590
CHAPTER 5
THE INTEGRAL
y
1
θ
x
1
b
87.
Suppose that f and g are continuous functions such that, for all a,
Z
a
!a
f .x/ dx D
Z
a
g.x/ dx
!a
Give an intuitive argument showing that f .0/ D g.0/. Explain your idea with a graph.
Ra
Ra
SOLUTION Let !a f .x/ dx D !a g .x/ dx. Consider what happens as a decreases in size, becoming very close to zero.
Intuitively, the areas of the functions become .a ! .!a//.f .0// D 2a.f .0// and .a ! .!a//.g.0// D 2a.g.0//. Because we know
these areas must be the same, we have 2a.f .0// D 2a.g.0// and therefore f .0/ D g.0/.
88. Theorem 4 remains true without the assumption a " b " c. Verify this for the cases b < a < c and c < a < b.
Rc
Rb
Rc
SOLUTION The additivity property of definite integrals states for a " b " c, we have a f .x/ dx D a f .x/ dx C b f .x/ dx.
Rc
Ra
Rc
" Suppose that we have b < a < c. By the additivity property, we have
b f .x/ dx D b f .x/ dx C a f .x/ dx. Therefore,
Rc
Rc
Ra
Rb
Rc
a f .x/ dx D b f .x/ dx ! b f .x/ dx D a f .x/ dx C b f .x/ dx.
Rb
Ra
Rb
" Now suppose that we have c < a < b. By the additivity property, we have
c f .x/ dx D c f .x/ dx C a f .x/ dx.
Rc
Ra
Rb
Rb
Rb
Rc
Therefore, a f .x/ dx D ! c f .x/ dx D a f .x/ dx ! c f .x/ dx D a f .x/ dx C b f .x/ dx.
" Hence the additivity property holds for all real numbers a, b, and c, regardless of their relationship amongst each other.
5.3 The Fundamental Theorem of Calculus, Part I
Preliminary Questions
1. Suppose that F 0 .x/ D f .x/ and F .0/ D 3, F .2/ D 7.
(a) What is the area under y D f .x/ over Œ0; 2! if f .x/ # 0?
(b) What is the graphical interpretation of F .2/ ! F .0/ if f .x/ takes on both positive and negative values?
SOLUTION
(a) If f .x/ # 0 over Œ0; 2!, then the area under y D f .x/ is F .2/ ! F .0/ D 7 ! 3 D 4.
(b) If f .x/ takes on both positive and negative values, then F .2/ ! F .0/ gives the signed area between y D f .x/ and the x-axis.
2. Suppose that f .x/ is a negative function with antiderivative F such that F .1/ D 7 and F .3/ D 4. What is the area (a positive
number) between the x-axis and the graph of f .x/ over Œ1; 3!?
Z 3
SOLUTION
f .x/ dx represents the signed area bounded by the curve and the interval Œ1; 3!. Since f .x/ is negative on Œ1; 3!,
1
Z 3
f .x/ dx is the negative of the area. Therefore, if A is the area between the x-axis and the graph of f .x/, we have:
1
AD!
3.
(a)
(b)
(c)
Z
3
1
f .x/ dx D ! .F .3/ ! F .1// D !.4 ! 7/ D !.!3/ D 3:
Are the following statements true or false? Explain.
FTC I is valid only for positive functions.
To use FTC I, you have to choose the right antiderivative.
If you cannot find an antiderivative of f .x/, then the definite integral does not exist.
SOLUTION
(a) False. The FTC I is valid for continuous functions.
(b) False. The FTC I works for any antiderivative of the integrand.
(c) False. If you cannot find an antiderivative of the integrand, you cannot use the FTC I to evaluate the definite integral, but the
definite integral may still exist.
The Fundamental Theorem of Calculus, Part I
S E C T I O N 5.3
4. Evaluate
SOLUTION
Z
9
2
f 0 .x/ dx where f .x/ is differentiable and f .2/ D f .9/ D 4.
Because f is differentiable,
Z
9
f 0 .x/ dx D f .9/ ! f .2/ D 4 ! 4 D 0.
2
Exercises
In Exercises 1–4, sketch the region under the graph of the function and find its area using FTC I.
1. f .x/ D x 2 , Œ0; 1!
SOLUTION
y
1
0.8
0.6
0.4
0.2
x
0.2
0.4
0.6
0.8
1
We have the area
AD
2. f .x/ D 2x ! x 2 , Œ0; 2!
Z
1
0
ˇ
1 3 ˇˇ1
1
x dx D x ˇ D :
3
3
0
2
SOLUTION
y
1
0.8
0.6
0.4
0.2
0.5
Let A be the area indicated. Then:
Z 2
Z
AD
.2x ! x 2 / dx D
0
3. f .x/ D
x !2 ,
2
0
2x dx !
