3.
FUTURE VALUE OF A CONTINUOUS INCOME STREAM
If f(t) is the rate of flow of a continuous income stream,
0 ≤ t ≤ T, and if the income is continuously invested at a
rate r, compounded continuously, then the FUTURE VALUE, FV,
at the end of T years is given by:
FV =
T
r(T-t)dt =
!0 f(t)e
T
erT ! f(t)e-rtdt
0
The future value of a continuous income stream is the total
value of all money produced by the continuous income stream
(income and interest) at the end of T years.
4.
p
CONSUMERS' SURPLUS
If ( x , p ) is a point on the graph of
the price-demand equation p = D(x)
for a particular product, then the
CONSUMERS' SURPLUS, CS, at a price
level of p is
CS =
x
!0
CS
–
p
[D(x) - p ]dx
p = D(x)
which is the area between p = p and
p = D(x) from x = 0 to x = x .
Consumers' surplus represents the
x
–x
total savings to consumers who are
willing to pay more than p for the product but are still able to
buy the product for p .
5.
PRODUCERS' SURPLUS
If ( x , p ) is a point on the graph of the price-supply
equation p = S(x), then the PRODUCERS' SURPLUS, PS, at a
price level of p is
PS =
x
!0 [ p
- S(x)]dx
which is the area between p = p and p = S(x) from x = 0 to
x = x
p
p = S(x)
–
p
PS
–
x
426
CHAPTER 7
x
ADDITIONAL INTEGRATION TOPICS
Producers' surplus
represents the total gain
to producers who are
willing to supply units at
a lower price than p but
are still able to supply
units at p .
6.
EQUILIBRIUM PRICE AND EQUILIBRIUM QUANTITY
If p = D(x) and p = S(x) are the price-demand and the pricesupply equations, respectively, for a product and if ( x , p )
is the point of intersection of these equations, then p is
called the EQUILIBRIUM PRICE and x is called the EQUILIBRIUM
QUANTITY.
!
1
# 1&
1
1
1
dt = - e-2t = - e-2 - %" ( ≈ 0.43
0
$ 2'
2
2
1.
"0 e-2t
3.
"0 e4(2-!t)dt
2
!
=
!
2
2
· e-4t dt = e8 "0 e-4t
!
# 1
&2
= e8 %" e"4t (
$ 4
'0
!
) 1
# 1 &,
= e8 +" e"8 " %" (.
$ 4'* 4
!
!
1
1
= - + e8 ≈ 744.99
4
4
"0 e!8
!
!
5.
8 0.06(8-t)
e
dt =
0
"
!
"
!
7.
!
8 0.48
e
·
0
!
20
"0
e0.08t e0.12(20-t)dt =
8
e-0.06t dt = e0.48 "0 e-0.06t dt
!
#
&8
1
e"0.06t (
= e0.48 %"
$ 0.06
'0
!
e0.48 -0.48
= [e
– 1]
0.06
!
! 0.48
e
"1
=
≈ 10.27
0.06
!
20
e0.08t · e2.4 · e-0.12t dt
!
20
= e2.4 "0 e-0.04t dt
!
!
!
"0
#
& 20
1
e"0.04t (
= e2.4 %"
$ 0.04
' 0
!
e2.4
= (e-0.8 – 1)
0.04
!
!
1.6
2.4
"e
+ e
=
≈ 151.75
0.04
EXERCISE 7-2
427
9.
30
"0
!
30
500e0.02t e0.09(30-t)dt = 500 "0 e0.02t · e2.7 · e-0.09t dt
30
= 500e2.7 "0 e-0.07t dt
#
& 30
1
e"0.07t (
= 500e2.7 %"
!
$ 0.07
' 0
2.7
500e
= !
(e-2.1 – 1)
0.07
!500(e2.7 " e0.6)!
=
≈ 93,268.66
0.07
11. (A)
8 0.07(8-t)
!0 e
dt =
8 !0.56 - 0.07
!0 e
dt =
8
!
-0.07
= e0.56
dt =
! e
= (B)
0
0.56
e
0.07
8 0.56
!0 e
· e-0.07dt
e0.56 -0.07t 8
e
0
!0.07
[e-0.56 - 1] ≈ 10.72
e0.07t 8
8
0.56
0.07
0.56
(
e
e
)
dt
=
(
e
)
t
!0
0
0.07 0
1
= 8e0.56 [e0.56 - 1] ≈ 3.28
0.07
8
8
(C) e0.56 ! e-0.07dt ≈ 10.72 as in (A)
0
#
2
%
,
13. f(x) = $(x + 2)2
%&
0
x " 0
x < 0
2
dx
+ 2)2
!2
(x + 2)!1 6
6
= 2
=
0
(x + 2) 0
!1
1
3
= - + 1 =
= 0.75
4
4
Thus, Probability (0 ≤ x ≤ 6) = 0.75
(A) Probability (0 ≤ x ≤ 6) =
!
