Solutions of Lectures of Calculus

Contents

1. Chapter 1 1
2. Chapter 2 12
3. Chapter 3 23
4. Chapter 4 33
5. Chapter 5 44
6. Chapter 6 49
7. Chapter 7 56
8. Chapter 8 71
9. Chapter 9 84
10. Chapter 10 99
11. Chapter 11 102
12. Chapter 12 112

iii
 1. CHAPTER 1 1

1. Chapter 1

1.1. Exercises 1.1.

x3 +8
Exercise 1.1.1. Suppose f (x) = x+2
, …nd lim f (x).
x! 2

Sol.

(x + 2) (x2 2x + 4)
lim f (x) = lim = lim x2 2x + 4 = 4:
x! 2 x! 2 x+2 x! 2

jxj
Exercise 1.1.2. Suppose f (x) = x
, …nd lim f (x).
x!0

Sol.
jxj 1 ;8 x > 1
f (x) = = . Since lim+ f (x) = 1 6= 1 = lim f (x); lim f (x)
x 1; 8 x < 1 x!0 x!0 x!0
dose not exist.

Exercise 1.1.3. Suppose f (x) = 3jxj , …nd lim f (x).

x! 1

Sol.

lim f (x) = 3j 1j
= 31 = 3:
x! 1

3
Exercise 1.1.4. Suppose f (x) = , …nd lim f (x).
x!3

Sol.

3
lim f (x) = :
x!3

1.2. Exercises 1.2.

Exercise 1.2.1. Prove that lim (1 x) = 3.
x! 2

Sol.
Consider j(1 x) 3j = j x 2j = jx + 2j. Given " > 0, take = " > 0 such
that if 0 < jx ( 2)j = jx + 2j < , then j(1 x) 3j = jx + 2j < = ". Hence
lim (1 x) = 3.
x! 2

Exercise 1.2.2. Prove that lim (4x 5) = 7.

x!3
2 CONTENTS

Sol.
Consider j(4x 5) 7j = j4x 12j = 4 jx 3j. Given " > 0, take = 4" > 0 such
"
that if 0 < jx 3j < , then j(4x 5) 7j = 4 jx 3j < 4 = 4 4
= ". Hence
lim (4x 5) = 7.
x!3

x2 4
Exercise 1.2.3. Prove that lim = 4.
x!2 x 2

Sol.
Consider
x2 4 (x2 4) 4 (x 2) x2 4x + 4 (x 2)2
4 = = = = jx 2j ;
x 2 x 2 x 2 x 2
when x =
6 2. Given " > 0, take = " > 0 such that if 0 < jx 2j < , then
x2 4 2
x 2
4 = jx 2j < = ". Hence lim xx 24 = 4.
x!2

1
Exercise 1.2.4. Prove that lim = 12 .
x!2 x

Sol.
Consider x1 21 = 22xx = jx2jxj2j , so if 0 < jx 2j < 1 implies 1 < x < 3, then
1
jxj
< 1. Given " > 0, take = min f1; 2"g > 0 such that if 0 < jx 2j < , then
1 1 jx 2j jx 2j 2"
= < < = ":
x 2 2 jxj 2 2 2
1
Hence lim = 12 .
x!2 x

Exercise 1.2.5. Prove that lim (x2 + x 4) = 16.

x!4

Sol.
Consider
x2 + x 4 16 = x2 + x 20 = j(x 4) (x + 5)j ;
so if 0 < jx 4j < 1 implies 3 < x < 5, then jx + 5j < 10. Given " > 0, take
"
= min 1; 10 > 0 such that if 0 < jx 4j < , then
"
x2 + x 4 16 = j(x 4) (x + 5)j < 10 jx 4j < 10 10 = ":
10
Hence lim (x2 + x 4) = 16.
x!4

Exercise 1.2.6. Prove that lim f (x) = 0 if and only if lim jf (x)j = 0.
x!c x!c
 1. CHAPTER 1 3

Sol.
proof of ())
Given any " > 0, since lim f (x) = 0, 9 > 0 such that if 0 < jx cj < ,
x!c
then jf (x) 0j < ". So if 0 < jx cj < , then jf (x) 0j = jf (x)j = jjf (x)jj =
jjf (x)j 0j < ". Hence lim jf (x)j = 0.
x!c
proof of (()
Given any " > 0, since lim jf (x)j = 0, 9 > 0 such that if 0 < jx cj < ,
x!c
then jjf (x)j 0j < ". So if 0 < jx cj < , then jjf (x)j 0j = jjf (x)jj = jf (x)j =
jf (x) 0j < ". Hence lim f (x) = 0.
x!c

Exercise 1.2.7. True or false:

(a) If lim f (x) = L, then lim jf (x)j = jLj.
x!c x!c
(b) If lim jf (x)j = L, then lim f (x) = L or L.
x!c x!c
Sol.
(a) True.
Given any " > 0, since lim f (x) = L, 9 > 0 such that if 0 < jx cj < , then
x!c
jf (x) Lj < ". So if 0 < jx cj < , then jjf (x)j jLjj < jf (x) Lj < ". Hence
lim jf (x)j = jLj.
x!c

(b) False.
1 ;8 x 0
Consider the function f (x) = , then jf (x)j = 1; 8 x 2 R. So
1; 8 x < 0
lim jf (x)j = 1, but lim f (x) does not exist.
x!0 x!0

x; x2Q;
Exercise 1.2.8. Suppose f (x) = prove that lim f (x) = 0.
0; x2
=Q; x!0

Sol.
Consider jf (x) 0j = jf (x)j jxj ; 8 x 2 R. Given " > 0, take = " such that if
0 < jx 0j = jxj < , then jf (x) 0j jxj < = ". Hence lim f (x) = 0.
x!0

1 ; x2Q;
Exercise 1.2.9. Suppose f (x) = prove that lim f (x) does not
1; x2
=Q; x!0
exist.
You may use the fact about the density property of rational(irrational) numbers:
for any a; b 2 R and a < b, then 9 r 2 Q ( 9 q 2
= Q ) such that a < r < b ( a < q < b ).
Sol.
Suppose lim f (x) exists, that is, lim f (x) = L for some L 2 R, then for any
x!0 x!0
" > 0, there exists > 0 such that if 0 < jx 0j = jxj < , then jf (x) Lj < ".
4 CONTENTS

So if 0 < jxj < , then L " < f (x) < L + ". Now we choose 21 > 0, 9 0 > 0
such that if 0 < jxj < 0 , then L 12 < f (x) < L + 21 . By the density property of
rational numbers and the density property of irrational numbers, we can …nd a rational
number x1 2 Q such that 0 < jx1 j < 0 and an irrational number x2 2 = Q such that
0 < jx2 j < 0 . Hence we have L 12 < f (x1 ) < L + 21 and L 12 < f (x2 ) < L + 12 .
However, since f (x1 ) = 1 and f (x2 ) = 1, the length of interval L 12 ; L + 12 is
(L + 12 ) L 21 = 1, but jf (x1 ) f (x2 )j = 2. This is a contradiction ( f (x1 )
and f (x2 ) could not belong to the interval L 21 ; L + 12 at the same time ). Hence
lim f (x) does not exist.
x!0

jxj
Exercise 1.2.10. Prove that lim does not exist.
x!0 x

Sol.
jxj
Let f (x) = x
. Suppose lim f (x) exists, that is, lim f (x) = L, for some L 2 R,
x!0 x!0
then for any " > 0, there exists > 0 such that if 0 < jx 0j = jxj < , then
jf (x) Lj < ". So if < x < 0 and 0 < x < , then L " < f (x) < L + ". Now
1
we choose 2 > 0, 9 0 > 0 such that if for any 0 < x < 0 and 0 < x < 0 , then
L 12 < f (x) < L + 12 . Choose a number a with 0 < a < 0 and a number b with
1
0 < b < 0 such that L 2
< f (a) < L + 12 and L 12 < f (b) < L + 12 . Notice that
f (a) = jaj
a
= aa = 1 and f (b) = jbjb = bb = 1, the length of interval L 12 ; L + 12 is
(L + 12 ) L 12 = 1, but jf (a) f (b)j = 2. This is a contradiction(f (a) and f (b)
could not belong to the interval L 21 ; L + 12 at the same time). Hence lim f (x)
x!0
does not exist.

