Chapter 6
Further Topics in Integration
6.1
Definite Integration
1. Z
1
(x4 + 3x3 + 1)dx =
0
3.
1
39
x5
3x4
|
+
+x | =
0
20
5
4
5
Z
(2 + 2t + 3t2 )dt
2
=
5.
2
1
1
+ 2 dx
x x
1
8
1 |3
x + ln |x| −
= + ln 3.
x |1
3
Z
=
3
|5
(2t + t2 + t3 )| = 144.
1+
1
11. Let u = 6t + 1. Then du = 6dt or dt = du.
6
When t = 0, u = 1, and when t = 4, u = 25.
Hence,
Z 4
Z
1 25 −1/2
1
√
dt =
u
du
6 1
6t + 1
0
u1/2 |25
4
=
= .
|
1
3
3
13. Let u = t4 + 2t2 + 1. Then du = (4t3 + 4t)dt or
1
(t3 + t)dt = du.
4
When t = 0, u = 1, and when t = 1, u = 4.
Hence
7.
Z
−1
−3
−1
=
0
t+1
dt
t3
=
Z
(t−2 + t−3 )dt
1
1 |−1
2
− − 2 |
= .
−3
t
2t
9
=
15. Let u = x − 1. Then du = dx and x = u + 1.
When x = 2, u = 1, and when x = e + 1, u = e.
1
9. Let u = 2x − 4. Then du = 2dx or dx = du.
2
When x = 1, u = −2, and when x = 2, u = 0.
Hence,
Z 2
Z
1 0 4
4
(2x − 4) dx =
u du
2 −2
1
=
p
(t3 + t) t4 + 2t2 + 1dt
Z
1 4 1/2
u3/2 |4
u du =
4 1
6 |1
1
7
(8 − 1) = .
6
6
−3
=
1
Z
Z
Hence
=
e+1
Z2 e
1
x
dx
x−1
Z e
u+1
1
du =
1+
du
u
u
1
|e
= (u + ln |u|)| = e.
u5 |0
16
=
.
|
−2
10
5
1
165
166
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
1
dx. When x = 1,
x
2
u = 0, and when x = e , u = 2. Hence
17. Let u = ln x. Then du =
Z
e2
1
19.
(ln x)2
dx =
x
Z
I=
Z
2
u2 du =
0
u3 |2
8
=
3 |0
3
Summing up these infinitesimal elements in
4 < x < 9 leads to the integral
Z 9
Z 9
√
xdx =
x1/2 dx
A =
=
1
4
4
xe−x dx
38
2x3/2 |9
=
square units
3 |4
3
0
.
g(x) = e−x
G(x) = −e−x
f (x) = x
f 0 (x) = 1
|1
= −xe−x | +
I
0
Z
1e−x dx
0
1
|
= −(x + 1)e−x | = 1 − 2e−1
0
21. The element of area has a height y and a base
dx. Thus
dA = ydx = (4 − 3x)dx.
Summing up these infinitesimal elements leads
to the integral
A =
=
=
Z
4/3
(4 − 3x)dx
0
3x2 |4/3
4x −
|0
2
4(4) 3(42 )
8
−
= square units
3
2(32 )
3
23. The element of area has a height y and a base
dx. Thus
dA = ydx = 5dx.
Summing up these infinitesimal elements in
−2 < x < 1 leads to the integral
Z 1
A =
(5)dx
−2
1
=
|
5x|
= 15 square units
−2
27. The element of area has a height y and a base
dx. Thus
dA = ydx = ex dx.
Summing
up these infinitesimal elements in
1
= − ln 2 < x < 0 leads to the integral
ln
2
A =
Z
|0
ex dx = ex |
− ln 2
− ln 2
− ln 2
= 1−e
= 1 − eln(1/2)
1
=
square units
2
29. The region is split into two subregions on either
side of the vertical line x = 1.
Z 1
x
x−
dx
A =
8
0
Z 2
x
1
+
−
dx
x2
8
1
7x2 |1
1 x2 |2
=
−
+
16 |0
x 16 |1
3
=
square units.
4
31. The element of area has a height 9x − x3 and a
base dx. Thus
dA = (9x − x3 )dx
y = x3 intersects y = 9x at (0, 0) and (3, 27).
A =
Z
=
3
(9x − x3 )dx
2
9x
x4 |3
−
2
4 |0
81
square units
4
0
=
25. The element of area has a height y and a base
dx. Thus
√
dA = ydx = xdx.
0
6.2. APPLICATIONS TO BUSINESS AND ECONOMICS
33. (a)
I=
Z
0
1
p
1 − x2 dx
is the area under the
p
y = 1 − x2 ≥ 0
on 0 ≤ x ≤ 1, that is one-quarter of the
π
area of a circle of radius 1. Thus I = .
4
Z 2p
(b)
2
I=
2x − x dx
167
3. Let C(q) denote the total cost of producing q
units. Then the marginal cost is
dC
= 6(q − 5)2 ,
dq
and the increase in cost is
Z 13
6(q − 5)2 dq
C(13) − C(10) =
10
|13
= 2(q − 5)3 | = $774.
1
is the area under the curve
p
p
y = −x2 + 2x = 1 − (x − 1)2
10
5. Let N (t) denote the number of bushels that are
produced over the next t days. Then
dN
on 1 ≤ x ≤ 2, that is one-quarter of the
= 0.3t2 + 0.6t + 1,
π
dt
area of a circle of radius 1. Thus I = .
4
and the increase in the crop over the next five
r
35.
Z 4.2
days is
2x2
A = 2
− 2dx
Z 5
5
2.34
! N (5) − N (0) =
Z 2.97 r 2
(0.3t2 + 0.6t + 1)dt
2x
0
+
− 2 − (x3 − 8.9x2 6.7x − 27) dx
5
|5
2.6
= (0.1t3 + 0.3t + t)| = 25
0
= 2.037
6.2
Applications to Business
and Economics
1. Let V (x) denote the value of the machine after
x years. Then
dV
= 220(x − 10),
dx
and the amount by which the machine
depreciates during the 2nd year is
Z 2
220(x − 10)dx
V (2) − V (1) =
1
2
2
x
|
= 220
− 10x | = −$1, 870
1
2
where the negative sign indicates that the value
of the machine has decreased.
bushels. If the price remains fixed at $3 per
bushel, the corresponding increase in the value
of the crop is $75.
7. Let Q(t) denote the production after t hours.
Then t = 0 at 8:00 a.m., t = 2 at 10:0 a.m., and
t = 4 at noon. Thus,
Z 4
Q(t) = 100
te−0.5t dt
2
Note: This problem can be solved by
integration by parts, discussed in section 5, or
you can use your graphing utility.
f (t) = t
f 0 (t) = 1
g(t) = e−0.5t
G(t) = −2e−0.5t
Z 4
|4
100 −2te−0.5t | + 2
e−0.5t dt
2
0
|4
= 100 −8e−2 + 4e−1 − 4e−0.5t |
Q(t) =
2
= 131.90 units
168
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
9. (a) The first plan generates profit at the rate
of
P10 (t) = 130 + t2
hundred dollars per year and the second
generates profit at the rate of
11. (a) The first plan generates profit at the rate
of
P10 (t) = 90e0.1t
thousand dollars per year and the second
generates profit at the rate of
P20 (t) = 140e0.07t
P20 (t) = 306 + 5t
hundred dollars per year. The second plan
will be the more profitable until
thousand dollars per year. The second
plan will be the more profitable until
P10 (t) = P20 (t),
P10 (t) = P20 (t),
that is, until 130 + t2 = 306 + 5t or t = 16
years.
(b) For 0 ≤ t ≤ 16, the rate at which the profit
generated by the second plan exceeds that
of the first plan is
that is, until 90e0.1t = 140e0.07t or
t = 14.73 years.
(b) For 0 ≤ t ≤ 14.73, the rate at which the
profit generated by the second plan
exceeds that of the first plan is
P20 (t) − P10 (t).
P2 (t) − P1 (t).
Hence the net excess profit generated by
the second plan over the 16−year period is
the definite integral
Z 16
[P2 (t) − P1 (t)]dt
Hence the net excess profit generated by
the second plan over the 14.73−year
period is the definite integral
Z 14.73
[P2 (t) − P1 (t)]dt
0
=
=
=
Z
0
16
[306 + 5t − (130 + t2 )]dt
0
t3 |16
5t2
−
176t +
2
3 |0
2, 090.67 hundred dollars.
(c) In geometric terms, the net excess profit
generated by the second plan is the area of
the region between the curves y = P2 (t)
and y = P1 (t) from t = 0 to t = 16.
=
14.73
Z
[140e0.07t − (90e0.1t )]dt
0
=
=
|14.73
2000e0.07t − 900e0.1t |
0
582.23 thousand dollars.
(c) In geometric terms, the net excess profit
generated by the second plan is the area of
the region between the curves y = P2 (t)
and y = P1 (t) from t = 0 to t = 14.73.
