7 ADDITIONAL INTEGRATION TOPICS
EXERCISE 7-1
Things to remember:
1.
AREA BETWEEN TWO CURVES
If f and g are continuous and f(x) ≥ g(x) over the interval
[a, b], then the area bounded by y = f(x) and y = g(x), for
a ≤ x ≤ b, is given exactly by:
y
A =
y = f(x)
b
"a
[f(x) ! g(x)]dx .
A
b
a
2.
y = g(x)
x
GINI INDEX OF INCOME CONCENTRATION
If y = f(x) is the equation of a Lorenz curve, then the
1
Gini Index = 2 " [x ! f(x)]dx .
0
1. A =
b
3. A =
!a g(x)dx
b
"a [!h(x)]dx)
5. Since the shaded region in Figure (c) is below the x-axis, h(x) ≤ 0.
b
!a h(x)dx
Thus,
represents the negative of the area of the region.
y=x+4
y
7. y = x + 4; y = 0 on [0, 4]
4
"1
%4
A = ! (x + 4)dx = $ x 2 + 4x ' 0
0
#2
&
4
= (8 + 16) – 0 = 24
x
4
!
y
9. y = -2x – 5; y = 0 on [-1, 2]
A
2
= - "1 (-2x – 5)dx
2
= "1 (2x + 5)dx = (x2 + 5x) 2
"1
#
#
!
!
–1
x
2
–3
= 14 – (1 – 5) = 18
y = -2x - 5
!
EXERCISE 7-1
415
y
2
11. y = x
A
–3
– 20; y = 0 on [-3, 0]
0
= - "3 (x2 – 20 )dx
#
1
0
= "3 (20 – x2)dx = %20x "
$
3
#
#
&
x 3( 0
' "3
= 0 - (-60 + 9) = 51
!
x
3
–20
y
!
! 3]
13. y = -x2 + 10; y != 0 on [-3,
# 1
&
A = ∫(-x2 + 10)dx = %" x 3 + 10x ( 3
$ 3
' "3
= (-9 + 30) – (9 – 30) = 42
10
x
–3
!
!
3
y
3
15. y = x
+ 1; y = 0 on [0, 2]
# 1
&
A = ! (x3 + 1)dx = %" x 4 + x ( 2
0
$ 4
' 0
= 4 + 2 = 6
2
2
1
x
2
–1
–2
!
!
y
17. y = x(1 – x); y = 0 on [-1, 0]
A
0
= - "1
#
0
= - "1 (x – x2)dx
0
= "1 (x2 – x)dx
#1
&
1
= % x 3 " x 2( 0
$3
' "1
2
#
x(1 – x)dx
#
!
!
1
# 1
1&
= 0 - %" " (
$ 3
2'
5
!
=
≈ 0.833
6
!
!
y
!
19. y = -ex; y = 0 on [-1,
1]
1
1
1
A = - # "1 -ex dx = # "1 ex dx
!
= ex 1 = e – e-1
–1
!
!
1
; y = 0
x
e 1
A = "1
dx =
x
= ln e - ln
21. y =
!
!
416
!
CHAPTER 7
!
on [1, e]
ln x e
1
x
1
"1
≈ 2.350
x
–1
y
3
2
1
1 = 1
!
ADDITIONAL
INTEGRATION TOPICS
1
e
x
y
x
23. y = -e-2x; y = 0 on [0, 1]
1
A = - "0 -e-2x dx =
!
1
dx
1
1
1
= - e-2x 1 = - e-2 +
0
2
2
2
≈ 0.432
!
25. A =
1
"0 e-2x
–1
!
b
#a ["f(x)]dx
!
!
27. a =
!
f(x)dx +
c
!b
d
d
!c [-f(x)]dx
c
b
29. A = ! [f(x) - g(x)]dx 31. A = ! [f(x) - g(x)]dx + ! [g(x) - f(x)]dx
c
a
b
!
33. Find the x-coordinates of the points of intersection of the two
curves on [a, d] by solving the equation f(x) = g(x), a ≤ x ≤ d,
to find x = b and x = c. Then note that f(x) ≥ g(x) on [a, b],
g(x) ≥ f(x) on [b, c] and f(x) ≥ g(x) on [c, d].