Z
2
0
Œ1; 2!
1
1.5
2
x
ˇ2
ˇ
!
"
ˇ
1 ˇ2
8
4
x 2 dx D x 2 ˇˇ ! x 3 ˇˇ D .4 ! 0/ !
!0 D :
3
3
3
0
0
SOLUTION
y
1.0
0.8
0.6
0.4
0.2
x
1.0
1.2
1.4
1.6
1.8
2.0
We have the area
AD
#
4. f .x/ D cos x, 0;
SOLUTION
!
2
$
Z
2
1
x
!2
ˇ2
x !1 ˇˇ
1
1
dx D
ˇ D! C1D :
!1 ˇ
2
2
1
591
592
CHAPTER 5
THE INTEGRAL
y
1
0.8
0.6
0.4
0.2
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
x
Let A be the shaded area. Then
AD
Z
!=2
0
0
In Exercises 5–42, evaluate the integral using FTC I.
Z 6
5.
x dx
3
SOLUTION
6.
Z
9
Z
6
x dx D
3
2 dx
0
SOLUTION
7.
Z
1
0
8.
2
SOLUTION
9.
2
0
10.
2
!2
Z
0
3
12.
1
!1
13.
1
0
Z
ˇ1
ˇ
.4x ! 9x 2 / dx D .2x 2 ! 3x 3 /ˇˇ D .2 ! 3/ ! .0 ! 0/ D !1.
0
2
!3
u2 du D
ˇ
1 3 ˇˇ2
1
1
35
u ˇ D .2/3 ! .!3/3 D
.
3
3
3
3
!3
Z
2
0
ˇ2
ˇ
.12x 5 C 3x 2 ! 4x/ dx D .2x 6 C x 3 ! 2x 2 /ˇˇ D .128 C 8 ! 8/ ! .0 C 0 ! 0/ D 128.
0
Z
!
" ˇ2
!
" !
"
ˇ
1
1
1
.10x 9 C 3x 5 / dx D x 10 C x 6 ˇˇ D 210 C 26 ! 210 C 26 D 0:
2
2
2
!2
!2
2
Z
0
3
.2t 3 ! 6t 2 / dt D
.5u4 C u2 ! u/ du
SOLUTION
Z
Z
.2t 3 ! 6t 2 / dt
SOLUTION
Z
0
.10x 9 C 3x 5 / dx
SOLUTION
11.
0
.12x 5 C 3x 2 ! 4x/ dx
SOLUTION
Z
ˇ9
ˇ
2 dx D 2x ˇˇ D 2.9/ ! 2.0/ D 18.
9
u2 du
!3
Z
ˇ
1 2 ˇˇ6
1
1
27
x ˇ D .6/2 ! .3/2 D
.
2
2
2
2
3
.4x ! 9x 2 / dx
SOLUTION
Z
Z
ˇ!=2
ˇ
cos x dx D sin x ˇˇ
D 1 ! 0 D 1:
4p
Z
1
!1
!
"ˇ0
!
"
ˇ
1 4
81
27
t ! 2t 3 ˇˇ D .0 ! 0/ !
! 54 D
.
2
2
2
3
.5u4 C u2 ! u/ du D
y dy
!
"ˇ1
!
" !
"
ˇ
1
1
1 1
1 1
8
u5 C u3 ! u2 ˇˇ D 1 C !
! !1 ! !
D .
3
2
3 2
3 2
3
!1
0
SOLUTION
Z
4p
0
y dy D
Z
4
0
y 1=2 dy D
ˇ
2 3=2 ˇˇ4
2
2
16
y ˇ D .4/3=2 ! .0/3=2 D
.
3
3
3
3
0
The Fundamental Theorem of Calculus, Part I
S E C T I O N 5.3
14.
Z
8
x 4=3 dx
1
SOLUTION
15.
Z
1
Z
8
ˇ
3 7=3 ˇˇ8
3
381
x ˇ D .128 ! 1/ D
.
7
7
7
1
x 4=3 dx D
1
t 1=4 dt
1=16
SOLUTION
16.
Z
1
SOLUTION
17.
3
dt
t2
1
SOLUTION
18.
Z
4
SOLUTION
19.
1=16
1
1=2
Z
Z
1
Z
4
3
dt
D
t2
1
4
x
1
8
dx
x3
Z 1
1=2
Z
!1
1
dx
x3
Z !1
!2
SOLUTION
21.
Z
2
1
!2
22.
9
1
23.
27
1
3
1
!4
ˇ3
ˇ
1
2
t !2 dt D !t !1 ˇˇ D ! C 1 D .