Exercise 1.2.11. Does lim sin( x1 ) exist? If so, …nd it.

x!0

Sol.
Let f (x) = sin( x1 ) and let an = (4n+1)
2 2
and bn = (4n+3) ; n 2 N: Then f (an ) =
1 1
sin( an ) = 1 and f (bn ) = sin( bn ) = 1:
Suppose lim f (x) exists, that is, lim f (x) = L for some L 2 R. Then for = 12 > 0,
x!0 x!0
9 > 0 such that if 0 < jx 0j = jxj < , then L 12 < f (x) < L + 12 . Now choose
n0 2 N such that (4n+3)2
> (4n+3)
2
> 1 ;then we have 0 < jan0 j < and 0 < jbn0 j < .
Hence L 12 < f (an0 ) < L + 12 and L 12 < f (bn0 ) < L + 12 . However, since f (an0 ) = 1
and f (bn0 ) = 1, the length of interval L 21 ; L + 12 is (L + 21 ) L 12 = 1, but
jf (an0 ) f (bn0 )j = 2. This is a contradiction ( f (an0 ) and f (bn0 ) could not belong to
the interval L 12 ; L + 21 at the same time ). Hence lim f (x) does not exist.
x!0

1.3. Exercises 1.3.

 1. CHAPTER 1 5

Exercise 1.3.1. Suppose f (x) g(x), 8 x 2 R, and lim f (x) = L, lim g(x) = M ,
x!c x!c
prove that L M .
Sol.
L M
Suppose L > M . Since lim f (x) = L, for 2
> 0, there exists 1 > 0, such
x!c
L M
that if 0 < jx cj < 1, then jf (x) Lj < 2
.
So if 0 < jx cj < 1 , then
L+M 3L M L M
2
< f (x) < 2
. And since lim g(x) = M , for 2 > 0, there exists 2 > 0,
x!c
such that if 0 < jx cj < 2 , then jg(x) M j < L 2M . So if 0 < jx cj < 2 ,
then L+3M
2
< g(x) < L+M
2
. Let = minf 1 ; 2 g > 0, so if 0 < jx cj < , then
g(x) < L+M2
< f (x). This is a contradiction that f (x) g(x), 8 x 2 R. Hence
L M.

Exercise 1.3.2. Evaluate the following limits,

(a) lim (2x3 + 4x2 x + 6).
x!2
x3 3x+7
(b) lim .
x! 1 1 2x
2
(c) lim ( x 1
).
x!1 xp 1 x 1

(d) lim x 4 .
x!16 x 16
(e) lim (x + 3)20 .
x! 4

Sol.
(a) Since lim x3 = 8, lim x2 = 4 and lim x = 2, by sum rule and constant multiple,
x!2 x!2 x!2
we have

lim (2x3 + 4x2 x + 6) = 2 lim x3 + 4 lim x2 lim x + 6

x!2 x!2 x!2 x!2
= 2 8+4 4 2 + 6 = 36:

(b) Since lim (x3 3x + 7) = 9 and lim (1 2x) = 3 6= 0, by quotient rule, we

x! 1 x! 1
have

x3 3x + 7 9
lim = = 3:
x! 1 1 2x 3
(c)

x2 1 x2 1 (x + 1) (x 1)
lim ( ) = lim = lim
x!1 x 1 x 1 x!1 x 1 x!1 x 1
= lim (x + 1) = 1 + 1 = 2:
x!1
6 CONTENTS

(d)
p p
x 4 x 4 1 1 1
lim = lim p p = lim p = = :
x!16 x 16 x!16 ( x + 4) ( x 4) x!16 x+4 4+4 8

(e) Since lim (x + 3) = 1, by product rule, we have

x! 4
20
lim (x + 3) 20
= lim (x + 3) = ( 1)20 = 1:
x! 4 x! 4

Exercise 1.3.3. Suppose f (x) and g (x) are real function on R, c 2 R. True or
false:
(a) If lim f (x) and lim g(x) both do not exist, then lim(f (x)+g (x)) does notexist.
x!c x!c x!c
(b) If lim f (x) = L, for some L 2 R, but lim g(x) does not exist, thenlim(f (x) +
x!c x!c x!c
g (x)) does not exist.
1
(c) If lim f (x) = 0 and f (x) 6= 0, for all x 2 R, then lim f (x) does not exist.
x!c x!c
(d) If lim f (x) = L, for some 0 6= L 2 R, and lim g(x) = 0, and g(x) 6= 0, forall
x!c x!c
x 2 R, then lim fg(x)
(x)
does not exist.
p
x!c
(e) If lim f (x) = L, for some L 2 R, then lim f (x) = L2 .
x!c x!c

Sol.
(a) False.
1; x 0 1; x 0
Consider fhe functions f (x) = and g(x) = . Nei-
1; x<0 1; x<0
ther lim f (x) nor lim g(x) exists, but lim (f (x) + g(x)) = lim 0 = 0.
x!0 x!0 x!0 x!0

(b) True.
Suppose lim(f (x)+g (x)) = M exists. Since lim f (x) = L exists, by sum rule,
x!c x!c
we have lim g(x) = lim ((f (x) + g (x)) f (x)) = lim(f (x)+g (x)) lim f (x) = M L
x!c x!c x!c x!c
exists. This is a contradiction that lim g(x) does not exist. Hence lim(f (x) + g (x))
x!c x!c
does not exist.

(c) True.
1
Suppose lim f (x) = L exists. Since lim f (x) = 0 exists, by product rule, we have
x!c x!c
1 1
1 = lim 1 = lim f (x) f (x) = lim f (x) lim f (x) = 0 L = 0. This is a contradiction
x!c x!c x!c x!c
1
that 0 6= 1. Hence lim f (x) does not exist.
x!c
 1. CHAPTER 1 7

(d) True.
Suppose lim fg(x)
(x)
= M exists. Since lim g(x) = 0 exists, by product rule, we
x!c x!c
f (x)
have L = lim f (x) = lim g(x)
g(x) = lim fg(x)
(x)
lim g(x) = M 0 = 0. This is a
x!c x!c x!c x!c
contradiction that L 6= 0. Hence lim fg(x)
(x)
does not exist.
x!c

(e) True. p p p
Since lim f (x) = L exists and we have f (x) = f (x) f (x), by product rule,
x!c
p p p p
we get lim f (x) = lim f (x) f (x) = lim f (x) lim f (x) = L L = L2 .
x!c x!c x!c x!c

1.4. Exercises 1.4.

p
Exercise 1.4.1. Prove that f (x) = x is continuous on [0; 1). (Hint: f (x) is
right continuous at x = 0.)
Sol.
Claim 1: f (x) is continuous on (0; 1).
For any c 2 (0; 1), consider
p p p p
p p j x cj j x + cj jx cj jx cj
jf (x) f (c)j = x c = p p = p p p :
j x + cj j x + cj c
p
Given " > 0, take 1 = c" > 0; then if jx cj < ; we have
jx cj 1 1 p
jf (x) f (c)j p <p =p c" = ":
c c c
Hence f (x) is continuous on (0; 1).