Thus,
Area =
b
!a [f(x)
35. A = A1 + A2 =
=
0
"!2
0
"!2
- g(x)]dx +
-xdx +
xdx +
c
[g(x) - f(x)]dx +
!b
d
!c [f(x)
- g(x)]dx
1
!0 -(-x)dx
1
!0 xdx
x2 0
x2 1
= +
2 !2
2 0
"
" 12
%
(!2)2 %'
+ $
! 0'
= - $0 !
2 &
#
#2
&
1
5
= 2 +
= 2.5
=
2
2
37. A = A1 + A2 =
=
2
-(x2 - 4)dx +
!0
2
!0 (4
- x2)dx +
3
2
!2 (x
3
2
!2 (x
- 4)dx
- 4)dx
"
" x3
% 3
x 3 %' 2
$
$
'
!
4
x
= 4x !
+
3& 0
#
# 3
& 2
8
"
" 27
"8
= # 8 ! $% + #
! 12$% ! # ! 8$%
3
3
3
16
39
16
23
= 13 ≈ 7.667
=
!
=
3
3
3
3
EXERCISE 7-1
417
39. A = A1 + A2 =
0
"!2
(x2 - 3x)dx +
y
=
10
6
4
A1 2
41. A =
2
"!1
"!2
2
- 3x)dx
(x2 - 3x)dx +
A
12 2 3 4 5 6
x
[12 - (-2x + 8)]dx =
2
"!1
(2x + 4)dx
! 2x 2
$ 2
2
+ 4x & !1 = (x2 + 4x) !1
= #
" 2
%
= (4 + 8) - (1 - 4)
= 12 + 3 = 15
43. A =
3%
"
2
$ 12x ! 3x ' 2
(12
3
x
)
dx
=
"!2
3 & !2
#
2
2
= (12x - x3) !2
= (12·2 - 23) - [12·(-2) - (-2)3]
= 16 - (-16) = 32
45. (3, -5) and (-3, -5) are the points of
intersection.
A =
3
"!3
[4 - x2 - (-5)]dx
3%
"
2
$ 9x ! x ' 3
(9
x
)
dx
=
"!3
3 & !3
#
#
#
33 &(
("3)3 &(
" % 9("3) "
= %9 ! 3 "
3'
3 '
$
$
= 18 + 18 = 36
=
418
2
!0
(3x - x2)dx
3 2$ 0
1 3$ 2
"1
"3
= # x3 !
x % !2 + # x 2 !
x % 0
3
2
2
3
8
" 8
"
= 0 - #!
! 6$% + #6 ! $% - 0 = 12
3
3
8
–3 –2 –1
–2
0
2
!0 -(x
3
CHAPTER 7
ADDITIONAL INTEGRATION TOPICS
47. A =
2
"!1
[(x2 + 1) - (2x - 2)]dx
# x3
&2
2
2
%
(
x
2
x
+
3)
dx
=
"!1
% 3 " x + 3x (( !1
$
'
#8
& # 1
&
= % " 4 + 6( " %" " 1 " 3(
$3
' $ 3
'
= 3 - 4 + 6 + 1 +! 3 = 9
=
!
49. A =
2
) 0.5x
# 1 &,
" %" (.dx
$ x '2 " 0.5x
1%
+ ' dx
= ! $e
1 #
x&
0.5
x
!e
$ 2
!
+ ln x & 1
= #
" 0.5
%
! = 2e + ln 2 - 2e0.5 ≈ 2.832
51. y =
2
!1 +*e
9 " x 2 ; y = 0 on [-3, 3]
area =
3
# "3
!
9 " x 2 dx = area of semicircle of radius 3
9
= π ≈ 14.137
2
!
53. y! = - 16 " x 2 ; y = 0 on [0, 4]
area =
4
"0
!
16 " !
x 2 dx = area of quarter circle of radius 4
1
=
· 16π = 4π ≈ 12.566
4
! !