6.2. APPLICATIONS TO BUSINESS AND ECONOMICS
13. (a) The machine generates revenue at the rate
of
R0 (t) = 7, 250 − 18t2
15. (a) The campaign generates revenue at the
rate of
R(t) = 5, 000e−0.2t
dollars per week and accumulates expenses
at the rate of $676 per week.
The campaign will be profitable as long as
R(t) is greater than 676, that is, until
dollars per year and results in costs that
accumulate at the rate of
C 0 (t) = 3, 620 + 12t2
5, 000e−0.2t = 676
dollars per year.
The use of the machine will be profitable
as long as the rate at which revenue is
generated is greater than the rate at which
costs accumulate, that is, until
R0 (t) = C 0 (t),
t=−
(b) The difference R(x) − C(x) represents the
rate of change of the net earnings
generated by the machine.
Hence, the net earnings over the next 11
years is the definite integral
Z 11
[R(t) − C(t)]dt
0
=
=
[5, 000e−0.2t − 676]dt
|10
|10
= −25, 000e−0.2t | − 676t| = $14, 857
0
[(7, 250 − 18t2 ) − (3, 620 + 12t2 )]dx
11
(3, 530 − 30t2 )dt
0
=
10
0
11
0
Z
Z
0
0
=
ln(676/5, 000)
≈ 10 weeks
0.2
(b) For 0 ≤ t ≤ 10, the difference R(t) − 676 is
the rate of change with respect to time of
the net earnings generated by the
campaign. Hence, the net earnings during
the 10 week period is the definite integral
Z 10
[R(t) − 676]dt
7, 250 − 18t2 = 3, 620 + 12t2 or t = 11 years
Z
169
(c) In geometric terms, the net earnings in
part (b) is the area between the curve
y = R(t) and the horizontal line y = 676
from t = 0 to t = 10.
|11
(3, 630t − 10t3 )| = $26, 620
0
(c) In geometric terms, the net earnings in
part (b) is the area of the region between
the curves y = R0 (t) and y = C 0 (t) from
t = 0 to t = 11.
17. Recall that P dollars invested at an annual
interest rate of 6 % compounded continuously
will be worth
P e0.06t
dollars t years later.
To approximate the future value of the income
stream, divide the 5-year time interval
170
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
0 ≤ t ≤ 5 into n equal sub-intervals of length
∆t years and let tj denote the beginning of the
j th sub-interval.
Then, the money deposited during the j th
sub-interval is 2, 400∆t.
This money will remain in the account
approximately 5 − tj years hence. The future
value of the money deposited during the j th
sub-interval is
0.06(5−tj )
2, 400e
∆t.
|10
P (t) = −10e−0.1t (10, 000 + 500t)|
0
+5, 000
lim
n→∞
= −10[15, 000e−1 − 10, 000]
|10
−50, 000e−0.1t |
0
=
$76, 424.11.
23. (a) If the consumers’ demand function is
=
D(q) =
2, 400e0.06(5−tj ) ∆t
j=1
0.06(5−t)
2, 400e
dt
0
=
2, 400e0.3
5
Z
e−0.06t dt
0
=
19. Recall that the present value of B dollars
payable t years from now with an annual
interest rate of 12 % compounded continuously
is
Be−0.06t .
Divide the interval 0 ≤ t ≤ 5 into n equal
sub-intervals of length ∆t years.
Then, the income during the j th sub-interval is
1, 200∆t and the present value of this income is
1, 200e−0.12tj ∆t.
Hence, the present value of the investment is
n→∞
=
n
X
=
300
Z
5
(0.1q + 1)−2 dq
0
|5
= −3, 000(0.1q + 1)−1 |
0
1
= −3, 000
− 1 = $1, 000.
1.5
1, 200e−0.12tj ∆t
j=1
1, 200
Z
5
e−0.12t dt
0
1, 200e−0.12t |5
|0 = $4, 511.88.
−0.12
Z 10
P (t) =
(10, 000 + 500t)e−0.1t dt.
= −
21.
0
2, 400 0.3 −0.06t |5
e e
|0 = $13, 994.35.
−0.06
lim
300
(0.1q + 1)2
dollars per unit, the total amount that
consumers are willing to spend to get 5
units is the definite integral
Z 5
D(q)dq
5
Z
e−0.1t dt
0
The future value of the income stream is
n
X
10
Z
0
f (t) = 10, 000 + 500t
f 0 (t) = 500
−0.1t
g(t) = e
G(t) = −10e−0.1t
(b) The total willingness to spend in part (a)
is the area of the region under the demand
curve from q = 0 to q = 5.
25. (a) If the consumers’ demand function is
D(q) =
300
4q + 3
6.2. APPLICATIONS TO BUSINESS AND ECONOMICS
dollars per unit, the total amount that
consumers are willing to spend to get 10
units is the definite integral
Z 10
D(q)dq
171
(b) The total willingness to spend in part (a)
is the area of the region under the demand
curve from q = 0 to q = 15.
29. (a) The consumers’ demand function is
0
=
300
D(q) = 150 − 2q − 3q 2
10
Z
(4q + 3)−1 dq
0
=
300
|10
ln |4q + 3|| = 199.70
0
4
dollars per unit. For the market price of 6
units
p0 = 150 − 12 − 108 = 30
Thus the consumer’s surplus is
Z 6
S(q) =
(150 − 2q − 3q 2 )dq − (30)(6)
0
=
150(6) − 62 − 63 − 180 = $468.
(b) The total willingness to spend in part (a)
is the area of the region under the demand
curve from q = 0 to q = 10.
27. (a) If the consumers’ demand function is
D(q) = 50e−0.04q
dollars per unit, the total amount that
consumers are willing to spend to get 15
units is the definite integral
Z 15
D(q)dq
0
=
50
Z
15
e−0.04q dq
0
50 −0.04q |15
e
|0
0.04
1, 250(1 − 0.5488) = $563.99.
= −
=
(b) The consumer’s surplus in part (a) is the
area of the region under the demand curve
from q = 0 to q = 6 from which the actual
spending is subtracted.
31. (a) The consumers’ demand function is
D(q) = 75e−0.04q dollars per unit.
The market price for 3 units is
D(3) = 75e−0.12 = $66.52
Thus the consumer’s surplus is
Z 3
S(q) = 75
e−0.04q dq − 3(66.52)
0
75 −0.04q |3
= −
e
|0 − 199.56 = $12.46.
0.04
172
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
37. (a) D(q) = p0 if
400
= 20, 20 = 0.5q + 2
0.5q + 2
or q = 36.
(b) The consumer’s surplus is
I
(b) The consumer’s surplus in part (a) is the
area of the region under the demand curve
from q = 0 to q = 3.
Z
36
400
dq − 36 × 20
0.5q
+2
0
|36
= 800 ln |0.5q + 2|| − 720 = $1, 122.07.
0
=
33. The producer’s supply function is
S(q) = 0.5q + 15 dollars per unit.
p0 = S(5) = 2.5 + 15 = 17.5
The producer’s surplus for q0 = 5 is
Z 5
P S = (5)(17.5) −
(0.5q + 15)dq = $6.25
0
39. (a) The dollar price per unit is p = 124 − 2q
and the cost function is
C(q) = 2q 3 − 59q 2 + 4q + 7, 600.
The profit function is
(124 − 2q)q
−(2q 3 − 59q 2 + 4q + 7, 600)
= −2q 3 + 57q 2 + 120q − 7, 600.
P 0 (q) = −(6q 2 − 114q − 120) = 0
P (q) =
35. The producer’s supply function is
S(q) = 17 + 11e0.01q dollars per unit.
p0 = S(7) = 17 + 11e0.07 = 28.80
The producer’s surplus for q0 = 7 is is
Z 7
P S = (7)(28.80) −
(17 + 11e0.01q )dq = $2.84
0
(b) Profit is maximized when
p
57 ± 572 + 6(120)
if q =
= 20
6
(since q > 0).
R00 (q) = −12q + 114 and R(20) < 0,
so q = 20 produces a maximum.
(c) The corresponding consumer’s surplus is
S(q)
=
Z
20
(124 − 2q)dq − 20(124 − 40)
0
=
|20
(124q − q 2 )| − (20)(84) = $400.
0
6.3. ADDITIONAL APLICATIONS OF DEFINITE INTEGRATION
41. The supply function for a certain commodity is
S(p) =
173
where R(0) = 0 = 300(0) + C implies
C = 0. Then
q+1
3
R(36) = 300[18(36)+0.2(36)3/2 ] = $207, 360
and the demand function is
D(q) =
(b) Divide the interval 0 ≤ t ≤ 36 into n equal
subintervals of length ∆t years.
Then, the quantity of oil during the j th
subinterval is 300∆t and the revenue
generated by that much oil is
p
(18 + 0.3 tj )300∆t
16
− 3.
q+2
(a) The supply equals the demand if
q+1
16
−3=
,
q+2
3
48 − 9q − 18 = q 2 + 3q + 2,
Hence, the total revenue is
q 2 + 12q − 28 = 0,
√
or q = −6 ± 36 + 28 = 2 (since q > 0.)
lim
n→∞
(b) The corresponding consumer’s surplus is
Z 2
16
CS =
− 3 dq − 2(1)
q+2
0
=
PS
=
=
2(1)
0
√
(18 + 0.3 t)300dt
0
0
2
j=1
p
(18 + 0.3 tj )300∆t
= $207, 360
|2
(16 ln |q + 2| − 3q)| − 2 = $3.09.