3
3
1
ˇ
1 !3 ˇˇ4
1
1
21
dx D ! x ˇ D ! .4/!3 C D
.
3
3
3
64
1
8
dx D
x3
Z
1
8x
1=2
Z
2
1
!3
Z
9
1
dx D !4x
ˇ1
ˇ
ˇ
!2 ˇ
1=2
D !4 C 16 D 12.
ˇ
1
1 !2 ˇˇ!1
1
1
3
dx D ! x ˇ D ! .!1/!2 C .!2/!2 D ! .
2
2
2
8
x3
!2
.x 2 ! x !2 / dx D
!
t !1=2 dt
SOLUTION
Z
Z
.x 2 ! x !2 / dx
SOLUTION
Z
ˇ
2 7=2 ˇˇ1
2
254
t ˇ D .1 ! 128/ D !
.
7
7
7
4
t 5=2 dt D
SOLUTION
20.
ˇ
4 5=4 ˇˇ1
4
1
31
t ˇ
D !
D
.
5
5 40
40
1=16
t 1=4 dt D
x !4 dx
1
Z
1
t 5=2 dt
4
Z
Z
" ˇ2 !
" !
"
ˇ
1 3
8
1
1
11
x C x !1 ˇˇ D
C
!
C1 D
:
3
3
2
3
6
1
ˇ9
ˇ
t !1=2 dt D 2t 1=2 ˇˇ D 2.9/1=2 ! 2.1/1=2 D 4.
1
t C1
p dt
t
SOLUTION
Z
1
24.
Z
1
8=27
27
t C1
p dt D
t
Z
27
.t
1
D
!
10t 4=3 ! 8t 1=3
dt D
t2
Z
10t 4=3 ! 8t 1=3
dt
t2
1=2
Ct
!1=2
/ dt D
" ˇ27
ˇ
2 3=2
1=2 ˇ
t
C 2t
ˇ
3
1
"
p
8
C 2 D 60 3 ! :
3
!
" !
p
2 p
2
.81 3/ C 6 3 !
3
3
SOLUTION
Z
1
8=27
1
8=27
D .30t
.10t !2=3 ! 8t !5=3 / dt
1=3
C 12t
!2=3
ˇ1
ˇ
/ˇˇ
8=27
D .30 C 12/ ! .20 C 27/ D !5:
593
594
25.
THE INTEGRAL
CHAPTER 5
Z
3!=4
sin " d"
!=4
SOLUTION
26.
Z
Z
3!=4
!=4
4!
sin x dx
2!
SOLUTION
27.
Z
Z
!=2
cos
0
SOLUTION
28.
Z
Z
4!
2!
p
p
ˇ3!=4
p
ˇ
2
2
sin " d" D ! cos " ˇˇ
D
C
D 2:
2
2
!=4
ˇ4!
ˇ
sin x dx D ! cos x ˇˇ D !1 ! .!1/ D 0:
2!
!
1
" d"
3
!=2
0
5!=8
!
! "ˇ!=2
1
1 ˇˇ
3
cos " d" D 3 sin
" ˇ
D .
3
3
2
0
cos 2x dx
!=4
SOLUTION
Z
Z
5!=8
!=4
p
ˇ5!=8
ˇ
1
1
5#
1
#
2 1
ˇ
cos 2x dx D sin 2x ˇ
D sin
! sin D !
! .
2
2
4
2
2
4
2
!=4
%
#&
sec2 3t !
dt
6
0
ˇ
!
"
Z !=6
%
%
#&
1
# &ˇˇ!=6
1 p
1
4
SOLUTION
sec2 3t !
dt D tan 3t !
D
3
C
p
D p .
ˇ
6
3
6
3
3
3 3
0
0
Z !=6
30.
sec " tan " d"
29.
!=6
0
SOLUTION
31.
Z
Z
!=6
0
!=10
p
ˇ!=6
ˇ
#
2 3
ˇ
sec " tan " d" D sec " ˇ
D sec ! sec 0 D
! 1.
6
3
0
csc 5x cot 5x dx
!=20
SOLUTION
32.
Z
!=14
Z
!=10
!=20
ˇ!=10
p &
ˇ
1
1%
1 p
csc 5x cot 5x dx D ! csc 5x ˇˇ
D ! 1 ! 2 D . 2 ! 1/.
5
5
5
!=20
csc2 7y dy
!=28
SOLUTION
33.
Z
1
Z
!=14
!=28
e x dx
0
SOLUTION
34.
Z
5
3
35.
3
1
0
e !4x dx
SOLUTION
Z
Z
Z
5
3
e 1!6t dt
0
SOLUTION
36.