Claim 2: f (x) is right continuous

p at x
p= 0.
Consider jf (x) f (0)j = j x 0j = x. Given " > 0, take 2 = "2 > 0; then if
0 x 0 < ; we have
p p p
jf (x) f (0)j = x < = "2 = ":
Hence f (x) is right continuous at x = 0.
p
Hence by claim 1 and claim 2, f (x) = x is continuous on [0; 1).

Exercise 1.4.2. Determine the continuity of the following functions at the indi-
cated points. p
(a) f (x) = p 2x 5, c = 4.
3
(b) f (x) = xx+4
7
, c = 7.
8 CONTENTS

p
4 x; x 4;
(c) f (x) = c = 4.
x 4; x > 4;
(d) f (x) = jxj, c = 0.
Sol.
(a) First, we consider
p p p p
p p 2x 5 3 2x 5 + 3
jf (x) f (4)j = 2x 5 3 = p p
2x 5+ 3
j(2x 5) 3j 2
= p p p jx 4j ;
2x 5+ 3 3
p
5 3
for all x 2
. Given " > 0, take = 2
" > 0 such that if jx 4j < then
p
2 2 2 3
jf (x) p jx
f (4)j 4j < p = p " = ":
3 3 3 2
Hence f (x) is continuous at x = 4.

(b) Since x = 7 is not in the domaian of the function f (f is not de…ned at x = 7),
f (x) is not continuous at x = 7.
p
(c) First, we consider jf (x) f (4)j. If x 4, jf (x) f (4)j = 4 x 0 =
p
4 x . If x > 4, jf (x) f (4)j = j(x 4) 0j = jx 4j. Given " > 0, take
= minf"2 ; "g > 0 such that if jx 4j < , then jf (x) f (4)j < ". Hence f (x) is
continuous at x = 4.

(d) First, we consider jf (x) f (0)j = jjxj 0j = jjxjj = jxj, for all x 2 R. Given
" > 0, take = " > 0 such that if jx 0j < then jf (x) f (0)j = jxj < = ". Hence
f (x) is continuous at x = 0.

Exercise 1.4.3. Investigate the continuity of the following functions,

4 +3x 7
(a) f (x) = 2x sin x
.
(b) f (x) = [x], where [ ] is the Gauss symbol.
Sol.
(a) We have known the polynomial 2x4 + 3x 7 is continuous on R, the function
sin x is also continuous on R and sin x = 0 if x = n for all n 2 Z. By quotient rule,
4 +3x 7
f (x) = 2x sin x
is continuous on Rnfn jn 2 Zg.

(b) Since lim+ f (x) = c and lim f (x) = c 1 for all c 2 Z. And lim f (x) = [c] =
x!c x!c x!c
f (c) for all c 2 RnZ, hence f (x) is continuous on RnZ.
 1. CHAPTER 1 9

cx + 1; x 2;
Exercise 1.4.4. Supoose f (x) = determine the value of c such
cx2 3; x > 2;
that the function f (x) will be continuous at x = 2.
Sol.
If we want f (x) is continuous at x = 2, it must be lim f (x) = lim+ f (x). Where
x!2 x!2
lim f (x) = lim (cx + 1) = 2c + 1 and lim+ f (x) = lim+ (cx2 3) = 4c 3. So,
x!2 x!2 x!2 x!2
2c + 1 = 4c 3 implies c = 2.

1.5. Exercises 1.5.

Evaluate the following limits:
sin 3x
Exercise 1.5.1. lim 4x
.
x!0

Sol.

sin 3x sin 3x 3x 3 sin 3x 3 3

lim = lim = lim = 1= :
x!0 4x x!0 3x 4x 4 x!0 3x 4 4

sin 3x
Exercise 1.5.2. lim .
x!0 sin 4x

Sol.

sin 3x sin 3x 4x 3 3 sin 3x 4x

lim = lim = lim lim
x!0 sin 4x x!0 3x sin 4x 4 4 x!0 3x x!0 sin 4x

3 3
= 1 1= :
4 4

x
Exercise 1.5.3. lim .
x!0 tan x

Sol.

x x cos x x
lim = lim = lim lim cos x = 1 1 = 1:
x!0 tan x x!0 sin x 1 x!0 sin x x!0

1 cos x
Exercise 1.5.4. lim .
x!0 2 sin x

Sol.
10 CONTENTS

1 cos x 1 1 cos x x
lim = lim
x!0 2 sin x x!0 2 x sin x
1 1 cos x x 1
= lim lim = 0 1 = 0:
2 x!0 x x!0 sin x 2

sin x
Exercise 1.5.5. lim .
x!0 2x+tan x

Sol.
Consider
2x + tan x 2x tan x x 1
lim = lim + = lim 2 +
x!0 sin x x!0 sin x sin x x!0 sin x cos x
x 1
= 2 lim + lim = 2 1 + 1 = 3 6= 0:
x!0 sin x x!0 cos x
By reciprocal rule, we have
sin x 1 1 1
lim = lim 2x+tan x = 2x+tan x = :
x!0 2x + tan x x!0 lim sin x 3
sin x x!0

1 cos x
Exercise 1.5.6. lim x2
.
x!0

Sol.

1 cos x (1 cos x) (1 + cos x) 1 cos2 x

lim = lim = lim
x!0 x2 x!0 x2 (1 + cos x) x!0 x2 (1 + cos x)
!
2
sin2 x sin x 1
= lim 2 = lim
x!0 x (1 + cos x) x!0 x 1 + cos x
2
sin x 1 1 1
= lim lim = 12 = :
x!0 x x!0 1 + cos x 2 2

cos(2x) 1
Exercise 1.5.7. lim x2
.
x!0

Sol.

cos (2x) 1 2 sin2 x 1

1 2 sin2 x
lim = lim = lim
x!0 x2 x!0 x2 x!0 x2
2
sin x
= lim 2 = 2:
x!0 x
 1. CHAPTER 1 11

1.6. Exercises 1.6.

Exercise 1.6.1. Suppose that f (x) = x3 + x + 1, prove that there exists c 2 R
such that f (c) = 100.
Sol.
Since f (0) = 1 < 100, f (10) = 1011 > 100 and f (x) is continuous on [1; 10], by
intermediate value theorem, there exists c 2 (1; 10) such that f (c) = 100.

Exercise 1.6.2. Suppose that f : [0; 1] ! [0; 1] is a continuous function, prove

that there exists c 2 [0; 1] such that f (c) = c.
Sol.
Case1: f (0) = 0 or f (1) = 1.
In this case, it’s nothing further to prove.
Case2: f (0) 6= 0 and f (1) 6= 1.
Since f (0) 6= 0 and f (1) 6= 1, 0 < f (0) 1 and 0 f (1) < 1. Let g(x) = f (x) x,
8 x 2 [0; 1]. Since f (x) is continuous on [0; 1], so is g(x). We have g(0) = f (0) 0 =
f (0) > 0, g(1) = f (1) 1 < 0 and g(x) is continuous on [0; 1]. By intermediate value
theorem, there exists c 2 (0; 1) such that g(c) = 0. Where 0 = g(c) = f (c) c implies
f (c) = c for c 2 (0; 1).
12 CONTENTS

2. Chapter 2
2.1. Exercises 2.1.
2x ; x 1;
Exercise 2.1.1. Suppose f (x) = …nd f 0 (1).
x2 + 1; x < 1;
Sol.
f (1+h) f (1)
By de…nition, f 0 (1) = lim h
; so we have
h!0

f (1 + h) f (1) 2 (1 + h) 2 2h
lim+ = lim+ = lim+ = lim+ 2 = 2
h!0 h h!0 h h!0 h h!0

and
f (1 + h) f (1) (1 + h)2 + 1 2 h2 + 2h
lim = lim = lim = lim+ (h + 2) = 2:
h!0 h h!0 h h!0 h h!0