55. y = - 4 " x 2 ; y =
4 " x 2 , on [-2, 2]
2
2
2
area = # "2 [ 4 " !
x 2 - (- 4 " x 2 )]dx = # "2 2 4 " x 2 dx = 2 # "2
= 2 area of semi circle of radius 2
!
= area !of circle of radius 2 = 4π ≈ 12.566
!
!
! graphs
! y != 2x 2 + 3x -!2
57. The
of y = 3 - 5x - 2x 2 and
are shown at the right. The x-coordinates of the
points of intersection are: x1 = -2.5, x2 = 0.5.
A =
=
0.5
"!2.5
4 " x 2 dx
!
[(3 - 5x - 2x2 ) - (2x2 + 3x - 2)]dx
#
4 3 & 0.5
2
2
"!2.5 (5 - 8x - 4x )dx = %$5x " 4x " 3 x (' !2.5
= 1.333 + 16.667 = 18
0.5
x1
x2
!
EXERCISE 7-1
419
1
are shown below.
x
The x-coordinates of the points of intersection are: x1 = 0.5, x2 = 4.
59. The graphs of y = -0.5x + 2.25 and y =
A =
4
!0.5
# 1
= %"
$ 4
!= [-4
)
# 1 &,
+("0.5x + 2.25) " % (. dx
$ x '*
& 4
9
x 2 + x " ln x ( 0.5
'
4
+ 9 - ln 4] - [-0.0625 + 1.125 - ln(0.5)]
x1
≈ 1.858
x2
!
y
= ex and
61. The graphs of y
are shown at the right.
A =
-x
y = e , 0 ≤ x ≤ 4,
4
40
30
4
x
-x
x
-x
!0 (e - e )dx = (e + e ) 0
= e4 + e-4 - (1 + 1)
≈ 52.616
20
–2
63. The graphs are given at the right. To find the points of
intersection, solve:
x3 = 4x
x3 - 4x = 0
x(x2 - 4) = 0
x(x + 2)(x - 2) = 0
Thus, the points of intersection are
(-2, -8), (0, 0), and (2, 8).
A = A1 + A2 =
0
"!2
(x3 - 4x)dx +
2
!0 (4x
" x4
"
% 0
= $
! 2x2 ' !2 + $ 2x 2
# 4
&
#
#("2)4
&
2
= 0 - %
" 2("2) ( +
%$ 4
('
= -4 + 8 + 8 - 4 = 8
!
- x3)dx
x4 %' 2
4 & 0
#
24 &
2
%2(2 ) "
( - 0
4 ('
%$
65. The graphs are given at the right.
!
!
To find the points
of intersection,
solve:
3
2
x - 3x - 9x + 12 = x + 12
x3 - 3x2 - 10x = 0
x(x2 - 3x - 10) = 0
x(x - 5)(x + 2) = 0
x = -2, x = 0, x = 5
Thus, (-2, 10), (0, 12), and (5, 17) are
the points of intersection.
420
CHAPTER 7
ADDITIONAL INTEGRATION TOPICS
A
10
2
4
x
A = A1 + A2
0
= "
[x3 - 3x2 - 9x + 12 - (x + 12)]dx
!2
5
=
0
"!2
(x3 - 3x2 - 10x)dx +
+ 12 - (x3 - 3x2 - 9x + 12)]dx
!0 [x
+
5
3
!0 (-x
+ 3x2 + 10x)dx
" x4
" x4
% 0
% 5
= $
! x 3 ! 5x 2 ' !2 + $ !
+ x 3 + 5x 2' 0
# 4
&
# 4
&
#("2)4
&
# !54
&
= -%
+ 53 + 5 " 52(
" ("2)3 " 5("2)2( + %
$ 4
'
%$ 4
('
375
407
= 8 +
= 101.75
=
4
4
67.! The graphs are given at the right. To find the points of
intersection, solve:
x 4 - 4x 2 + 1
x 4 - 5x 2 + 4
(x 2 - 4)(x 2 - 1)
x
=
=
=
=
x2 - 3
0
0
-2, -1, 1, 2
A2
!1
[(x2 - 3) - (x4 - 4x2 + 1)]dx +
!1
(-x4 + 5x2 - 4)dx +
=
"!2
=
"!2
1
"!1
A3
A1
A = A1 + A2 + A3
(x4 -
1
4
2
2
"!1 [(x - 4x + 1) - (x - 3)]dx
2
+ ! [(x2 - 3) - (x4 - 4x2 + 1)]dx
1
2
2
5x + 4)dx + ! (-x4 + 5x2 - 4)dx
1
" x5
" x5
% !1
5 3
= $!