Z
2
3
Z
3
0
e 4t !3 dt
ˇ!=14
ˇ
1
1
#
1
#
1
csc2 7y dy D ! cot 7y ˇˇ
D ! cot C cot D .
7
7
2
7
4
7
!=28
ˇ1
ˇ
e x dx D e x ˇˇ D e ! 1.
0
ˇ5
ˇ
1
1
1
e !4x dx D ! e !4x ˇˇ D ! e !20 C e !12 .
4
4
4
3
ˇ3
ˇ
1
1
1
1
e 1!6t dt D ! e 1!6t ˇˇ D ! e !17 C e D .e ! e !17 /.
6
6
6
6
0
S E C T I O N 5.3
SOLUTION
37.
Z
10
2
38.
!4
!12
39.
1
0
e
2
Z
10
2
dx
x
SOLUTION
Z
3
Z
dt
t C1
Z
!12
1
40.
4
1
0
dt
5t C 4
Z
4
SOLUTION
1
41.
Z
0
!2
ˇ
1 4t !3 ˇˇ3
1 9 1 5
dt D e
ˇ D 4e ! 4e .
4
2
ˇ10
ˇ
dx
D ln jxjˇˇ D ln 10 ! ln 2 D ln 5.
x
2
!4
SOLUTION
Z
4t !3
dx
x
SOLUTION
Z
Z
The Fundamental Theorem of Calculus, Part I
ˇ!4
ˇ
dx
1
D ln jxjˇˇ
D ln j!4j ! ln j!12j D ln D ! ln 3.
x
3
!12
ˇ1
ˇ
dt
D ln jt C 1jˇˇ D ln 2 ! ln 1 D ln 2.
t C1
0
ˇ4
ˇ
dt
1
1
1
1 24
D ln j5t C 4jˇˇ D ln 24 ! ln 9 D ln .
5t C 4
5
5
5
5
9
1
.3x ! 9e 3x / dx
SOLUTION
0
!2
!
.3x ! 9e 3x / dx D
"ˇ0
ˇ
3 2
x ! 3e 3x ˇˇ D .0 ! 3/ ! .6 ! 3e !6 / D 3e !6 ! 9.
2
!2
"
1
dx
x
2
"
!
"ˇ6
Z 6!
ˇ
1
1 2
SOLUTION
xC
dx D
x C ln jxj ˇˇ D .18 C ln 6/ ! .2 C ln 2/ D 16 C ln 3.
x
2
2
2
42.
Z
6!
Z
xC
In Exercises 43–48, write the integral as a sum of integrals without absolute values and evaluate.
Z 1
43.
jxj dx
!2
SOLUTION
44.
Z
Z
5
1
!2
jxj dx D
Z
0
!2
.!x/ dx C
Z
0
j3 ! xj dx
0
1
ˇ
ˇ
!
"
1 ˇ0
1 ˇ1
1
1
5
x dx D ! x 2 ˇˇ C x 2 ˇˇ D 0 ! ! .4/ C D :
2
2
2
2
2
!2
0
SOLUTION
Z
45.
Z
3
!2
!
" ˇ3 !
" ˇ5
ˇ
ˇ
1
1 2
.x ! 3/ dx D 3x ! x 2 ˇˇ C
x ! 3x ˇˇ
2
2
0
3
0
3
!
"
!
" !
"
9
25
9
13
D 9!
!0C
! 15 !
!9 D
:
2
2
2
2
5
0
j3 ! xj dx D
Z
3
.3 ! x/ dx C
Z
5
jx 3 j dx
SOLUTION
Z
3
!2
jx 3 j dx D
Z
0
!2
.!x 3 / dx C
Z
3
0
ˇ
ˇ
1 ˇ0
1 ˇ3
x 3 dx D ! x 4 ˇˇ C x 4 ˇˇ
4
4
!2
0
1
97
1
D 0 C .!2/4 C 34 ! 0 D
:
4
4
4
595
596
46.
THE INTEGRAL
CHAPTER 5
Z
3
0
jx 2 ! 1j dx
SOLUTION
Z
47.
Z
!
3
0
!
"ˇ
!
" ˇ3
ˇ
1 3 ˇˇ1
1 3
jx ! 1j dx D
.1 ! x / dx C
.x ! 1/ dx D x ! x ˇ C
x ! x ˇˇ
3
3
0
1
0
1
!
"
!
"
1
1
22
D 1!
! 0 C .9 ! 3/ !
!1 D
:
3
3
3
Z
2
1
Z
2
3
2
jcos xj dx
0
SOLUTION
48.
Z
Z
5
0
!
jcos xj dx D
0
Z
!=2
cos x dx C
0
jx 2 ! 4x C 3j dx
Z
!
!=2
ˇ!=2
ˇ!