6
!0 f(x)dx
=
6
!0 (x
2
dx
(x + 2)2
!2 12
1
1
3
=
= - +
≈ 0.11
=
6
x + 2
7
4
28
(B) Probability (6 ≤ x ≤ 12) =
(C)
12
!6
f(x)dx =
f(x)
0.5
0
428
CHAPTER 7
x
6
12
ADDITIONAL INTEGRATION TOPICS
12
!6
15. We want to find d such that
Probability (0 ≤ x ≤ d) =
d
!0 f(x)dx
d
d + 2
d
0.2d
d
Now,
d
=
!0 (x
d
!0 f(x)dx
= 0.8:
2
d
2
!2
d
dx = =
+
1
=
2
+ 2)
d + 2
x + 2 0
d + 2
= 0.8
= 0.8d + 1.6
= 1.6
= 8 years
!0.01t
#
if t " 0
17. f(t) = $0.01e
0
otherwise
%
(A) Since t is in months, the probability of failure during the
warranty period of the first year is
12
Probability (0 ≤ t ≤ 12) =
!0
f(t)dt =
12
!0
0.01 -0.01t 12
e
0
!0.01
=
0.01e-0.01tdt
= -1(e-0.12 - 1) ≈ 0.11
24
24
-0.01t
dt = -1e-0.01t 12
!12 0.01e
= -1(e-0.24 - e-0.12) ≈ 0.10
(B) Probability (12 ≤ t ≤ 24 =
19. Probability (0 ≤ t ≤ ∞) = 1 =
But,
!
"0
f(t)dt =
12
!0
f(t)dt +
!
!
"0 f(t)dt
"12
f(t)dt
Thus, Probability (t ≥ 12) = 1 - Probability (0 ≤ t ≤ 12)
≈ 1 - 0.11 = 0.89
21. f(t) = 2500
Total income =
5
5
!0 2500dt = 2500t 0
= $12,500
f(t)
23.
2500
y = f(t)
Total Income
0
5
t
If f(t) is the rate of flow of a
continuous income stream, then the total
income produced from 0 to 5 years is the
area under the curve y = f(t) from t = 0
to t = 5.
25. f(t) = 400e0.05t
Total income =
3
!0
400e0.05t dt =
400 0.05t 3
e
0
0.05
= 8000(e0.15 - 1) ≈ $1295
EXERCISE 7-2
429
f(t)
27.
If f(t) is the rate of flow of a
continuous income stream, then the total
income produced from 0 to 3 years is the
area under the curve y = f(t) from t = 0
to t = 3.
y = f(t)
400
Total
Income
0
t
3
29. f(t) = 2,000e0.05t
The amount in the account after 40 years is given by:
40
40
2,000e0.05tdt = 40,000e0.05t 0 = 295,562.24 - 40,000 ≈ $255,562
Since $2,000 × 40 = $80,000 was deposited into the account, the
interest earned is:
$255,562 - $80,000 = $175,562
!0
31. f(t) = 1,650e-0.02t, r = 0.0625, T = 4.
4
FV = e0.0625(4) ! 1,650e-0.02te-0.0625tdt
=
0
0.25 4 -0.0825t
1,650e
e
dt
0
"0.0825t #4
0.25 e
%
!
= 1,650e
"0.0825 %$
= 20,000(e0.25 - e0.08) ≈ $7,218.
0
33. Total Income =
!
4
!0 1,650e
-0.02t
4
1650 "0.02t #
dt =
e
%$
"0.02
0
= -82,500(e-0.08 - 1) ≈ $6,343.
From Problem 31,
Interest earned = $7,218 - $6,343 = $875
!
35. Clothing store: f(t) = 12,000, r = 0.1, T = 5.
5
5
FV = e0.1(5) ! 12,000e-0.1tdt = 12,000e0.5 ! e-0.1tdt
0
0.5
0
12,000e
= -120,000e0.5(e-0.5 - 1)
!0.1
= 120,000(e0.5 - 1) ≈ $77,847
=
5
e-0.1t 0
Computer store: g(t) = 10,000e0.05t, r = 0.1, T = 5.