+
x ! 4x ' !2 + $
!
# 5
3
&
# 5
#1
&
# 32
&
5
40
+ 4( - %
"
+ 8( +
= % "
$5
'
$5
'
3
3
" x5
% 1
% 2
5 3
5 3
x + 4x ' !1 + $ !
+
x ! 4x ' 1
3
&
# 5
3
&
#1
&
# 1
&
5
5
+ 4( - %" +
" 4(
% "
$5
'
$ 5
'
3
3
# 32
&
# 1
&
40
5
+
" 8( - %" +
" 4( = 8
+ %"
$ 5
'
$ 5
'
3
3
!
!
!
!
69. The graphs are given below. The x-coordinates of the points of
intersection are: x1 = -2, x2 = 0.5, x3 = 2
!
!
A = A1 + A2
=
0.5
"!2
[(x3 - x2 + 2) - (-x3 + 8x - 2)]dx
+
=
0.5
"!2
2
!0.5
[(-x3 + 8x - 2) - (x3 - x2 + 2)]dx
(2x3 - x2 - 8x + 4)dx +
2
!0.5
(-2x3 + x2 + 8x - 4)dx
EXERCISE 7-1
421
#1
& 0.5
# 1
1
= % x 4 " x 3 " 4x 2 + 4x ( !2 + %" x 4 +
$2
'
$ 2
3
#1
&
#
&
1
8
"
" 1 + 2( - %8 +
" 16 " 8(
= %
$ 32
'
$
'
24
3
#
&
8
+ 16 " 8( + %"8 !+
$
'
3
1
1
= 18 +
! ! ≈ 17.979
16
12
!
!
!
!
& 2
1 3
x + 4x 2 " 4x ( 0.5
'
3
# 1
&
1
+
+ 1 " 2(
%"
$ 32
'
24
A2
x1
x2 x3
A1
71. The graphs are given at the right. The x-coordinates of the points of
intersection are: x1 ≈ -1.924, x2 ≈ 1.373
A =
1.373
"!1.924
[(3 - 2x) - e-x]dx
1.373
= (3x - x 2 + e-x) -1.924
≈ 2.487 - (-2.626) = 5.113
73. The graphs are given at the right. The x-coordinates of the points of
intersection are: x1 ≈ -2.247, x2 ≈ 0.264, x3 ≈ 1.439
A = A1 + A2 =
0.264
"!2.247
[ex - (5x - x3)]dx
+
1.439
!0.264
[5x - x3 - ex]dx
5
1
1 4
"
"5
0.264
1.439
= # e x ! x 2 + x 4$% !2.247 + # x 2 !
x ! e x $% 0.264
2
4
2
4
≈ (1.129) - (-6.144) + (-0.112) - (-1.129) = 8.290
75. y = e-x; y = ln x ; 2 ≤ x ≤ 5
The graphs of y1 = e-x and y2 =
ln x are
2
!
!
0
7
0
Thus, A =
422
CHAPTER!7
5
!2 (
A2
ln x - e-x)dx ≈ 3.166
ADDITIONAL INTEGRATION TOPICS
x1
A1
x2 x3
2
77. y = ex ; y = x + 2
The graphs of y1 = ex
2
and y2 = x + 2 are
10
-2
2
0
The curves intersect at
x ! "0.588
x ! 1.057
and
y ! 1.412
y ! 3.057
79.
10
!5
R(t)dt =
1.057
"!0.588
10 "
!5
10
%
1
100
+ 10' dt = 100 !
dt +
$
5 t + 10
#t + 10
&
10
2
(x + 2 - ex )dx ≈ 1.385
10
!5
10dt
10
= 100 ln(t + 10) 5
+ 10t 5
! = 100 ln 20 - 100 ln 15 + 10(10 - 5)
= 100 ln 20 - 100 ln 15 + 50 ≈ 79
The total production from the end of the fifth year to the end of the
tenth year is approximately 79 thousand barrels.