ˇ
ˇ
.! cos x/ dx D sin x ˇˇ
! sin x ˇˇ
0
!=2
D 1 ! 0 ! .!1 ! 0/ D 2:
SOLUTION
Z
5
0
jx 2 ! 4x C 3j dx D
D
Z
Z
5
j.x ! 3/.x ! 1/j dx
0
1
0
.x 2 ! 4x C 3/ dx C
Z
1
3
!.x 2 ! 4x C 3/ dx C
Z
5
3
.x 2 ! 4x C 3/ dx
" ˇ1 !
" ˇ3 !
" ˇ5
ˇ
ˇ
ˇ
1 3
1 3
1 3
2
2
2
ˇ
ˇ
D
x ! 2x C 3x ˇ !
x ! 2x C 3x ˇ C
x ! 2x C 3x ˇˇ
3
3
3
0
1
3
!
"
!
" !
"
1
1
125
D
! 2 C 3 ! 0 ! .9 ! 18 C 9/ C
!2C3 C
! 50 C 15 ! .9 ! 18 C 9/
3
3
3
!
D
28
:
3
In Exercises 49–54, evaluate the integral in terms of the constants.
Z b
49.
x 3 dx
1
SOLUTION
50.
Z
a
SOLUTION
51.
b
b
1
x 3 dx D
x 4 dx
b
Z
Z
Z
a
b
x 4 dx D
x 5 dx
ˇ
&
1 4 ˇˇb
1
1
1% 4
x ˇ D b 4 ! .1/4 D
b ! 1 for any number b.
4
4
4
4
1
ˇ
1 5 ˇˇa
1
1
x ˇ D a5 ! b 5 for any numbers a, b.
5
5
5
b
1
SOLUTION
52.
Z
x
!x
Z
b
1
ˇ
1 6 ˇˇb
1
1
1
x dx D x ˇ D b 6 ! .1/6 D .b 6 ! 1/ for any number b.
6
6
6
6
1
5
.t 3 C t/ dt
SOLUTION
53.
Z
5a
a
Z
dx
x
SOLUTION
Z
5a
a
x
!x
.t 3 C t/ dt D
!
"ˇ
!
" !
"
1 4 1 2 ˇˇx
1 4 1 2
1 4 1 2
t C t ˇ D
x C x !
x C x
D 0:
4
2
4
2
4
2
!x
ˇ5a
ˇ
dx
D ln jxjˇˇ D ln j5aj ! ln jaj D ln 5.
x
a
The Fundamental Theorem of Calculus, Part I
S E C T I O N 5.3
54.
Z
b
b2
597
dx
x
SOLUTION
55. Calculate
Z
ˇb 2
ˇ
dx
D ln jxjˇˇ D ln jb 2 j ! ln jbj D ln jbj.
x
b
b2
b
Z
3
f .x/ dx, where
!2
f .x/ D
(
12 ! x 2
x3
for x " 2
for x > 2
SOLUTION
Z
3
!2
56. Calculate
Z
f .x/ dx D
Z
2
!2
f .x/ dx C
Z
3
2
f .x/ dx D
Z
2
!2
.12 ! x 2 / dx C
Z
3
x 3 dx
2
ˇ
!
"ˇ
1 3 ˇˇ2
1 4 ˇˇ3
D 12x ! x ˇ C x ˇ
3
4
!2
2
!
" !
"
1 3
1
1
1
D 12.2/ ! 2 ! 12.!2/ ! .!2/3 C 34 ! 24
3
3
4
4
128
65
707
C
D
:
3
4
12
D
2!
f .x/ dx, where
0
f .x/ D
(
cos x
cos x ! sin 2x
for x " #
for x > #
SOLUTION
Z
2!
0
f .x/ dx D
57. Use FTC I to show that
SOLUTION
We have
Z
1
!1
Z
!
0
f .x/ dx C
Z
2!
!
f .x/ dx D
Z
0
!
cos x dx C
ˇ!
!
"ˇ2!
ˇ
ˇ
1
D sin x ˇˇ C sin x C cos 2x ˇˇ
2
0
!
!!
" !
""
1
1
D .0 ! 0/ C
0C
! 0C
D 0:
2
2
Z
2!
!
.cos x ! sin 2x/ dx
x n dx D 0 if n is an odd whole number. Explain graphically.
Z
1
!1
x n dx D
ˇ
x nC1 ˇˇ1
.1/nC1 .!1/nC1
D
!
:
ˇ
n C 1 !1
nC1
nC1
Because n is odd, n C 1 is even, which means that .!1/nC1 D .1/nC1 D 1. Hence
.1/nC1 .!1/nC1
1
1
!
D
!