5
5
FV = e0.1(5) ! 10,000e0.05te-0.1tdt = 10,000e0.5 ! e-0.05tdt
0
0.5
0
10,000e
= -200,000e0.5(e-0.25 - 1)
!0.05
= 200,000(e0.5 - e0.25) ≈ $72,939
The clothing store is the better investment.
=
430
CHAPTER 7
5
e-0.05t 0
ADDITIONAL INTEGRATION TOPICS
37. Bond: P = $10,000, r = 0.08, t = 5.
FV = 10,000e0.08(5) = 10,000e0.4 ≈ $14,918
Business: f(t) = 2000, r = 0.08, T = 5.
5
5
0
0
FV = e0.08(5) ! 2000e-0.08tdt = 2000e0.4 ! e-0.08tdt
2000e0.4
5
e-0.08t 0 = -25,000e0.4(e-0.4 - 1)
!0.08
= 25,000(e0.4 - 1) ≈ $12,296
The bond is the better investment.
=
39. f(t) = 9,000, r = 0.0695, T = 8.
8
FV = e0.0695(8) ! 9000e-0.0695tdt
0
=
8
9000e0.556 e-0.0695t
0
!
8
9000e0.556 "0.0695t #
%
=
e
"0.0695
%$
0
-0.556
≈ -225,800.78(e
- 1) ≈ $96,304.
The relationship between present value (PV) and future value (FV) at
a continuously compounded
interest rate r (expressed as a decimal)
!
for t years is:
FV = PVert or PV = FVe-rt
Thus, we have:
PV = 96,304e-0.0695(8) = 96,304e-0.556 ≈ $55,230
41. f(t) = k, rate r (expressed as a decimal), years T:
T
T -rt
ke rT -rt T
rT
-rt
rT
FV = e ! ke dt = ke ! e dt =
e
0
0
0
!r
k
k
= - erT(e !rT ! 1) = r (erT - 1)
r
1
x, p = 150
20
1
First, find x : 150 = 400 x
20
x = 5000
5000 #
5000 #
&
1
1 &
400 "
x " 150(dx = !
x ( dx
CS = !
%250 "
%
0
0
$
'
$
20
20 '
#
1 2 & 5000
x (0
= %250x "
= $625,000
$
'
40
43. D(x) = 400 -
!
!
45.
p = D(x)
CS
p = 150
! The shaded area is the consumers' surplus
and represents the total savings to
consumers who are willing to pay more than
$150 for a product but are still able to
buy the product for $150.
x = 5,000
EXERCISE 7-2
431
47. p = S(x) = 10 + 0.1x + 0.0003x 2 , p = 67.
First find x : 67 = 10 + 0.1 x + 0.0003 x 2
0.0003 x 2 + 0.1 x - 57 = 0
"0.1 + 0.01 + 0.0684
x =
0.0006
!0.1 + 0.28
=
= 300
0.0006
PS =
=
300
!0 [67
300
!0 (57
- (10 + !
0.1x + 0.0003x2 )]dx
- 0.1x - 0.0003x2 )dx
300
= (57x - 0.05x 2 - 0.0001x 3 ) 0
= $9,900
49.
p = 67
PS
p = S(x)
The area of the region PS is the producers'
surplus and represents the total gain to
producers who are willing to supply units
at a lower price than $67 but are still
able to supply the product at $67.
x = 300
51. p = D(x) = 50 - 0.1x; p = S(x) = 11 + 0.05x
Equilibrium price: D(x) = S(x)
50 - 0.1x = 11 + 0.05x
39 = 0.15x
x = 260
Thus, x = 260 and p = 50 - 0.1(260) = 24.
CS =
260
!0
[(50 - 0.1x) - 24]dx =
260
!0
(26 - 0.1x)dx
260
= (26x - 0.05x 2 ) 0
= $3,380
PS =
260
!0
[24 - (11 + 0.05x)]dx =
260
!0
[13 - 0.05x]dx
260
p = D(x)
= (13x - 0.025x 2 ) 0
= $1,690
CS
p = 24
PS
p = S(x)
x = 260
432
CHAPTER 7
ADDITIONAL INTEGRATION TOPICS
53. D(x) = 80e-0.001x and S(x) = 30e0.001x
Equilibrium price: D(x) = S(x)
p = D(x)
CS
80e-0.001x = 30e0.001x
p = 49
8
PS
e0.002x =
3
p = S(x)
" 8%
0.002x = ln $ '
# 3&
" 8%
x = 490
ln$ '
# 3&
≈ 490
x =
0.002
!
Thus, p = 30e0.001(490) ≈ 49.