Hence
f (1 + h) f (1) (1 + h)2 + 1 2
lim+ = lim = 2 = f 0 (1):
h!0 h h!0 h

x ; x 2 Q; x2 ; x 2 Q;
Exercise 2.1.2. Suppose f (x) = and g(x) =
0; x2 = Q; 0 ;x 2 = Q:
(a) Is f di¤erentiable at x = 0 ?
(b) Is g di¤erentiable at x = 0 ?
Sol.
(a) Let
f (x) 1; x 2 Q n f0g;
k(x) = =
x 0; x 2
= Q:
Then by the analogously argumentation of exercise 1.2.9, lim k(x) does not exist.
x!0
f (0+h) f (0) f (h) 0 f (h)
So we have lim h
= lim h
= lim = lim k(x) does not exist.
h!0 h!0 h!0 h h!0
Hence f is not di¤erentiable at x = 0.
g(0+h) g(0) g(h) 0 g(h) g(h) h2
(b) Compute lim h
= lim h
= lim . Where 0 = h
h!0 h!0 h!0 h h h

and lim 0 = lim h = 0, by pingching theorem lim g(h) = 0. Hence g is di¤erentiable

h!0 h!0 h!0 h
at x = 0.

Exercise 2.1.3. Determine the value of P and Q such that the function f (x) =
x2 2 ; x 2;
is di¤erentiable at x = 2.
P x2 + Qx ; x > 2;
 2. CHAPTER 2 13

Sol.
f (2+h) f (2)
Since f (x) is di¤erentiable at x = 2, lim h
exists, that is
h!0

f (2 + h) f (2) f (2 + h) f (2)
lim+ = lim ;
h!0 h h!0 h
where
f (2 + h) f (2) (2 + h)2 2 2 h2 + 4h
lim = lim = lim = lim (h + 4) = 4
h!0 h h!0 h h!0 h h!0

and
f (2 + h) f (2) P (2 + h)2 + Q (2 + h) 2
lim+ = lim+
h!0 h h!0 h
2
P h + 4P h + 4P + Qh + 2Q 2
= lim+
h!0 h
4P + 2Q 2
= lim+ P h + 4P + Q + :
h!0 h
4P + Q = 4; P = 23 ;
Hence we have which impies
4P + 2Q = 2; Q = 2:

x2 sin( x1 ) ; x 6= 0;
Exercise 2.1.4. Is f (x) = di¤erentiable at x = 0 ?
0 ; x = 0;
Sol.
Since
f (0 + h) f (0) f (h) 0 f (h)
lim = lim = lim
h!0 h h!0 h h!0 h
2 1
h sin h 1
= lim = lim h sin ;
h!0 h h!0 h
1 1
and since 0 h sin h
h and lim 0 = lim h = 0, by pingching theorem lim h sin h
=
h!0 h!0 h!0
0. Hence f is di¤erentiable at x = 0.

Exercise 2.1.5. Is f (x) = x jxj di¤erentiable at x = 0 ?

Sol.
Since
f (0 + h) f (0) f (h) 0 f (h) h jhj
lim = lim = lim = lim = lim jhj = 0
h!0 h h!0 h h!0 h h!0 h h!0

exists, f is di¤erentiable at x = 0.
14 CONTENTS

Exercise 2.1.6. Find the tangent line of the graph y = x3 +2x+1 passing through
the point (x; y) = (1; 4).
Sol.
Let y = f (x) = x3 + 2x + 1, we are going to …nd f 0 (1). Since

0 f (1 + h) f (1) (1 + h)3 + 2 (1 + h) + 1 4
f (1) = lim = lim
h!0 h h!0 h
3 2
h + 3h + 5h
= lim = lim h2 + 3h + 5 = 5;
h!0 h h!0

the tangent line of the graph y = x3 + 2x + 1 passing through the point (1; 4) is
y 4 = 5 (x 1).

2.2. Exercises 2.2.

Exercise 2.2.1. Di¤erentiate the following functions.
(a) f (x) = 4.
(b) f (x) = x1 .
(c) f (x) = 4x2 .
(d) f (x) = (x6 + 52 x3 )( 13 x4 38 x2 ).
(e) f (x) = 4x24x+3
+11x+2
.
x2 +x+1
(f) f (x) = x6 +x4 +x2 .
(g) f (x) = x(x + 1)(x + 2)(x + 3).
Sol.
(a)

f 0 (x) = (4)0 = 0:

(b)

0
1 1 0 1
f 0 (x) = = x =x 2
= :
x x2

(c)

0
f 0 (x) = 4x2 = 8x:

(d)
 2. CHAPTER 2 15

0
0 5 1 4 8 2
f (x) = x + x3
6
x x
2 3 3
0 0
5 1 4 8 2 5 1 4 8 2
= 6
x + x3 x x + x6 + x3 x x
2 3 3 2 3 3
15 1 4 8 2 5 4 3 16
= 6x5 + x2 x x + x6 + x3 x x :
2 3 3 2 3 3

(e)

0
0 4x + 3
f (x) = 2
4x + 11x + 2
0
(4x + 3)0 (4x2 + 11x + 2) (4x + 3) (4x2 + 11x + 2)
=
(4x2 + 11x + 2)2
4 (4x2 + 11x + 2) (4x + 3) (8x + 11)
= :
(4x2 + 11x + 2)2

(f)

0
0 x2 + x + 1
f (x) =
x 6 + x4 + x2
0 0
( x2 + x + 1) (x6 + x4 + x2 ) ( x2 + x + 1) (x6 + x4 + x2 )
=
(x6 + x4 + x2 )2
( 2x + 1) (x6 + x4 + x2 ) ( x2 + x + 1) (6x5 + 4x3 + 2x)
= :
(x6 + x4 + x2 )2

(g)
16 CONTENTS

f 0 (x)
= (x(x + 1)(x + 2)(x + 3))0
= (x)0 (x + 1)(x + 2)(x + 3) + x ((x + 1)(x + 2)(x + 3))0
= (x + 1)(x + 2)(x + 3)
+x (x + 1)0 (x + 2)(x + 3) + (x + 1) ((x + 2)(x + 3))0
= (x + 1)(x + 2)(x + 3)
+x ((x + 2)(x + 3) + (x + 1) ((x + 2)0 (x + 3) + (x + 2)(x + 3)0 ))
= (x + 1)(x + 2)(x + 3)
+x ((x + 2)(x + 3) + (x + 1) ((x + 3) + (x + 2)))
= (x + 1)(x + 2)(x + 3) + x(x + 2)(x + 3)
+x (x + 1) (x + 3) + x (x + 1) (x + 2):

(h)

0 0 0 0
0 1 2 3 1 2 3
f (x) = + 2+ 3 = + +
x x x x x2 x3
0 0 3 0
= x 1 + 2x 2 + 3x = x 2
4x 3
9x 4

1 4 9
= 2 3
:
x x x4

Exercise 2.2.2. If g(x) is continuous at x = 0 and f (x) = xg(x), prove that f (x)
is di¤erentiable at x = 0.
Sol.
Since g is continuous at x = 0; that is, lim g(h) = g(0); we have
h!0

f (0 + h) f (0) f (h) 0 g(0) hg(h)

f 0 (0) = lim = lim = lim = lim g(h) = g(0):
h!0 h h!0 h h!0 h h!0

Hence f is di¤erentiable at x = 0.