D 0:
nC1
nC1
nC1 nC1
Graphically speaking, for an odd function such as x 3 shown here, the positively signed area from x D 0 to x D 1 cancels the
negatively signed area from x D !1 to x D 0.
y
1
−0.5
−1
0.5
−0.5
−1
0.5
1
x
598
THE INTEGRAL
CHAPTER 5
58.
Plot the function f .x/ D sin 3x ! x. Find the positive root of f .x/ to three places and use it to find the area under the
graph of f .x/ in the first quadrant.
SOLUTION The graph of f .x/ D sin 3x ! x is shown below at the left. In the figure below at the right, we zoom in on the positive
root of f .x/ and find that, to three decimal places, this root is approximately x D 0:760. The area under the graph of f .x/ in the
first quadrant is then
Z
0:760
.sin 3x ! x/ dx D
0
!
"ˇ0:760
ˇ
1
1
! cos 3x ! x 2 ˇˇ
3
2
0
1
1
1
D ! cos.2:28/ ! .0:760/2 C $ 0:262
3
2
3
y
0.5
x
− 0.2
0.2 0.4 0.6 0.8
x
1
0.756
0.758
0.76
0.762
0.764
−0.5
59. Calculate F .4/ given that F .1/ D 3 and F 0 .x/ D x 2 . Hint: Express F .4/ ! F .1/ as a definite integral.
SOLUTION
By FTC I,
F .4/ ! F .1/ D
Z
4
x 2 dx D
1
43 ! 13
D 21
3
Therefore F .4/ D F .1/ C 21 D 3 C 21 D 24.
60. Calculate G.16/, where dG=dt D t !1=2 and G.9/ D !5.
SOLUTION
By FTC I,
G.16/ ! G.9/ D
Z
16
9
t !1=2 dt D 2.161=2 / ! 2.91=2 / D 2
Therefore G.16/ D !5 C 2 D !3.
Z 1
61.
Does
x n dx get larger or smaller as n increases? Explain graphically.
0
Let n # 0 and consider
possibility.) Now
SOLUTION
Z
0
1
x n dx D
!
R1
0
x n dx. (Note: for n < 0 the integrand x n ! 1 as x ! 0C, so we exclude this
"ˇ1 !
" !
"
ˇ
1
1
1
1
x nC1 ˇˇ D
.1/nC1 !
.0/nC1 D
;
nC1
n
C
1
n
C
1
n
C
1
0
R1
which decreases as n increases. Recall that 0 x n dx represents the area between the positive curve f .x/ D x n and the x-axis over
the interval Œ0; 1!. Accordingly, this area gets smaller as n gets larger. This is readily evident in the following graph, which shows
curves for several values of n.
y
1
1/4
1/2
1
2
4
0
8
1
x
62. Show that the area of the shaded parabolic arch in Figure 1 is equal to four-thirds the area of the triangle shown.
y
a
a+b
2
b
x
FIGURE 1 Graph of y D .x ! a/.b ! x/.
The Fundamental Theorem of Calculus, Part I
S E C T I O N 5.3
SOLUTION
599
We first calculate the area of the parabolic arch:
Z
b
a
.x ! a/.b ! x/ dx D !
Z
b
a
.x ! a/.x ! b/ dx D !
!
Z
b
a
.x 2 ! ax ! bx C ab/ dx
"ˇb
ˇ
1 3 a 2 b 2
D!
x ! x ! x C abx ˇˇ
3
2
2
a
&ˇb
1 % 3
ˇ
D!
2x ! 3ax 2 ! 3bx 2 C 6abx ˇ
a
6
%
&
1
D ! .2b 3 ! 3ab 2 ! 3b 3 C 6ab 2 / ! .2a3 ! 3a3 ! 3ba2 C 6a2 b/
6
&
1%
D ! .!b 3 C 3ab 2 / ! .!a3 C 3a2 b/
6
&
1%
1
D ! a3 C 3ab 2 ! 3a2 b ! b 3 D .b ! a/3 :
6
6
The indicated triangle has a base of length b ! a and a height of
!
"!
" !
"
aCb
aCb
b!a 2
!a
b!
D
:
2
2
2
Thus, the area of the triangle is
!
"
1
b!a 2
1
.b ! a/
D .b ! a/3 :
2
2
8
Finally, we note that
1
4 1
.b ! a/3 D % .b ! a/3 ;
6
3 8
as required.
Further Insights and Challenges
63. Prove a famous result of Archimedes (generalizing Exercise 62): For r < s, the area of the shaded region in Figure 2 is equal
to four-thirds the area of triangle 4ACE, where C is the point on the parabola at which the tangent line is parallel to secant line
AE.
(a) Show that C has x-coordinate .r C s/=2.