Z
36
Z
n
X
(computed in part (a) above.)
q+1
dq = $0.67
3
(c) Writing exercise—
Answers will vary.
45. During a time interval dt the revenue generated
is f (t)dt dollars which has a present value of
f (t)e−rt dt at the interest rate r.
The present value corresponding to N years is
(c)
P (t)
=
Z
=
Z
N
f (t)e−rt dt
0
43. (a) Let t denote the time from now in months
and R(t) the total revenue generated. The
price is
√
P (t) = 18 + 0.3 t
dollars per barrel and the revenue is
generated at a rate of
√
dR
= 300(18 + 0.3 t). Then
dt
R(t)
√
300(18 + 0.3 t)dt
=
Z
=
300(18t + 0.2t3/2 ) + C
10
1, 750e−0.095t dt = 11, 296.88
0
6.3
Additional Aplications of
Definite Integration
1. The average value is
Z 4
1
1 x2 |4
fav =
xdx =
=2
4−0 0
4 2 |0
174
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
9. The average rate during the first 3 months is
Z 3
1
(700 − 400e−0.5t )dt
Qav =
3−0 0
1
|3
=
(700t + 800e−0.5t )|
0
3
= 492.83 ≈ 493 letters per hour.
f (t) = e−0.2t
11.
3. The average value is
fav
Z
Of the 200 present members, 200f (8) will still
be in the club in 8 months.
Of the 10 new members picked up t months
from now, 10f (8 − t) will still be members in 8
months. Thus
Z 8
P (t) = 200e−1.6 +
10e−0.2(8−t) dt
0
=
1
0 − (−4)
=
(x + 2)3 |0
4
= .
12 |−4
3
(x + 2)2 dx
−4
0
=
200e−1.6 + 10e−1.6
Z
8
e0.2t dt
0
=
|8
40.379 + 50e−1.6 e0.2t | = 80.28
0
The club will consist of approximately 80
members in 8 months.
5. The average value is
fav
=
1
2 − (−1)
= −
Since f (t) = e−t/10
13.
Z
2
e−2t dt
−1
e−2t |2
= 1.2285.
6 |−1
is the fraction of members active after t
months, and since there were 8, 000 charter
members, the number of charter members still
active at the end of 10 months is
8, 000f (10) = 8, 000e−1
Now, divide the interval 0 ≤ t ≤ 10 into n equal
subintervals of length ∆t months and let tj
denote the beginning of the j th subinterval.
During this j th subinterval, 200∆t new
members join, and at the end of the 10 months
(10 − tj months later), the number of these
retaining membership is
200f (10 − tj ) = 200e−(10−tj )/10 ∆t
7. The average temperature between 9:00 a.m.
and noon is
Z 12
1
fav =
(−0.3t2 + 4t + 10)dt
12 − 9 9
1
|12
=
(−0.1t3 + 2t2 + 10t)| = 18.7◦ C
9
3
Hence the number of new members still active
10 months from now is approximately
lim
n→∞
=
Z
0
n
X
200e−(10−tj )/10 ∆t
j=1
10
200e−(10−t)/10 dt
6.3. ADDITIONAL APLICATIONS OF DEFINITE INTEGRATION
Hence, the total number N of active members
10 months from now is
Z 10
N = 8, 000e−1 +
200e−(10−t)/10 dt
0
=
−1
2, 943 + 200e
Z
10
t/10
e
dt
175
Applying integration by parts,
Z 3
re−0.1r dr
0
|3
= −10re−0.1r | −
0
0
=
−1
2, 943 + 200e
t/10
(10e
|10
)| = 4, 207
0
15. Divide the interval 0 ≤ r ≤ 3 into n equal
subintervals of length ∆r, and let rj denote the
beginning of the j th subinterval.
This divides the circular disc of radius 3 into n
concentric circles, as shown in the figure.
If ∆r is small, the area of the j th ring is
2πrj ∆r
where 2πrj is the circumference of the circle of
radius rj that forms the inner boundary of the
ring and ∆r is the width of the ring.
Then, since D(r) = 5, 000e−0.1r is the
population density (people per square mile) r
miles from the center, it follows that the
number of people in the j th ring is
D(rj )(area of the j th ring)
= 5, 000e−0.1rj (2πrj ∆r)
=
=
−0.1r
(−10re
3
Z
(−10)e−0.1r dr
0
|3
− 100e−0.1r )|
0
−0.3
(−30e
−0.3
− 100e
) − (−100) = 3.6936
Hence, the total number of people within 10
miles of the center of the city is
Z 3
N = 10, 000π
re−0.1r dr
0
=
116, 038
(Your answer may differ slightly due to
round-off errors.)
17. Let S(r) = k(R2 − r2 ) denote the speed of the
blood in centimeters per second at a distance r
from the central axis of the artery of (fixed)
radius R.
The area of a small circular ring at a distance
rj is (approximately) 2πrj ∆r square
centimeters, so the amount of blood passing
through the ring is
V (r)
=
2πrj ∆r[k(R2 − rj2 )]
=
2πk(R2 rj − rj3 )∆r
cubic centimeters per second.
Hence, the total quantity of blood flowing
through the artery per second is
lim
n→∞
Hence, if N is the total number of people
within 3 miles of the center of the city,
N
=
=
lim
n→∞
Z
n
X
5, 000e−0.1rj (2πrj ∆r)
j=1
3
5, 000(2π)re−0.1r dr
0
=
10, 000π
Z
0
=
n
X
j=1
Z
R
(rR2 − r3 )dr
2 2
πkR4
r4 |R
R r
=
2πk
−
|
0
2
2
4
2πk
0
=
The area of the artery is πR2 and the average
velocity of the blood through the artery is
3
re−0.1r dr
2πk(R2 rj − rj3 )∆r
Vave =
πkR4 /2
kR2
=
2
πR
2
176
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
The maximum speed for the blood occurs at
r = 0, so S(0) = kR2 . Thus
amount left 140 years from now (i.e., after
140 − tj years) is approximately
1
S(0)
2
Vave =
(500∆t)A(140 − tj ) = 500A(140 − tj )∆t.
Hence, the total waste after 140 years is
19. Let s(t) denote the distance traveled after t
minutes.
ds
= 3t2 + 2t + 5
dt
3
N
2
s(t) = t + t + 5t + s0
=
=
n
X
lim
n→∞
Z
500A(140 − tj )∆t
j=1
140
500A(140 − t)dt
0
s(2) = 8 + 4 + 10 + s0
= 500
s(1) = 1 + 1 + 5 + s0
Z
140
e−(140−t)k dt
0
s(2) − s(1) = 15 meters
= 500e−140k
Z
140
ekt dt
0
21. (a) The average speed is
=
Sav =
N
Z
1
N
S(t)dt.
=
0
(b) The total distance is
d=
Z
Since k =
N
S(t)dt.
N
0
(c) The average speed is
total distance
.
number of hours
23. Radioactive material decays exponentially so
that if A(t) denotes the amount of radioactive
material present after t years, A(t) = A0 e−kt ,
where A0 is the amount present initially and k
is a positive constant.
Since the half-life is 28 years,
A0
= A(28) = A0 e−28k ,
2
1
ln 2
= − ln 2 or k =
2
28
Now divide the interval 0 ≤ t ≤ 140 into n
equal sub-intervals of length ∆t years and let tj
denote the beginning of the jth sub-interval.
During the jth sub-interval, 500∆t pounds of
radioactive material is produced, and the
−28k = ln
25.
500 −140k+kt |140
e
|0
k
500
(1 − e−140k ).
k
ln 2
it follows that
28
28(500)
[1 − e−(140 ln 2)/28 ]
ln 2
28(500)
=
[1 − e−5 ln 2 ]
ln 2
28(500)
=
[1 − 2−5 ] = 19, 567 pounds.
ln 2
c
C(t) =
(e−at − e−bt )
b−a
c
C 0 (t) =
(−ae−at + be−bt )
b−a
c
C 00 (t) =
(a2 e−at − b2 e−bt )
b−a
=
C 00 (t) = 0 when a2 e−at = b2 e−bt
a
a2
2
(a−b)t
=
e
or
t
=
ln
b
b2
a−b
2
Let b = 2a then t = ln 2
a
Z (2/a) ln 2
a
c −at
(e
− e−2at )dt
2 ln 2 0
a
c
|(2/a) ln 2
=
(−2e−at + e−2at )|
0
4a ln 2
6.3. ADDITIONAL APLICATIONS OF DEFINITE INTEGRATION
=
=
177
c
1
1
1
1
− +
− − +
2 ln 2
4a 32a
a 2a
9c
64a ln 2
27. Let f (t) denote the fraction of the membership
of the group that will remain active for at least
t years, P0 the initial membership, and r(t) the
rate per year at which additional members are
added to the group.