2.3. Exercises 2.3.

2.4. Exercises 2.4.

d
Exercise 2.4.1. dx
(2x 1)10 .
 2. CHAPTER 2 17

Sol.
Let u = 2x 1.
d d 10 du10 du
(2x 1)10 = u =
dx dx du dx
d (2x 1)
= 10u9
dx
= 10 (2x 1)9 2x:

d 3x 4 2
Exercise 2.4.2. (
dx 5x+3
).
Sol.
3x 4
Let u = 5x+3
.
d 3x 4 2 d 2 du2 du d 3x
5x+3
4
( ) = u = = 2u
dx 5x + 3 dx du dx dx
3x 4 3 (5x + 3) 5 (3x 4)
= 2 :
5x + 3 (5x + 3)2

d
Exercise 2.4.3. dx
(x sin( 2 ) + cos( 2 ))(x cos( 2 ) sin( 2 )).
Sol.

d 2 2 2 2
(x sin( ) + cos( ))(x cos( ) sin( ))
dx
d 2 2 2 2
= (x sin( ) + cos( )) (x cos( ) sin( ))
dx
2 2 d 2 2
+(x sin( ) + cos( )) (x cos( ) sin( ))
dx
2 2 2 2 2 2
= sin( )(x cos( ) sin( )) + cos( )(x sin( ) + cos( )):

d 1 2 3
Exercise 2.4.4. dx x
+ x2
+ x3
.
Sol.

d 1 2 3 d
+ 2+ 3 = x 1 + 2x 2 + 3x 3
= x 2
4x 3
9x 4
dx x x x dx
1 4 9
= 2 3
:
x x x4
18 CONTENTS

d
Exercise 2.4.5. dx
(cos 2x 2 sin x).
Sol.

d d d
(cos 2x 2 sin x) = (cos 2x) (2 sin x) = 2 sin 2x 2 cos x:
dx dx dx

d sin2 x
Exercise 2.4.6. (
dx sin(x2 )
).

Sol.

d sin2 x d
dx
sin2 x (sin(x2 )) sin2 x d
dx
sin(x2 )
( ) =
dx sin(x2 ) (sin(x2 ))2
(2 sin x cos x) (sin(x2 )) sin2 x (cos (x2 ) 2x)
= :
sin2 (x2 )

d 1
Exercise 2.4.7. dx
(tan x 3
tan3 x + 15 tan(x5 )).
Sol.

d 1 1
(tan x tan3 x + tan(x5 ))
dx 3 5
1 1
= sec2 x 3 tan2 x sec2 x + sec2 x5 5x4 :
3 5

d
Exercise 2.4.8. dx
(sin(cos2 (tan3 (x4 )))).
Sol.

d
(sin(cos2 (tan3 (x4 ))))
dx
= cos cos2 tan3 x4 2 cos tan3 x4
sin tan3 x4 3 tan2 x4 sec2 x4 4x3 :

d
Exercise 2.4.9. ( 1 ).
dx cos3 x

Sol.
 2. CHAPTER 2 19

d d
d 1 dx
1 (cos3 x) 1 dx
cos3 x
( 3 ) =
dx cos x (cos3 x)2
3 cos2 x sin x 3 sin x
= 6
= :
cos x cos4 x

d
Exercise 2.4.10. dx
(sec2 ( x2 ) + csc2 ( x2 )).
Sol.

d x x
(sec2 ( ) + csc2 ( ))
dx 2 2
x x x
= 2 sec tan sec
2 2 2
1 x x x 1
+ 2 csc cot csc
2 2 2 2 2
x x x x
= tan sec2 cot csc2 :
2 2 2 2

2.5. Exercises 2.5.

dy
Exercise 2.5.1. x2 + 2xy y 2 = 2x, dx
=?
Sol.
x2 + 2xy y 2 = 2x

d d
) x2 + 2xy y2 = (2x)
dx dx
dy dy
) 2x + 2 y + x 2y =2
dx dx
dy
) (x =1 x y) y
dx
dy 1 x y
) = :
dx x y

dy
Exercise 2.5.2. y 2 = 2x, dx
=?
Sol.
y 2 = 2x
20 CONTENTS

d d
) y2 = (2x)
dx dx
dy
) 2y =2
dx
dy 1
) = :
dx y

x2 y2 dy
Exercise 2.5.3. 4
+ 9
= 1, dx
=?
Sol.
x2 y 2
+ =1
4 9

d x2 y 2 d
) + = (1)
dx 4 9 dx
dy
2x 2y dx
) + =0
4 9
2y dy x
) =
9 dx 2
dy 9x
) = :
dx 4y

p p p dy
Exercise 2.5.4. x+ y = 2, dx
=?
Sol. p
p p
x+ y = 2

d p p d p
) x+ y = 2
dx dx
1 1 dy
) p + p =0
2 x 2 y dx
1 dy 1
) p = p
2 y dx 2 x
p r
dy y y
) = p = :
dx x x

2 2 dy
Exercise 2.5.5. x 3 + y 3 = 4, dx
=?
 2. CHAPTER 2 21

Sol.
2 2
x3 + y 3 = 4

d 2 2 d
) x3 + y 3 = (4)
dx dx
2 1 2 1 dy
) x 3+ y 3 =0
3 3 dx
2 dy 2
) p = p
3 y dx
3 3 x3

p r
3 y
dy y
) = p = 3 :
dx 3
x x

d p p
Exercise 2.5.6. dx
(x + x + 3 x).

Sol.

d p p d 1 1 1 1 1 2
(x + x + 3 x) = x + x2 + x3 = 1 + x 2 + x 3
dx dx 2 3
1 1
= 1+ p + p :
2 x 3 3 x2
d 1 p1 1
Exercise 2.5.7. (
dx x
+ x
+ 3 x ).
p

Sol.

d 1 1 1 d 1 1 1 3 1 4
( +p +p ) = (x 1 + x 2 +x 3 )= x 2
x 2 x 3
dx x x 3
x dx 2 3
1 1 1
= 2
p p
3
:
x 2 x3 3 x4

q p
d p
Exercise 2.5.8. dx
x+ x + x.

Sol.
22 CONTENTS

r q
d p
x+ x+ x
dx
1
1
d 1 2
2
= x+ x+x 2
dx
1
1 1
1 1 2
2
1 1 2 1 1
= x+ x+x 2 1+ x+x 2 1+ x 2
2 2 2
!
1 1 1
= q p 1+ p p 1+ p :
p 2 x+ x 2 x
2 x+ x+ x

d
p
3
p p
Exercise 2.5.9. dx
x + x + 3 x.
Sol.
q
d 3 p p
x+ x+ 3x
dx
1
d 1 1 3
= x+x +x 2 3
dx
2
1 1 1 3 1 1 1 2
= x + x2 + x3 1+ x 2 + x 3
3 2 3
1 1 1
= q 1+ p + p :
3 p p 2
3
2 x 3 x2
3 (x + 2 x + 3 x)
p
3
p
3
d 2
Exercise 2.5.10. dx
3( cot x + cot8 x).
Sol.

d p 3
p
3
3( cot2 x + cot8 x)
dx
d 2 8
= 3 (cot x) 3 + (cot x) 3
dx
2 1 8 5
= 3 (cot x) 3 csc2 x + (cot x) 3 csc2 x
3 3
2
2 csc x p3
= p 3
8 csc2 x cot5 x:
cot x
 3. CHAPTER 3 23

3. Chapter 3
3.1. Exercises 3.1.
Exercise 3.1.1. Let f (x) = x + x1 . Show that f satis…es the conditions of the
mean-value theorem on the interval [1; 9], and …nd all numbers c which satisfy the
conditions of the mean-value theorem.
Sol.
Since f is di¤erentiable on R, f is continuous on [1; 9] and is di¤erentiable on
(1; 9). Since f 0 (x) = 1 x12 and f (9)9 f1 (1) = 89 , if c 2 (1; 9) and f 0 (c) = f (9)9 1f (1) , then
1 c12 = 98 :So c = 3. ( 3 2 = (1; 9) )

Exercise 3.1.2. Let f (x) = sin x. Show that f satis…es the conditions of Rolle’s
theorem on the interval [0; 1], and …nd all numbers c which satisfy the conditions of
the Rolle’s theorem.
Sol.
Since f is di¤erentiable on R, f is continuous on [0; 1] and is di¤erentiable on
(0; 1). Morover, f (0) = 0 = f (1): Since f 0 (x) = cos x, if c 2 (0; 1) and f 0 (c) = 0,
then cos c = 0:So c = 21 .