" 80e !0.001x
% 490
490
! 49x ' 0
CS = !
[80e-0.001x - 49]dx = $
!
0
# !0.001
&
= -80,000e-0.49 + 80,000 - 24,010 ≈ $6,980
"
490
30e 0.001x %' 490
PS = !
[49 - 30e0.001x]dx = $ 49x !
0
0.001 & 0
#
= 24,010 - 30,000(e0.49 - 1) ≈ $5,041
55. D(x) = 80 - 0.04x; S(x) =
Equilibrium price: D(x) =
80 - 0.04x =
Using a graphing utility,
x ≈ 614
Thus, p = 80 - (0.04)614
CS =
30e0.001x
S(x)
30e0.001x
we find that
≈ 55
614
[80 - 0.04x - 55]dx =
!0
614
!0
(25 - 0.04x)dx
614
= (25x - 0.02x 2 ) 0
≈ $7,810
PS =
614
!0
[55 - (30e0.001x)]dx =
(55 - 30e0.001x)dx
614
= (55x - 30,000e0.001x) 0
p = D(x)
p = 55
614
!0
CS
≈ $8,336
PS
p = S(x)
x = 614
EXERCISE 7-2
433
57. D(x) = 80e-0.001x; S(x) = 15 + 0.0001x 2
Equilibrium price: D(x) = S(x)
Using a graphing utility, we find that
x ≈ 556
Thus, p = 15 + 0.0001(556)2 ≈ 46
556
[80e-0.001x - 46]dx = (-80,000e-0.001x - 46x) 0
≈ $8,544
556
[46 - (15 + 0.0001x2)]dx =
CS =
!0
PS =
!0
556
556
!0
(31 - 0.0001x2)dx
0.0001 3 $ 556
"
= # 31x !
x % 0
3
p = D(x)
CS
≈ $11,507
p = 46
PS
p = S(x)
x = 556
59. (A) Price-Demand
Price-Supply
p = D(x)
p = S(x)
Graph the price-demand and price-supply models and find their
point of intersection.
6.8
0
30
Equilibrium quantity x = 21.457
Equilibrium price p = 6.51
6.3
(B) Let D(x) be the quadratic regression model
in part (A).
Consumers' surplus:
CS =
21.457
!0
[D(x) - 6.51]dx ≈ $1,774.
Let S(x) be the linear regression model in
part (A).
Producers' surplus
PS =
434
CHAPTER 7
21.457
!0
[6.51 - S(x)]dx ≈ $1,087
ADDITIONAL INTEGRATION TOPICS
EXERCISE 7-3
Things to remember:
1.
INTEGRATION-BY-PARTS FORMULA
∫u dv = uv - ∫v du
2.
INTEGRATION-BY-PARTS: SELECTION OF u AND dv
(a) The product udv must equal the original integrand.
(b) It must be possible to integrate dv (preferably by using
standard formulas or simple substitutions.)
(c) The new integral, ∫v du, should not be more complicated than
the original integral ∫u dv.
(d) For integrals involving x p eax, try
u = x p ; dv = eaxdx.
(e) For integrals involving x p (ln x)q, try
u = (ln x)q; dv = x p dx.
1. ∫xe3xdx
Let u = x and dv = e3xdx.
∫xe3xdx =
xe 3x
3
Then du = dx and v =
e3x
.
3
e3x
1
1
1
1
∫ 3 dx = 3 xe3x - 3 ∫e3xdx = 3 xe3x - 9 e3x + C
3. ∫x2 ln xdx
Let u = ln x and dv = x2dx.
x3
 
2
x
ln
xdx
=
(ln
x
)
∫
3
Then du =
x 3 dx
1
∫ 3 · x = 3 x3
x 3 ln x
1 x3
x 3 ln
=
+ C =
!
"
3
3 3
3
-
x3
dx
and v =
.
3
x
1 2
ln x ∫x dx
3
x
x3
+ C
!
9
5. ∫(x + 1)5(x + 2)dx
The better choice is u = x + 2, dv = (x + 1)5dx
The alternative is u = (x + 1)5, dv = (x + 2)dx, which will lead to an
integral of the form
∫(x + 1)4(x + 2)2dx.
1
Let u = x + 2 and dv = (x + 1)5dx. Then du = dx and v = (x + 1)6.
6
Substitute into the integration by parts formula:
1
1
∫(x + 1)5(x + 2)dx = 6 (x + 1)6(x + 2) - ∫ 6 (x + 1)6dx
1
1
= (x + 1)6(x + 2) (x + 1)7 + C
6
42
EXERCISE 7-3
435