Then, the size of the group N years from now is
the number of initial members still active plus
the number of new members still active.
Of the P0 initial members, f (N ) is the fraction
remaining active for N years.
Hence, the number of initial members still
active after N years is
1
Z
[x − x2 ]dx
2
1
x
x3 |1
2
=
−
|
0
3
2
3
= 2
0
=
L(x) = 0.7x2 + 0.3x
31.
G
P0 f (N ).
= 2
Z
1
[x − L(x)]dx
0
To find the number of new members still active
after N years, divide the interval 0 ≤ t ≤ N into
n equal sub-intervals of length ∆t years and let
tj denote the beginning of the j th sub-interval.
During the j th sub-interval, approximately
=
Z
1
(0.7x − 0.7x2 )dx
2
0.7
x
x3 |1
1.4
=
−
3
2
3 |0
2
0
=
r(tj )∆t
new members joined the group. Of these, the
fraction still active t = N (that is N − tj years
later) is f (N − tj ), and so the number of these
still active after N years is
lim
n→∞
=
Z
n
X
r(tj )f (N − tj )∆t
j=1
L(x) =
33.
N
r(t)f (N − t)dt.
ex − 1
e−1
0
Putting it all together, the total number of
active members N years from now is
Z N
P0 f (N ) +
r(t)f (N − t)dt.
0
L(x) = x2
29.
G
=
2
Z
0
1
[x − L(x)]dx
G
Z
1
[x − L(x)]dx
Z 1
ex − 1
= 2
x−
dx
e−1
0
2
1
2(e − 2)
x
1
|
= 2
−
(ex − x) | = 1 −
0
e−1
2
e−1
= 2
0
178
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
35. (a) Writing exercise —
Answers will vary.
41.
land owners
0
0
0.2
0.002
0.4
0.1
0.6
0.28
0.8
0.11
1.0
1.00
(b) If L(x) = x then the area between L(x)
and y = x is 0.
If L(x) = 0 then the area between L(x)
and y = 0 is 0.5 (from 0 and 1.
(c) Writing exercise —
Answers will vary.
land owners
0.1
0.001
0.3
0.005
0.5
0.18
0.7
0.58
0.9
0.21
37. The Gini index for computer engineers is
Z 1
G1 = 2
[x − x1.8 ]dx
0 2
x2.8 |1
x
= 2
−
2
2.8 |0
= 0.2857
The Gini index for stock brokers is
Z 1
G2 = 2
[0.75x − 0.75x2 ]dx
0
2
x
x3 |1
= 1.5
−
2
3 |0
= 0.25
The distribution of incomes amongst stock
brokers is more fairly distributed.
6.4
Improper Integrals
1.
Z
1
3.
Z
1
land owners
0.1
0.025
0.3
0.075
0.5
0.13
0.7
0.22
0.9
0.42
1
dx =
x3
=
39.
land owners
0
0
0.2
0.05
0.4
0.1
0.6
0.18
0.8
0.28
1.0
1.00
∞
∞
1
√ dx =
x
=
5.
Z
∞
lim
N →∞
Z
1
N
1
dx
x3
−1 |N
1
= .
|
2
1
N →∞ 2x
2
Z N
lim
x−1/2 dx
lim
N →∞
1
|N
lim 2x1/2 | = ∞.
N →∞
1
1
dx
2x
−1
3
Z N
1
= lim
dx
N →∞ 3
2x − 1
1
|N
=
lim ln |2x − 1|| = ∞.
3
2 N →∞
6.4. IMPROPER INTEGRALS
7.
179
∞
Z
1
dx
(2x − 1)2
3
Z N
lim
(2x − 1)−2 dx
=
N →∞
N
2
2
|
− xe−3x − e−3x |
0
N →∞
3
9
N
1
2
2
|
=
= lim − e−3x x +
|
0
N →∞
3
3
9
=
3
1
−1 |N
1
lim
=
.
|
3
2 N →∞ 2x − 1
10
=
9.
∞
Z
19.
N →∞
Z
10
=
5e
e−2x dx
5
5
|
lim e−2x | = .
0
N
→∞
2
2
13.
x2
dx
x3 + 2
1
Z
1 N 2 3
3x (x + 2)−1/2 dx
= lim
N →∞ 3 1
1
|N
=
lim 2(x3 + 2)1/2 | = ∞.
1
3 N →∞
15.
∞
Z
√
e−
√
Z
23.
∞
Z
=
=
=
=
2xe−3x dx
=
lim
N →∞
lim
N →∞
Z
∞
N
2xe−3x dx
0
2
2
|N
− xe−3x | +
0
3
3
Z
0
N
−3x
e
dx
0
= 5e10
∞
|N
lim ln | ln x|| = ∞
2
N →∞
x2 e−x dx
lim
N →∞
lim
!
lim
"
N →∞
N
Z
N →∞
x
0
=
e−x dx
0
N
0
dx
x
Z N √ 1
− x
√
= lim 2
e
dx
N →∞
2 x
1
√ |N
2
= −2 lim e− x | = .
1
N →∞
e
Z
N
Z
1
dx
x
ln
x
2
Z N
1
1
= lim
dx
N →∞ 2
ln x x
1
17.
0
N →∞
=
√
xe−x dx
0
|
5e10 lim (−xe−x − e−x )|
21.
∞
N
N →∞
=
x2
dx
(x3 + 2)2
1
Z
1 N 2 3
= lim
3x (x + 2)−2 dx
N →∞ 3 1
1
−1 |N
1
=
lim 3
= .
|
1
3 N →∞ x + 2
9
Z
N →∞
Z
|N
5e10 lim (−xe−x )| +
=
0
∞
Z
lim
N
N
11.
5xe10−x dx
0
5e−2x dx
5 lim
= −
∞
Z
0
=
lim
x2 e−x dx
0
|N
−x2 e−x | + 2
0
2 −x
−x e
xe−x dx
0
−x
− 2xe
!
N
Z
|N
|0 +
Z
N
−x
2e
dx
0
|N
lim [(−x2 − 2x − 2)e−x | = 2.
N →∞
0
25. To find the present value of the investment of
$2,400 per year for N years, divide the N −year
interval 0 ≤ t ≤ N into n equal sub-intervals of
length ∆t years, and let tj denote the beginning
of the j th sub-interval.
Then, during the j th sub-interval, the amount
generated is approximately 2, 400∆t and the
present value is
2, 400e−0.12tj ∆t .
#
180
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
Hence, the present value of an N −year
investment is
lim
n→∞
N →∞
=
2, 400e−0.12tj ∆t
(5, 000)e−0.1t dt
0
lim (−100, 000 − 5, 000t − 50, 000)
N →∞
0
−0.12t
2, 400e
dt.
=
0
To find the present value P of the total
investment, let N → ∞ to get
Z N
P = lim
2, 400e−0.12t dt
N →∞
N
Z
|N
e−0.1t |
j=1
N
Z
=
n
X
+ lim
0
2, 400
|N
= −
lim e−0.12t |
0
0.12 N →∞
2, 400
= −
lim (e−0.12N − 1) = $20, 000.
0.12 N →∞
|N
lim (−150, 000 − 5, 000t)e−0.1t | = $150, 000.
0
N →∞
29. To find the present value of an investment
generating f (t) = A + Bt dollars per year for N
years, divide the N −year time interval
0 ≤ t ≤ N into n equal sub-intervals of length
∆t years, and let tj denote the beginning of the
j th sub-interval.
Then, during the j th sub-interval, the amount
generated is approximately f (tj )∆t. Hence the
present value of an N −year investment is
27. To find the present value of an apartment
complex generating
lim
n→∞
f (t) = 10, 000 + 500t
dollars per year for N years, divide the N −year
interval 0 ≤ t ≤ N into n equal sub-intervals of
length ∆t years, and let tj denote the beginning
of the j th sub-interval.
Then, during the j th sub-interval, the amount
generated is approximately f (tj )∆t and at the
interest rate of 10 %, the present value is
−0.1tj
f (tj )e
∆t.
lim
n→∞
Z
=
n
X
=
f (tj )e−0.1tj ∆t
j=1
N
f (t)e−0.1t dt.
f (tj )e−rtj ∆t
j=1
N
(A + Bt)e−rt dt.
0
To find the present value P of the total income,
let N → ∞ to get
P
=
=
Hence, the present value of the apartment
complex over an N −year period is
Z
n
X
lim
N →∞
lim
N →∞
N
Z
(A + Bt)e−rt dt
0
"
B
1
|N
− (A + Bt)e−rt | +
0
r
r
Z
N
e−rt dt
0
N
1
B
|
− (A + Bt)e−rt − 2 e−rt |
0
N →∞
r
r
N
1
B −rt |
= − lim (A + Bt) +
e |
0
r N →∞
r
1
B
A
B
=
A+
= + 2.
r
r
r
r
=
lim
0
To find the present value P of the total income,
let N → ∞ to get
Z N
P = lim
(10, 000 + 500t)e−0.1t dt
N →∞
0
|N
= − lim 10(10, 000 + 500t)e−0.1t |
N →∞
0
31. To find the number of patients after N months,
divide the N −month time interval 0 ≤ t ≤ N
into n equal sub-intervals of length ∆t months,
and let tj denote the beginning of the j th
sub-interval.