Exercise 3.1.3. Prove that the equation 2x3 + 6x2 + 7x 10 = 0 has at most one
root.
Sol.
We are going to prove by contradiction.
Let f (x) = 2x3 + 6x2 + 7x 10. If the equation had at least two roots, then 9
a1 ; a2 ; a1 < a2 ; such that f (a1 ) = f (a2 ) = 0: Then since f is di¤erentiable on R, f
is continuous on [a1 ; a2 ] and is di¤erentiable on (a1 ; a2 ). Thus by Rolle’s theorem, 9
c 2 (a1 ; a2 ) such that f 0 (c) = 0: However,
f 0 (x) = 6x2 + 12x + 7 = 6 (x + 1)2 + 1 > 0; 8 x 2 R:
This leads to a contradiction.

Exercise 3.1.4. Prove that the equation x4 + 4x3 + 7x2 20x 1 = 0 has at most
two roots.
Sol.
We are going to prove by contradiction.
Let g(x) = x4 + 4x3 + 7x2 20x 1. If the equation had at least three roots,
then 9 a1 ; a2 ; a3 ; a1 < a2 < a3 ; such that g(a1 ) = g(a2 ) = g(a3 ) = 0: Then since
g is di¤erentiable on R, g is continuous on [a1 ; a2 ]; [a2 ; a3 ] and is di¤erentiable on
24 CONTENTS

(a1 ; a2 ); (a2 ; a3 ). Thus by Rolle’s theorem, 9 c1 2 (a1 ; a2 ) and 9 c2 2 (a2 ; a3 ) such

that g 0 (c1 ) = g 0 (c2 ) = 0: However, since
g 0 (x) = 4x3 + 12x2 + 14x 20 = 2 2x3 + 6x2 + 7x 10 ;
by exercise 3.1.3, g 0 (x) has at most one root. This leads to a contradiction.

Exercise 3.1.5. Prove that jtan a tan bj ja bj, 8 a; b 2 3

; 3
.
Sol.
If a = b, then jtan a tan bj = 0 = ja bj :
If a 6= b, then without loss of generality, we may assume a < b. Let f (x) = tan x.
Then since f is di¤erentiable on R, f is continuous on [a; b] and is di¤erentiable on
(a; b). Then by the mean value theorem, 9 c 2 (a; b) such that f 0 (c) = f (b)b af (a) ; that
is, sec2 c = tan bb tan
a
a
: Then since sec2 c 1, we have

ja bj = sec2 c jtan a tan bj jtan a tan bj :

3.2. Exercises 3.2.

Exercise 3.2.1. Find the intervals on which f is increasing and the intervals on
which f is
decreasing.
x3 x + 2; x 0;
(a) f (x) =
x2 2x + 2; x > 0:
x+3
(b) f (x) = x2 +2x+2 .
(c) f (x) = tan x 2 sec x, 0 x 2 , x 6= 2 , 32 .
Sol.
(a) Since lim f (x) = lim+ f (x) = 2 = f (2), f is continuous on R. Then since for
x!0 x!0
x < 0, f 0 (x) = 3x2 1, and for x > 0, f 0 (x) = 2x 2, we have
x p1 0 1
3 :
0
f (x) + 0 0 +
Thus by theorem 3.2.5, f is increasing on ( 1; p1 ] and [1; 1), and is decreasing
3
on [ p13 ; 0] and [0; 1], that is, f is decreasing on [ p1 ; 1]:
3

(b) Since x2 + 2x + 2 > 0, 8 x 2 R, f is continuous on R. Then since

(x2 + 2x + 2) 1 (x + 3) (2x + 2) x2 6x 4
f 0 (x) = = ;
(x2 + 2x + 2)2 (x2 + 2x + 2)2
 3. CHAPTER 3 25

we have
p p
x 3 5 3+ 5
:
f 0 (x) 0 + 0
p p
Thus by theorem
p 3.2.5, f ispincreasing on [ 3 5; 3 + 5] and is decreasing on
( 1; 3 5] and [ 3 + 5; 1):

(c) f is continuous on [0; 2 ), ( 2 ; 32 ) and ( 32 ; 2 ]. Then since

1 2 sin x
f 0 (x) = sec2 x 2 sec x tan x = ;
cos2 x
we have
5 3
x 0 6 2 6 2
2
0 ;
f (x) + 0 0 + +
Thus by theorem 3.2.5, f is increasing on [0; 6 ]; [ 56 ; 32 ) and ( 32 ; 2 ], and is decreasing
on [ 6 ; 2 ) and ( 2 ; 56 ]:

p
Exercise 3.2.2. Prove that f (x) = cos x2 + x2 + 2 is increasing on 0; 2
.

Sol. p
Since f is di¤erentiable on R, f is continuous on [0; 2 ] and is di¤erentiable on
p
(0; 2 ). Then since
r
f 0 (x) = 2x sin x2 + 2x = 2x 1 sin x2 > 0; 8 x 2 (0; );
2
p
by theorem 3.2.5, f is increasing on [0; 2 ].

Exercise 3.2.3. Prove that tan x x2 , 8 x 2 0; 8

.

Sol.
Let f (x) = tan x x2 ; x 2 0; 8
. Then f is continuous on [0; 8 ] and is di¤eren-
tiable on (0; 8 ). Then since
2
f 0 (x) = sec2 x 2x 1 > 0; 8 x 2 (0; );
8 8
by theorem 3.2.5, f is increasing on [0; 8 ]. So we have f (x) f (0) = 0; 8 x 2 [0; 8 ];
that is, tan x x2 , 8 x 2 0; 8 .
26 CONTENTS

3.3. Exercises 3.3.

Exercise 3.3.1. Find the critical points and the local extreme points of f .
x
(a) f (x) = x2 +x+1 .
2 +1
xp
(b) f (x) = x ; x > 0.
(c) f (x) = sin2 x sin x + 1, 0 x 2 .
1
(d) f (x) = 1+sin 2 x, 0 x 2 .
Sol.
2
(a)Since x2 + x + 1 = x + 12 + 34 > 0, 8 x 2 R, f is di¤erentiable on R. Then
since

(x2 + x + 1) 1 x (2x + 1) x2 + 1 (x + 1) (x 1)
f 0 (x) = = = ;
(x2 + x + 1)2 (x2 + x + 1)2 (x2 + x + 1)2
the critical points are 1 and 1. Then since we have
x 1 1
0 ;
f (x) 0 + 0
by the …rst derivative test, 1 is the local maximum point and 1 is the local minimum
point.

(b) The domain of f is (0; 1) and f is di¤erentiable on (0; 1). Since

p p p
0
x 2x 2p1 x (x2 + 1) 3x2 1 ( 3x + 1)( 3x 1)
f (x) = p 2 = p = ;
x 2x x (x2 + x + 1)2
the critical point is p1 . Then since we have
3

x 0 p1
3 ;
0
f (x) 0 +
by the …rst derivative test, p1 is the local minimum point.
3

(c) f is di¤erentiable on (0; 2 ). Then since

f 0 (x) = 2 sin x cos x cos x = cos x(2 sin x 1);
5 3
the critical points are 6 ; 2
; 6
and 2
. Then since
f (x) = 2 cos2 x
00
2 sin2 x + sin x;
we have
5 3
x 6 2 6 2
00 ;
f (x) + +
3 3
by the second derivative test, 2 and 2
are the local maximum points and 6
and 2
are the local minimum points.
 3. CHAPTER 3 27

(d) Since 1 + sin2 x > 0, 8 x 2 (0; 2 ), f is di¤erentiable on (0; 2 ). Then since

2 sin x cos x
f 0 (x) = 2;
1 + sin2 x
3
the critical points are 2 ; and 2
. Then since we have
3
x 0 2 2
2
0 ;
f (x) 0 + 0 0 +
3
by the …rst derivative test, is the local maximum point and 2
and 2
are the local
minimum points.