81. To find the useful life, set R'(t) = C'(t) and solve for t:
9e-0.3t = 2
2
e-0.3t = 9
2
-0.3t = ln 9
-0.3t ≈ -1.5
t ≈ 5 years
5
!0
[R'(t) - C'(t)]dt] =
5
-0.3t
!0 [9e
5
- 2]dt
= 9 ! e-0.3tdt 0
9
-0.3t 5
2
dt
=
e
0
-0.3
!0
5
5
- 2t 0
= -30e-1.5 + 30 - 10
= 20 - 30e-1.5 ≈ 13.306
The total profit over the useful life of the game is approximately
$13,306.
EXERCISE 7-1
423
83. For 1935: f(x) = x2.4
1
1
0
0
Gini Index = 2 ! [x - f(x)] = 2 ! (x - x2.4)dx
" x2
x3.4 %' 1
$
!
= 2
3.4 & 0
# 2
#1
1 &
= 2% "
( ≈ 0.412
$2
3.4'
For 1947: g(x) = x1.6
1
1
0
0
Gini Index = 2 ! [x - g(!x)]dx = 2 ! (x - x1.6)dx
" x2
x 2.6 %' 1
$
!
= 2
2.6 & 0
# 2
#1
1 &
= 2% "
( ≈ 0.231
$2
2.6'
Interpretation: Income was more equally distributed in 1947.
85. For 1963: f(x) = x10
Gini Index = 2 !
1
0
!
1
[x - f(x)]dx = 2 ! (x - x10)dx
0
" x2
x11 %' 1
$
!
= 2
11 & 0
# 2
#1
1&
= 2% "
( ≈ 0.818
$ 2 11'
For 1983: g(x) = x12
1
1
Gini Index = 2 ! [x - g(x)]dx = 2 ! (x - x12)dx
0
0
!
" x2
x13 %' 1
$
!
= 2
13 & 0
# 2
#1
1&
= 2% "
( ≈ 0.846
$ 2 13'
Interpretation: Total assets were less equally distributed in 1983.
87. (A)
(B) Gini Index:
!
Lorenz curve:
f(x) = 0.3125x2 + 0.7175x - 0.015.
424
CHAPTER 7
ADDITIONAL INTEGRATION TOPICS
1
2 ! [x - f(x)]dx ≈ 0.104
0
89. W(t) =
10
!0
W'(t)dt =
10
!0
10 0.1t
e
0
0.3e0.1tdt = 0.3 !
dt
10
10
0.3
= 0.1 e0.1t 0
= 3e0.1t 0
= 3e - 3 ≈ 5.15
Total weight gain during the first 10 hours is approximately 5.15
grams.
91. V =
4
!2
4
4 1
15
dt = 15 !
dt = 15 ln t 2
2 t
t
!4
= 15 ln 4 - 15 ln 2 = 15 ln " #$ = 15 ln 2 ≈ 10
2
Average number of words learned during the second 2 hours is 10.
EXERCISE 7-2
Things to remember:
1. PROBABILITY DENSITY FUNCTION
A function f which satisfies the following three conditions:
a. f(x) ≥ 0 for all real x.
b. The area under the graph of f over the interval (-∞, ∞)
is exactly 1.
c. If [c, d] is a subinterval of (-∞, ∞), then the
probability that the outcome x of an experiment will be
in the interval [c, d], denoted Probability (c ≤ x ≤ d),
is given by
Probability (c ≤ x ≤ d) =
d
!c f(x)dx
f
f
Area = 1
c
d
!c f(x)dx
2.
d
= Probability (c ≤ x ≤ d)
TOTAL INCOME FOR A CONTINUOUS INCOME STREAM
If f(t) is the rate of flow of a continuous income stream,
then the TOTAL INCOME produced during the time period from
t = a to t = b is:
Total income =
b
!a f(t)dt
y
y = f(t)
Total income
a
b
t
EXERCISE 7-2
425