Then, the number of people starting treatment
during the j th sub-interval is approximately
#
6.4. IMPROPER INTEGRALS
181
10∆t. Of these, the number still receiving
treatment at time t = N (that is, N − tj
months later) is approximately 10f (N − tj )∆t.
Hence, the number of patients receiving
treatment at time t = N is
lim
n→∞
=
n
X
=
N →∞
0
N →∞
2
Z
I=
0
x2 |2
x
dx =
=1
2
4 |0
(b)
I
10f (N − t)dt
Z
=
3
x2 |2
=
4 |1
4
=
(c)
I=
N →∞
Z
1
0
N
et/20 dt
37. (a)
I
0
=
|N
lim 200e−N/20 et/20 | = 200 patients.
=
0
N →∞
33. To find the number of units of the drug in the
patient’s body after N hours, divide the
N −hour time interval 0 ≤ t ≤ N into n equal
sub-intervals of length ∆t hours, and let tj
denote the beginning of the j th sub-interval.
Then, during the j th sub-interval,
approximately 5∆t units of the drug are
received. Of these, the number remaining at
time t = N (that is, N − tj hours later) is
approximately 5f (N − tj )∆t.
Hence, the number of units of the drug in the
patient’s body at time t = N is
lim
n→∞
Z
n
X
(b)
x
x2 |1
1
dx =
=
|
0
2
4
4
I
=
=
(c)
I
Z 1
3
(4x − x2 )dx
32 0
3
5
x3 |1
=
2x2 −
32
32
3 |0
=
=
39. (a)
I
5f (N − tj )∆t
=
j=1
N
=
5f (N − t)dt
and the number Q of units in the patient’s
body in the long run is
Z N
Q = lim 5
f (N − t)dt
N →∞
lim 5
N →∞
0
Z
0
N
−(N −t)/10
e
dt
=
(b)
I
Z 4
3
(4x − x2 )dx
32 0
3
x3 |4
=1
2x2 −
32
3 |0
Z 2
3
(4x − x2 )dx
32 1
11
3
x3 |2
2
=
2x −
32
32
3 |1
0
=
Z
0
lim 10e−N/20
=
x
dx
2
1
and the number of patients receiving treatment
in the long run is
Z N
P = lim 10
e−(N −t)/20 dt
=
2
=
0
N →∞
et/10 dt
0
|N
lim 50e−N/10 et/10 | = 50 units.
10f (N − tj )∆t
j=1
N
Z
lim 5e−N/10
35. (a)
N
Z
=
=
1
10
Z
∞
e−x/10 dx
0
1
lim
10 N →∞
Z
N
e−x/10 dx
0
N
−x/10 |
− lim e
|
0
N →∞
1
10
Z
=1
2
e−x/10 dx
0
|2
= −e−x/10 | = 0.1813
0
182
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
(c)
I
=
∞
Z
1
10
e−x/10 dx
5
1
lim
10 N →∞
=
P (10 < x < 15)
N
Z
I
Z
1
4
=
∞
=
(b)
I
=
=
=
(b)
I
=
=
=
(c)
I
=
=
=
=
=
xe−x/2 dx
(c)
P (12 < x)
0
xe
∞
e−0.2x dx
|∞
= −e−0.2x | = 0.0907
12
= 0.2e−0.2x
Z 5
P (0 < x < 5) = 0.2
e−0.2x dx
f (x)
0
5
−0.2x |
−e
|0
=
(b)
P (6 ≤ x)
dx
= 0.6321
= 1 − P (0 ≤ x ≤ 6)
Z 6
= 1 − 0.2
e−0.2x dx
0
6
−0.2x |
e
|0
4
−x/2
Z
= 0.2
12
2
1
|N 

− 4e−x/2 | 
lim −
0
1 N/2
4 N →∞
e
2
1
(4 + 4) = 2
4
Z
= 0.7981
N
# 45. (a)
"
Z N
1
2x |N
−x/2
lim − x/2 | + 2
e
dx
0
4 N →∞
e
0


1
4
= 0.0855
e−0.2x dx
0
8
−0.2x |
−e
|0
=
g(x) = e−x/2
G(x) = −2e−x/2
f (x) = x
f 0 (x) = 1
e−0.2x dx
8
Z
= 0.2
0
1
lim
4 N →∞
=
15
= 0.6065
xe−x/2 dx
Z
Z
10
15
−0.2x |
−e
|10
e−x/10 dx
P (x < 8)
41. (a)
= 0.2
5
|N
− lim e−x/10 |
5
N →∞
=
f (x) = 0.2e−0.2x
43. (a)
= 1+
2
= 0.3012
Z Z 4
1 4
2x |4
47. The waiting period is represented by the
− x/2 | + 2
e−x/2 dx
2
4 2
e
2
function
4
1
4
2
−x/2 |
1
−2 2 −
− 4e
for 0 ≤ t ≤ 20
f (t) =
|2 = 0.3298
4
e
e
20
1
4
Z
∞
−x/2
xe
I
dx
1
20
=
6
1
lim
4 N →∞
Z
N
xe−x/2 dx
49.
I
∞
=
=
1
lim
3 N →∞
6
"
#
Z N
1
2x |N
−x/2
e
dx
lim − x/2 | + 2
6
4 N →∞
e
6
N
1
4
12
−x/2 |
lim − N/2 + 3 − 4e
|6
4 N →∞
e
e
0.1991
Z
1
3
20
Z
dt = 0.6
8
e−x/3 dx
3
N
Z
e−x/3 dx
3
|N
= − lim e−x/3 |
3
N →∞
−N/3
= − lim (e
N →∞
− e−1 ) = 0.3679
6.5. REVIEW PROBLEMS
51.
P (12) = 0.08
183
12
Z
1
3. Let u = 5x − 2. Then du = 5dx or dx = du.
5
When x = −1, u = −7, and when x = 2, u = 8.
Hence,
Z 2
30(5x − 2)2 dx
e−0.08t dt = 0.6171
0
The probability that the grenade is defective
after one year and the spy will expire is
1 − 0.6171 = 0.3829.
53. (a) The capitalized cost for the first machine is
Z ∞
M1 = 10, 000+
1, 000(1+0.06t)e−0.09t dt
−1
=
6
Z
−7
0
f (t) = 1, 000(1 + .06t) g(x) = e−.09t
f 0 (t) = 60
G(t) = −11.11e−.09t
+
0
−7
1
Z
Hence
x2 −1
2xe
= 1, 710.
dx =
666.67e−.09t dt = 29, 185.19
Z
0
eu du
−1
0
N
Z
|8
u2 du = 2u3 |
4. Let u = x2 − 1. Then du = 2xdx.
When x = 0, u = −1, and when x = 1, u = 0.
|N
M1 = 104 − lim 11, 111.11(1 + .06t)e−.09t |
N →∞
8
|0
= eu |
= e0 − e−1 = 0.6321.
−1
0
which increases beyond bound.
M2
=
8, 000 +
Z
∞
1, 100e−0.09t dt
0
=
|∞
8, 000 − lim 12, 222.22e−.09t |
=
20, 222.22
0
N →∞
0
Thus the second machine ought to be
purchased.
(b) Continuing as above Writing exercise —
Answers will vary.
6.5
Review Problems
Review Problems
1.
Z
5. Let u = x2 − 6x + 2. Then du = (2x − 6)dx or
1
(x − 3)dx = du.
2
When x = 0, u = 2, and when x = 1, u = −3.
Z 1
Hence
(x − 3)(x2 − 6x + 2)3 dx
1
(5x4 − 8x3 + 1)dx
=
1
2
Z
−3
u3 du =
2
u4 |−3
65
=
8 |2
8
6. Let u = x2 + 4x + 5. Then du = (2x + 4)dx or
3
(3x + 6)dx = du.
2
When x = −1, u = 2, and when x = 1, u = 10.
Z 1
3x + 6
Hence
dx
2 + 4x + 5)2
(x
−1
Z 10
3
=
u−2 du
2 2
3 |10
3
= − | =
2u 2
5
0
|1
= (x5 − 2x4 + x)| = (1 − 2 + 1) − 0 = 0
0
2.
Z
4
√
( x + x−3/2 )dx
−1
−1
1
3/2
=
7. Let g(x) = ex and f (x) = x.
Then, G(x) = ex and f 0 (x) = 1. Thus
Z 1
Z 1
1
x
x|
xe dx = xe | −
ex dx
2x
3
4
17
|
− 2x−1/2 | =
.