(b) Show that ABDE has area .s ! r/3 =4 by viewing it as a parallelogram of height s ! r and base of length CF .
(c) Show that 4ACE has area .s ! r/3 =8 by observing that it has the same base and height as the parallelogram.
(d) Compute the shaded area as the area under the graph minus the area of a trapezoid, and prove Archimedes’ result.
y
B
C
D
A
F
E
r
r+s
2
x
s
FIGURE 2 Graph of f .x/ D .x ! a/.b ! x/.
SOLUTION
(a) The slope of the secant line AE is
f .s/ ! f .r/
.s ! a/.b ! s/ ! .r ! a/.b ! r/
D
D a C b ! .r C s/
s!r
s!r
and the slope of the tangent line along the parabola is
f 0 .x/ D a C b ! 2x:
If C is the point on the parabola at which the tangent line is parallel to the secant line AE, then its x-coordinate must satisfy
a C b ! 2x D a C b ! .r C s/
or
xD
r Cs
:
2
600
CHAPTER 5
THE INTEGRAL
(b) Parallelogram ABDE has height s ! r and base of length CF . Since the equation of the secant line AE is
y D Œa C b ! .r C s/! .x ! r/ C .r ! a/.b ! r/;
the length of the segment CF is
!
"!
"
!
"
r Cs
r Cs
r Cs
.s ! r/2
!a
b!
! Œa C b ! .r C s/!
! r ! .r ! a/.b ! r/ D
:
2
2
2
4
Thus, the area of ABDE is
.s!r/3
.
4
(c) Triangle ACE is comprised of $ACF and $CEF . Each of these smaller triangles has height
Thus, the area of $ACE is
s!r
2
and base of length
.s!r/2
.
4
1 s ! r .s ! r/2
1 s ! r .s ! r/2
.s ! r/3
%
C
%
D
:
2 2
4
2 2
4
8
(d) The area under the graph of the parabola between x D r and x D s is
!
"ˇs
Z s
ˇ
1
1
.x ! a/.b ! x/ dx D !abx C .a C b/x 2 ! x 3 ˇˇ
2
3
r
r
1
1
1
1
D !abs C .a C b/s 2 ! s 3 C abr ! .a C b/r 2 C r 3
2
3
2
3
1
1
D ab.r ! s/ C .a C b/.s ! r/.s C r/ C .r ! s/.r 2 C rs C s 2 /;
2
3
while the area of the trapezoid under the shaded region is
1
.s ! r/ Œ.s ! a/.b ! s/ C .r ! a/.b ! r/!
2
h
i
1
D .s ! r/ !2ab C .a C b/.r C s/ ! r 2 ! s 2
2
1
1
D ab.r ! s/ C .a C b/.s ! r/.r C s/ C .r ! s/.r 2 C s 2 /:
2
2
Thus, the area of the shaded region is
!
"
!
"
1 2 1
1
1
1
1 2 1
1
1
.r ! s/
r C rs C s 2 ! r 2 ! s 2 D .s ! r/
r ! rs C s 2 D .s ! r/3 ;
3
3
3
2
2
6
3
6
6
which is four-thirds the area of the triangle ACE.
64. (a) Apply the Comparison Theorem (Theorem 5 in Section 5.2) to the inequality sin x " x (valid for x # 0) to prove that
x2
" cos x " 1
2
1!
(b) Apply it again to prove that
x!
x3
" sin x " x
6
.for x # 0/
(c) Verify these inequalities for x D 0:3.
SOLUTION
(a) We have
Z
x
0
ˇx
Z
ˇ
sin t dt D ! cos t ˇˇ D ! cos x C 1 and
0
0
x
t dt D
ˇ
1 2 ˇˇx
1
t ˇ D x 2 . Hence
2 0
2
! cos x C 1 "
Solving, this gives cos x # 1 !
x2
2 .
(b) The previous part gives us 1 !
x2
:
2
cos x " 1 follows automatically.
t2
2
" cos t " 1, for t > 0. Theorem 5 gives us, after integrating over the interval Œ0; x!,
x!
x3
" sin x " x:
6
(c) Substituting x D 0:3 into the inequalities obtained in (a) and (b) yields
0:955 " 0:955336489 " 1
respectively.
and
0:2955 " 0:2955202069 " 0:3;
The Fundamental Theorem of Calculus, Part I
S E C T I O N 5.3
601
65. Use the method of Exercise 64 to prove that
1!
x2
x2
x4
" cos x " 1 !
C
2
2
24
x!
x3
x3
x5
" sin x " x !
C
6
6
120
(for x # 0)
Verify these inequalities for x D 0:1. Why have we specified x # 0 for sin x but not for cos x?