Exercise 3.3.2. Suppose f is continuous on [a; b] and f (a) = f (b). Show that f
has at least a critical point in (a; b).
Sol.
If f is not di¤erentiable on (a; b), then f has a critical point in (a; b). Then there
is nothing further to prove.
If f is di¤erentiable on (a; b), then let g(x) = f (x) f (a). Then since f is
continuous on [a; b] and is di¤erentiable on (a; b), so is g. Moreover, we have g 0 (x) =
f 0 (x): Then since g(a) = f (a) f (a) = 0 and g(b) = f (b) f (a) = 0, by Rolle’s
theorem, 9 c 2 (a; b) such that g 0 (c) = 0: Hence we have f 0 (c) = 0, that is, c is a
critical point of f .

3.4. Exercises 3.4.

Exercise 3.4.1. Find the local and absolute extreme points of f .
(a) f (x) = x x1 , 1 x 10.
(b) f (x) = jx2 xj, 1 x 5.
(c) f (x) = xp+ sin x, 2 x 2 .
(d) f (x) = 1 + x2 .
Sol.
(a) Since f 0 (x) = 1+ x12 > 0; 8 x 2 (1; 10); there are no critical points and no local
99
extreme points. Then since f (1) = 0 and f (10) = 10 , 10 is the absolute maximum
point and 1 is the absolute minimum point.

(b) Since
8
< x2 x; 1 x < 0;
f (x) = jx(x 1)j = x x2 ; 0 x < 1;
: 2
x x; 1 x 5;
28 CONTENTS

we have 8
< 2x 1; 1 < x < 0;
f 0 (x) = 1 2x; 0 < x < 1;
:
2x 1; 1 < x < 5:
1
Hence the critical points are 0; 2
and 1. Then since we have
1
x 1 0 2
1 5
0
f (x) + 0 + ;
1
f (x) 2 0 2
0 20
by the …rst derivative test, 12 is the local maximum point, 0 and 1 are the local
minimum points, 5 is the absolute maximum point and 0 and 1 are the absolute
minimum point.

(c) Since f 0 (x) = 1 + cos x; the critical points are and . Then since we have
x 2 2
0
f (x) + 0 + 0 + ;
f (x) 2 2
by the …rst derivative test, and are saddle points, 2 is the absolute maximum
point and 2 is the absolute minimum point.
1
(d) Since f 0 (x) = 12 (1 + x2 ) 2 2x = p x
1+x2
;the critical point is 0. Then since we
have
x 0
0
f (x) 0 + ;
f (x) 1
0 is the local and absolute minimum point and there is no local and absolute maximum
point.

3.5. Exercises 3.5.

Exercise 3.5.1. Draw the grapf of y = f (x).
p
(a) f (x) = x x, 0 x 2.
(b) f (x) = x3 6x2 + 9x 2, 0 x 5.
(c) f (x) = 2x tan x, 2 < x < 2 .
(d) f (x) = sin2 x, 0 x .
(e) f (x) = sin2 x + 2 sin x 1, 0 x .
Sol. p
(a) f (x) = x x; 0 x 2:
 3. CHAPTER 3 29

y 1.2
1.0
0.8
0.6
0.4
0.2
0.0
-0.2 0.5 1.0 1.5 2.0
x
Since f 0 (x) = 1 2p1 x ; the critical point is 41 . Then since f 00 (x) = p1
4 x3
> 0; 8 x 2
(0; 2); there is no point of in‡ection. So we have
1
x 0 4
2
00
f (x) + + +
0 :
f (x) 0 + p
1
f (x) 0 2
2 2
Therefore, f is concave up on (0; 2), is decreasing on [0; 14 ], is increasing on [ 14 ; 2], 1
4
is
the local and absolute minimum point, and 2 is the absolute maximum point.

(b) f (x) = x3 6x2 + 9x 2, 0 x 5.

20
y
10

0
1 2 3 4 5
x
-10

Since f 0 (x) = 3x2 12x + 9 = 3(x 1)(x 3); the critical points are 1 and 3.
Then since f 00 (x) = 6x 12 = 6(x 2) ; f 00 (x) = 0 at 2. So we have
x 0 1 2 3 5
f 00 (x) 0 + + +
:
f 0 (x) + 0 0 +
f (x) 2 2 0 2 18
Therefore, f is concave down on (0; 2), is concave up on (2; 5), is decreasing on [1; 3],
is increasing on [0; 1] and [3; 5], 3 is the local minimum point, 1 is the local maximum
point, 0 and 3 are the absolute minimum points, and 5 is the absolute maximum
point.

(c) f (x) = 2x tan x, 2

< x < 2.
30 CONTENTS

6
y
4
2

-1.5 -1.0 -0.5 0.5 1.0 1.5

-2 x
-4
-6

Since f 0 (x) = 2 sec2 x; the critical points are 4

and 4
. Then since f 00 (x) =
2 sec2 x tan x ; f 00 (x) = 0 at 0. So we have
x 2 4
0 4 2
00
f (x) + + + 0
:
f 0 (x) 0 + + + 0
f (x) 1 2
+1 0 2
1 1
Therefore, f is concave down on (0; 2 ), is concave up on ( 2 ; 0), is decreasing on
[ 2 ; 4 ] and [ 4 ; 2 ], is increasing on [ 4 ; 4 ], 4 is the local minimum point, 4 is the
local maximum point, and there are no absolute extreme points.

(d) f (x) = sin2 x, 0 x .

1.0
y

0.5

0.0
0 1 2 3
x

Since f 0 (x) = 2 sin x cos x; the critical point is 2 . Then since f 00 (x) = 2 cos2 x
2 sin2 x ; f 00 (x) = 0 at 4 and 34 . So we have
3
x 0 4 2 4
.
00
f (x) + 0 0 +
:
f 0 (x) + + + 0
1 1
f (x) 0 2
1 2
0
Therefore, f is concave down on ( 4 ; 34 ), is concave up on (0; 4 ) and ( 34 ; ), is de-
creasing on [ 2 ; ], is increasing on [0; 2 ], 2 is the local maximum point, 0 and are
the absolute minimum points, and 2 is the absolute maximum point.

(e) f (x) = sin2 x + 2 sin x 1, 0 x .

 3. CHAPTER 3 31

2
y
1

0
1 2 3
x
-1

Since f 0 (x) = 2 sin x cos x + 2 cos x = 2 cos x(sin x + 1); the critical point is 2
.
Then since
f 00 (x) = 2 cos2 x 2 sin2 x 2 sin x
= 2 4 sin2 x 2 sin x
= 2(2 sin x 1)(sin x + 1);
5
f 00 (x) = 0 at 6
and 6
. So we have
5
x 0 6 2 6
.
00
f (x) + 0 0 +
:
f 0 (x) + + + 0
1 1
f (x) 1 4
2 4
1
Therefore, f is concave down on ( 6 ; 56 ), is concave up on (0; 6 ) and ( 56 ; ), is de-
creasing on [ 2 ; ], is increasing on [0; 2 ], 2 is the local maximum point, 0 and are
the absolute minimum points, and 2 is the absolute maximum point.