1
3
1
=
−1
|
(xex − ex )|
= 2e−1 = 0.7358
−1
184
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
1
dx. When x = e,
x
2
u = 1, and when x = e , u = 2. Hence,
Z e2
Z 2
1
1 |2
1
dx
=
u−2 du = − | =
2
1
x(ln
x)
u
2
e
1
8. Let u = ln x. Then du =
9. Let g(x) = x2 and f (x) = ln x. Then,
1
x3
G(x) =
and f 0 (x) = . Thus
3
x
Z e
x2 ln xdx
1
Z
x3
1 e 3 1
|e
=
ln x| −
x dx
1
3
3 1
x
Z
e
3
e
1
x
|
=
ln x| −
x2 dx
1
3
3 1
3
x
x3 |e
=
ln x −
3
9 |1
3
x
1 |e
2e3 + 1
ln x −
=
=
3
3 |1
9
10. Let g(x) = e0.2x and f (x) = 2x + 1. Then,
G(x) = 5e0.2x and f 0 (x) = 2. Thus
Z 10
(2x + 1)e0.2x dx
13.
∞
Z
3x
dx
+1
0
Z N
3 lim
x−1 (x2 + 1)−1 dx
=
N →∞
0
1
|N
3 lim
ln(x2 + 1)| = ∞.
0
N →∞ 2
=
14.
x2
∞
Z
3e−5x dx = 3 lim
N →∞
0
= −
15.
Z
Z
N
e−5x dx
0
3
3
|N
lim e−5x | = .
0
5 N →∞
5
∞
xe−2x dx
0
Z
N
xe−2x dx
Z
1 −2x |N
1 N −2x
= lim
− xe
e
dx
|0 + 2
N →∞
2
0
1
1
1
|N
= lim (− xe−2x − e−2x )| =
0
N →∞
2
4
4
=
lim
N →∞
16.
0
∞
Z
0
3
2x2 e−x dx
0
10
=
|
5(2x + 1)e0.2x |
=
0.2x
0
[5(2x + 1)e
11.
∞
Z
√
3
0
=
=
12.
lim
N →∞
− 10
10
Z
e0.2x dx
0
0.2x
− 50e
|10
]| = 55e2 + 45
0
1
dx
1 + 2x
(1 + 2x)−1/3 dx
0
3
|N
(1 + 2x)2/3 | = ∞.
0
N →∞ 4
∞
(1 + 2x)−3/2 dx
0
=
lim
N →∞
Z
N
(1 + 2x)−3/2 dx
0
|N
= − lim (1 + 2x)−1/2 | = 1.
N →∞
Z
0
lim
Z
=
17.
N
Z
=
0
∞
2 lim
N →∞
N
Z
3
x2 e−x dx
0
N
3|
1
2
2 lim (− )e−x | = .
0
N →∞
3
3
x2 e−2x dx =
lim
N →∞
Z
N
x2 e−2x dx
0
1 2 −2x |N
x e
|0
N →∞ 2
Z N
+ lim
xe−2x dx
= − lim
N →∞
0
1
|N
= − lim x2 e−2x |
0
N →∞ 2
1
|N
− lim xe−2x |
0
N →∞ 2
1
1
|N
− lim e−2x | =
0
N →∞ 4
4
6.5. REVIEW PROBLEMS
18.
Z
185
∞
1
dx
x(ln x)2
2
Z N
1
dx
= lim
N →∞ 2
(ln x)2 x
N
1
1
|
= lim
−
=
N →∞
ln x |2
ln 2
P (0 ≤ X ≤ 3) =
=
=
N →∞
2(3 − x)
dx
9
0
2
x2 |3
(3x − )|
9
2 0
2
9
(9 − − 0) = 1.
9
2
=
(b)
3
x+2
0
P (1 ≤ X ≤ 2)
Z 2
=
f (x)dx
dx
|N
lim (x − 3 ln |x + 2|)| = ∞.
1
0
N →∞
=
(Note: x grows much more quickly than ln x.)
20.
∞
Z N
3
3
x5 e−x dx = lim
x3 (x2 e−x )dx
N →∞ 0
0
"
#
Z N
3
1 3 −x3 |N
= lim
− x e
x2 e−x dx
|0 +
N →∞
3
0
N
23. (a)
3
3
1
1
1
|
= lim
− x3 e−x − e−x | =
0
N →∞
3
3
3
Z 4
21. (a) P (1 ≤ X ≤ 4) =
f (x)dx
Z
=
=
2
2(3 − x)
dx
9
1
2
x2 |2
3x −
9
2 |1
2
4
1
1
6−
− 3−
= .
9
2
2
3
Z
P (0 ≤ X)
4
1
=
P (2 ≤ X ≤ 3)
=
Z
=
Z
=
1
x |4
dx = | = 1.
3
3 1
lim
N →∞
P (X ≤ 2)
=
=
f (x)dx
(b)
3
22. (a)
1
x |3
1
dx = | = .
2
3
3
3
0.2e−0.2x dx
0
0
N →∞
−0.2N
lim (−e
N →∞
+ 1) = 1.
P (1 ≤ X ≤ 4)
Z 4
=
f (x)dx
1
Z
=
2
4
0.2e−0.2x dx
1
f (x)dx
|4
= −e−0.2x |
−∞
Z 2
1
N
3
2
Z
Z
|N
lim (−e−0.2x )|
=
2
(c)
f (x)dx
0
=
(b)
∞
Z
1
Z
=
3
Z
=
∞
x−1
dx
x+2
0
Z N lim
1−
Z
f (x)dx
0
=
19.
3
Z
1
1
x |2
1
dx = | = .
3
3 1
3
−0.8
= −e
(c)
+ e−0.2 = 0.3694.
P (5 ≤ x)
Z ∞
=
f (x)dx
5
186
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
=
lim
N →∞
Z
g(u) = u + 3 h(u) = e−u
g 0 (u) = 1
H(u) = −e−u
N
0.2e−0.2x dx
5
|N
= − lim e−0.2x |
gav
5
N →∞
lim [−e−0.2N + e−1 ] = 0.3679.
=
N →∞
28.
24. (a)
P (0 < X) = lim
N →∞
N
Z
0
f (x)
5
dx
(x + 5)2
fav
1 |N
=1
N →∞ x + 5 |0
= −5 lim
(b)
P (1 ≤ X ≤ 9) =
Z
9
1
5
dx
(x + 5)2
fav
9
= −
(c)
5 |
5
5
= −
= 0.4762
x + 5 |1
6 14
P (3 ≤ X)
Z N
= lim
5
dx
N →∞ 3
(x + 5)2
5 |N
= − lim
N →∞ x + 5 |3
5
=
= 0.625.
8
25.
y
fav
26.
√
= f (x) = x3 − 3x + 2x1/2
Z
√
1 8 3
=
(x − 3x + 2x1/2 )dx
7 1
"
#
√
1 x4
3x2
2 2x3/2 |8
=
−
+
|1 = 135.6629
7 4
2
3
f (t)
fav
p
3
= t 10 − 2t2
Z 1
=
t(10 − 2t2 )1/3 dt
0
= −
27.
1
|2
[−(u + 3)e−u − e−u ]| = 1.594
0
2
=
g(u)
gav
3
|1
(10 − 2t2 )4/3 | = 1.0396
0
16
= e−u (u + 3)
Z
1 2 −u
=
e (u + 3)du
2 0
√
1
= x ln x = x ln x
2
Z
1 5
=
x ln xdx
8 1
f (x) = ln x g(x) = x
1
x2
f 0 (x) =
G(x) =
x
2
2
1 x
x2 |5
=
= 1.7647
ln x −
8 2
4 |1
29. Recall that P dollars invested at an annual
interest rate of 8 percent compounded
continuously will be worth
P e0.08t
dollars after t years.
Let tj denote the time (in years) of the j th
deposit of $1,200. This deposit will remain in
the account for 5 − tj years and hence will grow
to
1, 200e0.08(5−tj )
dollars. At the end of 5 years the amount in the
account is
n
X
1, 200e0.08(5−tj ) .
j=1
Rewrite the sum as
n
X
1, 200e0.08(5−tj ) ∆t.
j=1
This sum can be approximated by the definite
integral
Z 5
1, 200e0.08(5−t) dt
0
=
0.4
1, 200e
Z
5
e−0.08t dt
0
=
1, 200e0.4 −0.08t |5
(e
)| = $7, 377.37.
0
−0.08
6.5. REVIEW PROBLEMS
187
30. Recall that the present value of B dollars
payable t years from now with an annual
interest rate of 7 percent compounded
continuously is
Be−0.07t .
Divide the interval 0 ≤ t ≤ 10 into n equal
sub-intervals of length ∆t years, and let tj
denote the beginning of the j th sub-interval.
During the j th sub-interval, the income will
equal (dollars per year)(number of
years)= 1, 000∆t and the present value of the
income will
−0.07tj
= 1, 000(∆t)e
−0.07tj
= 1, 000e
∆t.
Hence, over the entire 10 years, the present
value of the investment
=
n
X
−0.07tj
1, 000e
∆t.
from now (that is 10 − tj weeks later) the
number of these still on the market will be
8(∆t)f (10 − tj ).
Hence, the total number N of homes on the
market 10 weeks from now will be
approximately
N = 200f (10) +
0
1, 000 −0.07t |10
e
=
|0
−0.07
1, 000 −0.7
= −
(e
− 1) = $7, 191.64.