SOLUTION
cos x, yields:
By Exercise 64, t ! 16 t 3 " sin t " t for t > 0. Integrating this inequality over the interval Œ0; x!, and then solving for
1 2
1
x ! x 4 " 1 ! cos x "
2
24
1
1 ! x 2 " cos x " 1 !
2
x2
2 ,
These inequalities apply for x # 0. Since cos x, 1 !
Having established that
1!
and 1 !
x2
2
C
1 2
x
2
1 2
1
x C x4:
2
24
x4
24
are all even functions, they also apply for x " 0.
t2
t2
t4
" cos t " 1 !
C ;
2
2
24
for all t # 0, we integrate over the interval Œ0; x!, to obtain:
x!
x3
x3
x5
" sin x " x !
C
:
6
6
120
1
The functions sin x, x ! 16 x 3 and x ! 16 x 3 C 120
x 5 are all odd functions, so the inequalities are reversed for x < 0.
Evaluating these inequalities at x D 0:1 yields
0:995000000 " 0:995004165 " 0:995004167
0:0998333333 " 0:0998334166 " 0:0998334167;
both of which are true.
66. Calculate the next pair of inequalities for sin x and cos x by integrating the results of Exercise 65. Can you guess the general
pattern?
SOLUTION
Integrating
t!
t3
t3
t5
" sin t " t !
C
6
6
120
(for t # 0)
over the interval Œ0; x! yields
x2 x4
x2 x4
x6
!
" 1 ! cos x "
!
C
:
2
24
2
24
720
Solving for cos x yields
1!
x2
x4
x6
x2
x4
C
!
" cos x " 1 !
C
:
2
24 720
2
24
Replacing each x by t and integrating over the interval Œ0; x! produces
x!
x3
x5
x7
x3
x5
C
!
" sin x " x !
C
:
6
120 5040
6
120
To see the pattern, it is best to compare consecutive inequalities for sin x and those for cos x:
0 " sin x " x
x!
x!
x3
6
" sin x " x
x3
x3
x5
" sin x " x !
C
:
6
6
120
602
CHAPTER 5
THE INTEGRAL
Each iteration adds an additional term. Looking at the highest order terms, we get the following pattern:
0
x
!
x3
x3
D!
6
3Š
x5
5Š
We guess that the leading term of the polynomials are of the form
x 2nC1
:
.2n C 1/Š
.!1/n
Similarly, for cos x, the leading terms of the polynomials in the inequality are of the form
x 2n
:
.2n/Š
67. Use FTC I to prove that if jf 0 .x/j " K for x 2 Œa; b!, then jf .x/ ! f .a/j " Kjx ! aj for x 2 Œa; b!.
SOLUTION Let a > b be real numbers, and let f .x/ be such that jf 0 .x/j " K for x 2 Œa; b!. By FTC,
Z x
f 0 .t/ dt D f .x/ ! f .a/:
.!1/n
a
Since f
0 .x/
# !K for all x 2 Œa; b!, we get:
f .x/ ! f .a/ D
Z
x
a
Since f 0 .x/ " K for all x 2 Œa; b!, we get:
f .x/ ! f .a/ D
Z
f 0 .t/ dt # !K.x ! a/:
x
a
f 0 .t/ dt " K.x ! a/:
Combining these two inequalities yields
!K.x ! a/ " f .x/ ! f .a/ " K.x ! a/;
so that, by definition,
jf .x/ ! f .a/j " Kjx ! aj:
68. (a) Use Exercise 67 to prove that j sin a ! sin bj " ja ! bj for all a; b.
(b) Let f .x/ D sin.x C a/ ! sin x. Use part (a) to show that the graph of f .x/ lies between the horizontal lines y D ˙a.
(c)
Plot f .x/ and the lines y D ˙a to verify (b) for a D 0:5 and a D 0:2.
SOLUTION
(a) Let f .x/ D sin x, so that f 0 .x/ D cos x, and
jf 0 .x/j " 1
for all x. From Exercise 67, we get:
jsin a ! sin bj " ja ! bj:
(b) Let f .x/ D sin.x C a/ ! sin.x/. Applying (a), we get the inequality:
jf .x/j D jsin.x C a/ ! sin.x/j " j.x C a ! x/j D jaj:
This is equivalent, by definition, to the two inequalities:
!a " sin.x C a/ ! sin.x/ " a:
(c) The plots of y D sin.x C 0:5/ ! sin.x/ and of y D sin.x C 0:2/ ! sin.x/ are shown below. The inequality is satisfied in both
plots.
y
y
0.2
0.5
0.1
0.25
−4
−2
−0.25
−0.5
x
2
4
−4
−2
− 0.1
−0.2
x
2
4