a.  6dt
0
A(x) = 6x
x
b.   2dt
0
A(x) = -2x
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Fundamental Theorem / Antiderivatives
• Let f be a function of x. A function F is called
an antiderivative of f if F'(x) = f(x); that is, the
derivative of F is f.
• The Fundamental Theorem of Calculus
– For any continuous function f with input t, the
derivative of the accumulation function of f is the
function f in terms of x. In symbols, we write
d  x
 a f ( t )dt   f ( x )

dx 
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Fundamental Theorem / Antiderivatives
• Antiderivative Formulas
– Simple Power Rule for Antiderivatives
n1
x
n
 x dx  n  1  C, n  1
– Constant Multiplier Rule for Antiderivatives
 kf ( x)dx  k  f ( x)dx
– Sum Rule
 f ( x)  g( x)dx   f ( x)dx   g( x)dx
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Fundamental Theorem / Antiderivatives
• Antiderivative Formulas
– Natural Log Rule
1
 x dx  ln x  C
– eax Rule
1 ax
 e dx  a e  C
– Exponential Rule
x
b
x
 b dx  ln b  C
ax
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Fundamental Thm. / Antiderivatives: Example
An African country has an increasing population
but a declining birth rate. The rate of change in
the number of babies born each year is given by
b(t) = 87,000 - 1600t births per year t years from
the end of this year. The number of babies born
this year is 1,185,800. Find a function describing
the number of births each year t years from now.
1600 2
B( t )   b( t )dt  87,000 t 
t C
2
 87,000 t  800t 2  C births t years from now
 87,000 t  800t 2  1,185,800 births t years from now
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Antiderivatives: Exercise 6.4 #39
The rate of change of the growth rate of corn
during the first two months after it is planted may
be modeled by g'(d) = -0.00036d - 0.018 inches
per day squared where d is the number of days
after sprouting. Find the function for sweet corn
growth rate given it was growing 1.95 inches per
day on the day it sprouted.
 0.0036 d2
B( t )   g( t )dt 
 0.018 d  C
2
 0.00018 d2  0.018 d  C inches per day
 0.00018 d2  0.018 d  1.95 inches per day
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Definite Integrals
• If f is a continuous function from a to b and F
is any antiderivative of f, then

b
a
f ( x)  F(b)  F(a)
• The definite integral from a to b represents
the total change in F over the interval a to b
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Definite Integrals: Example
The marginal cost for toaster ovens may be
modeled by C'(x) = 47.638(0.0072)x dollars per
day. If the manufacturer wishes to increase
production from 300 to 500 ovens per day, how
much will the total cost change?

500
300
47.638(0.9972)x dx  C(500)  C(300)
 16,991.852(0.9972)
x
500
300
 16,991.852 (0.9972 )500  (16,991.852 (0.9972 )300 )
 $3,145
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Definite Integrals: Exercise 6.5 #22
The air speed of a small airplane during the first 25
seconds of takeoff and flight can be modeled by
v(t )  940,602t 2  19,269.3t  0.3 mph
t minutes after takeoff. Calculate and interpret the
following definite integral.

0.005
0

0.005
0
v( t ) dt
v(t ) dt  313,534t  9,634.65t  .3t
3
2
0.005
0
 313,534(0.005)3  9,634.65(0.005)2  .3(0.005) 
( 313,534(0)  9,634.65(0)  .3(0)
The plane traveled 0.200175 miles in the first 18
seconds (0.005 hours) of takeoff and flight.
3
2
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Improper Integrals
• Improper integrals are integrals of one of the
following three forms:





b
a
a
b



f ( x) dx,  f ( x) dx,  f ( x) dx
f ( x ) dx  lim

N
f ( x ) dx  lim
N

N
a
f ( x ) dx
b
 f ( x) dx
N
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Improper Integrals: Example


2
4.3e
0.06 x
dx
4.3 0.06 x N
lim
e
2
N  0.06
4.3 0.06( 2 ) 
 4.3 0.06(N)
lim 
e

e

N  0.06
 0.06


4.3 0.06(N)
4.3 0.12
lim
e
 lim
e
N  0.06
N 0.06
4.3 0.12
0
e
 63.562631
0.06
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Improper Integrals: Exercise 6.6 #23


10
3x
2

dx
 lim  3 x
N
1

N
0
3 
 3
 lim    (  ) 
N
10 
 N
3
 3
 lim     lim ( )
N
 N  N 10
3
0
10
 0.3
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