0.07
−0.2t
31. Let f (t) = e
denote the fraction of the
homes that will remain unsold for t weeks. Of
the 200 homes currently on the market, the
number that will still be on the market 10
weeks from now is
200f (10) = 200e−2 .
To find the number of additional homes on the
market 10 weeks from now, divide the interval
0 ≤ t ≤ 10 into n equal sub-intervals of length
∆t weeks and let tj denote the beginning of the
j th sub-interval.
During the j th sub-interval, 8∆t additional
homes are placed on the market, and 10 weeks
8f (10 − tj )∆t.
j=1
Now, as n increases without bound, this
approximation improves while the sum
approaches the corresponding integral. Hence,
Z 10
N = 200f (10) +
8f (10 − t)dt
= 200e−2 + 8
j=1
Now as n increases without bound, this
approximation improves and the sum
approaches the corresponding integral. Hence,
the present value is
Z 10
1, 000e−0.07t dt
n
X
Z
0
10
e−0.2(10−t) dt
0
= 200e−2 + 8e−2
Z
10
e0.2t dt
0
= 200e−2 +
8e−2 0.2t |10
e | = 61.65
0
0.2
or about 62 homes.
32. Let’s focus on one bicycle at a time. The price
per month is
√
P (x) = 80 + 3 x
so the monthly revenue for one bicycle is
Z
x3/2
R(x) = (80 + 3x1/2 )dx = 80x + 3
+C
3/2
The initial revenue, when x = 0, will be R(0).
Thus
x3/2
R(x) = 80x + 3
+ R(0)
3/2
Over 16 months, that is when x = 16,
R(16) = 1, 280 + 128 + R(0)
The revenue per bicycle is R(16) − R(0) and for
5,000 bicycles it is $5, 000 × 1, 408 = 7, 040, 000
188
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
33. The density at a point r miles from the center is
D(r) = 25, 000e−0.05r
people per square mile.
The area of the ring of thickness dr at r miles
from the center is 2πrdr, so
50, 000πre−0.05r dr
people live in that ring.
The total number of people is
Z 2
N (r) = 50, 000π
re−0.05r dr.
Hence, if N is the total number of people
within 3 miles of the center of the city,
N
=
lim
n→∞
1
g(r) = e−0.05r
G(r) = −20e−0.05r
f (r) = r
f 0 (r) = 1
=
n
X
3
Z
5, 000(2π)re−0.1r dr
0
=
10, 000π
2
N (r)
=
|
50, 000π[−20re−0.05r |
1
+20
2
Z
e−0.05r dr]
1
50, 000π[−20(1.8097 − 0.95123)
−400(−0.04639)]
= 218, 010 people.
=
5, 000e−0.1rj (2πrj ∆r)
j=1
3
Z
re−0.1r dr
0
Applying integration by parts,
Z 3
re−0.1r dr
0
|3
= −10re−0.1r | −
0
Z
3
(−10)e−0.1r dr
0
|3
= (−10re−0.1r − 100e−0.1r )|
0
34. Divide the interval 0 ≤ r ≤ 3 into n equal
subintervals of length ∆r, and let rj denote the
beginning of the j th subinterval.
This divides the circular disc of radius 3 into n
concentric circles, as shown in the figure.
If ∆r is small, the area of the j th ring is
=
(−30e−0.3 − 100e−0.3 ) − (−100) = 3.6936
Hence, the total number of people within 10
miles of the center of the city is
Z 3
N = 10, 000π
re−0.1r dr
0
2πrj ∆r
= 116, 038
where 2πrj is the circumference of the circle of
radius rj that forms the inner boundary of the
ring and ∆r is the width of the ring.
Then, since D(r) = 5, 000e−0.1r is the
population density (people per square mile) r
miles from the center, it follows that the
number of people in the j th ring is
(Your answer may differ slightly due to
round-off errors.)
D(rj )(area of the j th ring)
= 5, 000e−0.1rj (2πrj ∆r)
35. Let D(t) denote the demand for oil after t
years. Then
D0 (t) = D0 e0.1t and D0 = 30
D(t) =
30
Z
0
=
10
|10
e0.1t dt = 30 × 10e0.1t |
0
300(e − 1) = 515.48 billion barrels
6.5. REVIEW PROBLEMS
189
Then, P (t) = 16 + 0.08t and
36. Let D(t) denote the demand for the product.
Since the current demand is 5,000 and the
demand increases exponentially,
dollars
month
dollars barrels
=
= P (t)(900)
barrel month
= 900(16 + 0.08t).
Z
R(t) = 900 (16 + 0.08t)dt
dR
dt
D(t) = 5, 000e0.02t
units per year.
Let R(t) denote the total revenue t years from
now. Then the rate of change of revenue is
dR
dt
dollars
year
dollars units
=
unit year
= 400D(t) = 400(5, 000e0.02t )
= 2, 000, 000e0.02t .
=
=
=
900(16t + 0.04t2 ) + C.
Since R(0) = 0 it follows that C = 0 and the
appropriate particular solution is
R(t) = 900(16t + 0.04t2 ).
Since the well will run dry in 36 months, the
total future revenue will be
The increase in revenue over the next 2 years is
Z 2
R(2) − R(0) =
2, 000, 000e0.02t dt
R(36) = 900[16(36) + 0.04(36)2 ] = $565, 056.
0
=
=
|2
100, 000, 000e0.02t |
0
=
=
3
(3x2 + 2)dx
g(t) = e−0.5t
G(t) = −2e−0.5t
Z 4
|4
100 −2te−0.5t | + 2
e−0.5t dt
2
0
|4
−2
−1
100 −8e + 4e − 4e−0.5t |
2
131.90 units
38. Let R(t) denote the total revenue generated
during the next t months and P (t) the price of
oil t months from now.
|3
(x3 + 2x)|
=
40.
A=
−1
4
Z
1
41.
A =
Note: This problem can be solved by
integration by parts, discussed in section 5, or
you can use your graphing utility.
=
A =
−1
2
Q(t)
Z
$4, 081, 077.
37. Let Q(t) denote the production after t hours.
Then t = 0 at 8:00 a.m., t = 2 at 10:0 a.m., and
t = 4 at noon. Thus,
Z 4
Q(t) = 100
te−0.5t dt
f (t) = t
f 0 (t) = 1
39.
Z
= 36
1 |4
3
1
dx = − | =
2
1
x
x
4
2
(2 + x − x2 )dx
9
x2
x3 |2
=
2x +
−
|
−1
2
2
3
−1
=
42. Break R into two subregions R1 and R2 as in
the accompanying figure.
The area of R = that of R1 + that of
Z 4
Z 8
√
8
R2 =
xdx +
dx
x
0
4
2 3/2 |4
16
|8
=
x | + 8 ln |x|| =
+ 8 ln 2.
0
4
3
3
2
Z 1
3
x
x5 |1
4
43.
(x − x )dx =
=
−
|
0
10
2
5
0
190
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
44. Break R into two subregions R1 and R2 as in
the accompanying figure.
Area of R = area of R1 + area of R2
Z 2
Z 1
(7x − x2 )dx +
[(8 − x2 ) − x2 ]dx
=
1
0
=
=
Z
The consumers’ surplus in part (c) is
equal to the area of the region between the
demand curve and the horizontal line
p = 32.
1
Z
2
(7x − x2 )dx +
(8 − 2x2 )dx
0
1
2
13
2x3 |2
7x
x3 |1
+
(8x
−
) =
.
−
3 |1
2
2
3 |0
45. (a) The demand function is
D(q) = 50 − 3q − q 2
dollars per unit.
To find the number of units bought when
the price is p = 32, solve the equation
32 = D(q) for q to get
2
32 = 50 − 3q − q ,
(q + 6)(q − 3) = 0 or q = 3 units
(b) The amount that consumers are willing to
spend to get 3 units of the commodity is
Z 3
D(q)dq
46. Since the price of chicken t months after the
beginning of the year is
P (t) = 0.06t2 − 0.2t + 1.2
dollars per pound, the average price during the
first six months is
Z 6
1
(0.06t2 − 0.2t + 1.2)dx
6−0 0
1
|6
=
(0.02t3 − 0.1t2 + 1.2t)|
0
6
1
=
[0.02(6)3 − 0.1(6)2 + 1.2(6)] = $1.32
6
0
=
=
Z
3
(50 − 3q − q 2 )dq
0
3q 2
q 3 |3
= $127.50.
50q −
−
2
3 |0
per pound.
47. In N years the population of the city will be
Z N
P0 f (N ) +
r(t)f (N − t)dt
0
(c) When the market price is $32 per unit, 3
units will be bought and the consumer’s
surplus will be
Z 3
D(q)dq − (32)(3)
0
=
Z
where P0 = 100, 000 is the current population,
f (t) = e−t/20
is the fraction of the residents remaining for at
least t years, and
3
(50 − 3q − q 2 )dq − 96 = $31.5.
r(t) = 100t
0
(d) The consumer’s willingness to spend in
part b) is equal to the area under the
demand curve p = D(q) from q = 0 to
q = 3.
is the rate of new arrivals. Hence, in the long
run, the number of residents will be
Z N
lim [100, 000e−N/20 +
100te−(N −t)/20 dt]
N →∞
0
6.5. REVIEW PROBLEMS
=
0 + lim 100e−N/20
N →∞
191
N
Z
run is
tet/20 dt]
lim [(20, 000e−N/10)
0
N →∞
Z N
|N
= lim 100e−N/20 [20tet/20 − 400et/20 ]|
0
N →∞
=
−N/20
lim 100(20N − 400 + 400e
N →∞
+
1, 000e−(N −t)/10 dt]
0
) = ∞.
=
Thus the population will increase without
bound.
lim [(20, 000e−N/10 )
Z N
+1, 000e−N/10
et/10 dt]
N →∞
0
48. The probability density function is
=
f (x) = 0.4e−0.4x .
(a)
P (1 ≤ X ≤ 2) =
2
Z
0.4e−0.4x dx
1
|2
= −e−0.4x | = 0.2210
1
which corresponds to 22.10 %.
(b)
P (X ≤ 2)
=
Z
|N
0 + 10, 000 lim e−N/10 et/10 | = 10, 000.
50. To find the present value of the investment in
N years, divide the N −year interval 0 ≤ t ≤ N
into n equal sub-intervals of length ∆t years,
and let tj denote the beginning of the j th
sub-interval.
Then, during the j th sub-interval, the amount
generated is approximately f (tj )∆t and the
present value is
f (tj )e−0.1tj ∆t
2
−0.4x
0.4e
dx
0
|2
= −e−0.4x | = 0.5507
0
which corresponds to 55.07%.
Hence, the present value of an N −year
investment is
n
X
lim
f (tj )e−0.1tj ∆t
n→∞
(c)
which corresponds to 44.93%.
49. It was determined in exercise 37 of section 6.1
that the number of subscribers in N years will
be
Z
N
r(t)f (N − t)dt
P0 f (N ) +
f (t)e−0.1t dt.
0
To find the present value P of the total
investment, let
Z N
P = lim
f (t)e−0.1t dt
N →∞
=
0
where P0 = 20, 000 is the current number of
subscribers,
f (t) = e−t/10
j=1
N
Z
=
P (X ≥ 2) = 1−P (X ≤ 2) = 1−0.5507 = 0.4493
=
lim
N →∞
0
N
Z
(8, 000 + 400t)e−0.1t dt
0
|N
lim [−10(8, 000 + 400t)e−0.1t ]|
0
N →∞
+4, 000 lim
N →∞
is the fraction of subscribers remaining at least
t years, and
r(t) = 1, 000
is the rate at which new subscriptions are sold.
Hence, the number of subscribers in the long
0
N →∞
=
N
Z
e−0.1t dt
0
lim [−10(8, 000 + 400t)e−0.1t
N →∞
|N
−40, 000e−0.1t ]|
0
=
|N
lim (−120, 000 − 4, 000t)e−0.1t | = $120, 000.
N →∞
0
192
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
x
0.0
0.5
1.0
1.5
2.0
51. Let x denote the time (in minutes) between
your arrival and the next batch of cookies.
Then x is uniformly distributed with
probability density function
(
1
if
0 ≤ x ≤ 45
f (x) =
45
0 otherwise
Hence, the probability that you arrive within 5
minutes (before or after) the cookies were
baked is
P (0 ≤ X ≤ 5) + P (40 ≤ X ≤ 45)
= 2P (0 ≤ X ≤ 5)
Z 5
1
2x |5
2
= 2
dx =
= .
|
0
45
9
0 45
f (x)
1 ∗ ∗∗
2.08
2.5
2.849
3.162
total: 10.591
x
f (x)
0.0 1 ∗ ∗∗
0.25
1.8
0.5
2.08
0.75
2.303
1.0
2.5
1.25
2.681
1.5
2.849
1.75
3.009
2.0
3.162
total: 20.384
I4 = 0.5(10.591) = 5.296
(b)
I8 = 0.25(20.384) = 5.096
The leftmost point was not used in the
calculation. Since the graph of the
function increases, the approximating
rectangles, using the right endpoint of
52. Let x denote the time (in minutes) between the
each interval, lead to an overestimate.
arrivals of successive cars. Then the probability
The actual value of the integral, as
density function is
estimated by the graphing utility, is
I = 4.8698.
ln x
0.5e−0.5x
if
0≤x
55.
f (x) =
f (x) =
x
0
otherwise
Z e2
1
ln x
fav = 2
= 0.313
The probability that two cars will arrive at
e −1 1
x
least 6 minutes apart is
First turn on your graphics utility – that’s very
important.
P (6 ≤ X < ∞)
Z ∞
For the HP48G:
ln x
=
f (x)dx
Enter
in equation writer form and store in
6
x
Z N
EXPR.
= lim
0.5e−0.5x dx
Now numerically integrate from 1 to 7.389 to
N →∞ 6
get
2 which is divided by e2 − 1 for the result to
N
−0.5x |
be
0.313.
= lim (−e
)| = 0.0498.
6
N →∞
For the TI-85:
Press 2nd CALC and F5 for FNINT. Enter
53. (a) Z ∞
((ln x)/x, x, 1, e2 ) ENTER. The result is
|N
0.07e−0.07u du = − lim e−0.07u | = 0.7047 1.9999999997 (say 2) and divide by e2 − 1 just
5
N →∞
5
like above.
Z 15
Z 10
(b)
−0.07u
56.
0.07e
du = 0.1466
e−1.1x dx = 0.909 076
I1 =
10
54. (a)
√
1
f (x) = 2 x +
x+1
Z 2
I=
f (x)dx
0
I2 =
I3 =
Z
Z
0
0
50
e−1.1x dx = 0.909 0909
0
100
e−1.1x dx = 0.909 090 909
6.5. REVIEW PROBLEMS
I4 =
193
1,000
Z
e−1.1x dx = 0.909 090 909
0
This integral converges to 0.90909.
Hint: For
Z the HP48G Compose
For the numerator of x
Z e2
I1 =
x ln xdx
1
n
e−1.1x dx using the equation writer
f (n) =
f (x) = ln x
dx
f 0 (x) =
x
0
and store as a function. Then press f (10),
f (50), f (100), f (1, 000) and numerically
evaluate each result.
For the TI-85, use 2nd CALC F5, FNINT(),
ENTER, and 2nd ENTRY to reproduce the
previous command line.
Z 10
1
57.
dx = 1.7918
I1 =
x
+
2
0
Z 50
1
I1 =
dx = 3.2581
x+2
0
Z 100
1
I1 =
dx = 3.9318
x+2
0
Z 1,000
1
I1 =
dx = 6.2166
x
+
2
0
This integral does not behave as if it were
finite. It diverges.
58.
f (x) = 1 − x2 , a = 0 , and b = 1
Z 1
2
x3 |1
2
A=
(1 − x )dx = x −
|0 = 3
3
0
2
I1
=
1 2
|e
x ln x| −
1
2
x =
=
Z
1
e2
x
dx
2
2
x2 |e
3e4 + 1
= e4 − | =
4 1
4
3e4 + 1
≈ 4.911
4(e2 + 1)
Thus x =
For the numerator of y
I2
=
e2
Z
(ln x)2 dx
1
=
2
|e
x(ln x)2 |
1
−2
Z
e2
ln xdx
1
2
|e
= 4e2 − 2(x ln x − x)| = 2(e2 − 1)
1
Thus y =
Z
3 1
(x − x3 )dx
2 0
3 x2
3
x4 |1
=
−
|
0
2 2
8
4
g(x) = xdx
x2
G(x) =
2
2(e2 − 1)
≈ 0.762
2(e2 + 1)
60. The Gini index for high school teachers is
Z 1
G1 = 2
[x − 0.33x3 − 0, 67x4 ]dx ≈ 0.567
0
y
3 1
(1 − 2x2 + x4 )dx
4 0
3
2
2x3
x5 |1
=
x−
+
4
5
3
5 |0
Z
=
=
59.
y
=
A =
ln x
Z e2
ln xdx
1
2
=
|e
(x ln x − x)| = e2 + 1
1
The Gini index for real estate brokers is
Z 1
G2 = 2
[0.72x − 0.72x2 ]dx
0
2
x
x3 |1
= 1.44
−
2
3 |0
= 0.24
The distribution of incomes amongst real estate
brokers is more fairly distributed.
194
CHAPTER 6. FURTHER TOPICS IN INTEGRATION
y = −x3 − 2x2 + 5x − 2
61.
intersects y = x ln x
at (0.4062, −0.3660) and (1, 0) according to the
graphics utility.
Numerically integrate
Z 1
(−x3 − 2x2 + 5x − 2 − x ln x)dx
0.406
to get A = 0.1692
y=
62.
intersects
y=
x−2
x+1
p
25 − x2
at (−1.8204, 4.6568) and (−4.6568, 1.8204)
according to the graphics utility.
To find the area numerically integrate
Z −1.8204 √
x−2
25 − x2 −
dx
x+1
−4.6568
to get A = 2.9987