3.6. Exercises 3.6.

Exercise 3.6.1. Find the asymptotes of f .
x 2
(a) f (x) = 1+x 2.
1 x
(b) f (x) = 1+x .
Sol.
(a) Since 1 + x2 > 0; 8 x 2 R, there is no vertical asymptotes. Then since
lim x1 = 0 and lim x1 = 0, we have
x!1 x! 1

x2 1 1
lim 2
= lim 1 = =1
x!1 1 + x x!1 2 + 1 0+1
x
and
x2 1 1
=
lim lim 1 = = 1:
x! 1 1 + x2 x! 1 2 + 1 0 + 1
x
So the horizontal asymptotes is y = 1.
32 CONTENTS

(b) Since
1 x 2
lim + = lim + ( 1) = 1;
x! 1 1 + x x! 1 1 + x
1 x 2
lim = lim ( 1) = 1;
x! 1 1 + x x! 1 1+x
the vertical asymptotes is x = 1. Then since
1 x 1 x 0 1
lim = lim = = 1;
x!1 1 + x x!1 1 + x 0+1
1 x 1 x 0 1
lim = lim = = 1;
x! 1 1 + x x! 1 1 + x 0+1
the horizontal asymptotes is y = 1.
 4. CHAPTER 4 33

4. Chapter 4
4.1. Exercises 4.1.
P
Exercise 4.1.1. Given nk=1 k 2 = n(n+1)(2n+1)
, use the de…nition of the de…nite
R2 6
integral to prove that 1 x2 dx = 37 .
Sol.
Z 2
2 1 1 2 2 n
x2 dx = lim ((1 + ) + (1 + )2 + ::: + (1 + )2 )
1 n!1 n n n n
1X
n
2k k 2
= lim (1 + + 2)
n!1 n k=1
n n
1 2 n(n + 1) 1 n(n + 1)(2n + 1)
= lim (n + + 2 )
n!1 n n 2 n 6
n + 1 n(n + 1)(2n + 1) 2 7
= lim (1 + + 3
)=1+1+ = :
n!1 n 6n 6 3
P [n(n+1)]2
Exercise 4.1.2. Given nk=1 k 3 = , use the de…nition of the de…nite in-
R0 4
tegral to prove that 1 x3 dx = 14 .
Sol.
Z 0
( 1)0 1 2 n
x3 dx = lim (( 1 + )3 + ( 1 + )3 + ::: + ( 1 + )3 )
1 n!1 n n n n
1X
n
3k 3k 2 k 3
= lim ( 1+ + 3)
n!1 n
k=1
n n2 n
1 3 n(n + 1) 3 n(n + 1)(2n + 1) 1 [n(n + 1)]2
= lim
( n+ + )
n!1 n n 2 n2 6 n3 4
3(n + 1) 3n(n + 1)(2n + 1) [n(n + 1)]2
= lim ( 1 + + )
n!1 2n 6n3 4n4
3 1 1
= 1+ 1+ = :
2 4 4
Pn p
Exercise 4.1.3. Find lim ( p1 k).
n!1 n3 k=1

Sol.

n p Z
1 Xp 1X k
n 1 p 2
lim ( p k) = lim ( p )= xdx = :
n!1 n3 k=1 n!1 n n 0 3
k=1
34 CONTENTS

Pn 1
Exercise 4.1.4. Find lim (n k=1 (n+k)2 ).
n!1

Sol.

X
n
1 1X
n
n2 1X 1 2
n
lim (n ) = lim ( ) = lim ( ( ))
n!1
k=1
(n + k)2 n!1 n
k=1
(n + k)2 n!1 n
k=1
1 + nk
Z 1 1
1 2 1 1
= ( ) dx = = :
0 1+x 1+x 0 2

4.2. Exercises 4.2.

Rx p
Exercise 4.2.1. For x 2 R, set F (x) = 0
t 1 + sin tdt.
(a) Find F (0).
(b) Find F 0 (x).
(c) Find F 0 ( 2 ).
Sol. R0 p
(a) F (0) = 0 t 1 + sin tdt = 0:
p
(b) By the Fundamental Theorem of Calculus I, we have F 0 (x) = x 1 + sin x:
p p
2
(c) F 0 ( 2 ) = 2
1 + sin 2
= 2
:

Rx p
Exercise 4.2.2. For x 2 R, set F (x) = 1
t2 + 1dt.
(a) Find F ( 1).
(b) Find F 0 (x).
(c) Find F 0 ( 6).
(d) Find F 00 (x).
Sol. R 1p
(a) F ( 1) = 1 t2 + 1dt = 0:
p
(b) By the Fundamental Theorem of Calculus I, we have F 0 (x) = x2 + 1:
p p
(c) F 0 ( 6) = ( 6)2 + 1 = 37:
d
p
(d) F 00 (x) = x2 + 1 = p x :
dx x2 +1

Rxp
Exercise 4.2.3. For x 2 ( ; ), set F (x) =
2 2 0
sec tdt.
 4. CHAPTER 4 35

(a) Find F (0).

(b) Find F 0 (x).
(c) Find F 0 ( 4 ).
Sol. R0p
(a) F (0) = 0 sec tdt = 0.
p
(b) By the Fundamental Theorem of Calculus I, we have F 0 (x) = sec x:
p pp 1
(c) F 0 ( 4 ) = sec 4 = 2 = 24 :

Rx
Exercise 4.2.4. Let F (x) = 1 sin t2 dt:
p
(a) Find F 0 ( 2 ).
(b) Find F 00 (x).
Sol.
(a) By thepFundamental Theorem of Calculus I, we have F 0 (x) = sin x2 ; so
p
F 0 ( 2 ) = sin( 2 )2 = sin 2 = 1:
d
(b) F 00 (x) = dx
sin x2 = 2x cos x2 :

d
R sin x p
Exercise 4.2.5. Compute dx 0
t3 + 1dt.
Sol.
R sin x p R sin x p p
d
dx 0
d
t3 + 1dt = ( d sin x 0
t3 + 1dt)( d sin
dx
x
) = sin3 x + 1 cos x:

d
R ps p
t6 +1
Exercise 4.2.6. Compute ds 0 t2 +1
dt.
Sol.
R ps p R ps p p p
t6 +1 t6 +1
d
ds 0 t2 +1
dt = ( dpd s 0 t2 +1
dt)( ddss ) = s3 +1
s+1
1
p
2 s
:

d
R x2 sin t
Exercise 4.2.7. Compute dx 1 t
dt, for x 1.
Sol.
R x2 R x2
d sin t 2 sin x2 2 sin x2
dx 1 t
dt = ( dxd2 1
sin t
t
dt)( dx
dx
) = x2
2x = x
:

R x p R sec t
Exercise 4.2.8. Suppose f is continuous on R, let F (x) = 0
t( p2 f (u)du)dt.
(a) Compute F (0).
(b) Compute F 0 ( 4 ).
(c) Find F 00 (x).
Sol. R 0 p R sec t
(a) F (x) = 0 t( p2 f (u)du)dt = 0.
36 CONTENTS

p R sec x
(b) By the Fundamental Theorem of Calculus I, we have F 0 (x) = x( p2 f (u)du);
p R p2
so F 0 ( 4 ) = 4 ( p2 f (u)du) = 0:
d p
R sec x p 1
R sec x
(c) F 00 (x) = dx
x( p2 f (u)du) = xf (sec x) + p
2 x
p
2
f (u)du:

4.3. Exercises 4.3.

Compute the following integrals.
R 10
Exercise 4.3.1. 1 ( x12 + x2 )dx
Sol.
Z 10 10
1 1 1 3 3339
( 2 + x2 )dx = ( + x) = :
1 x x 3 1 10
R3 1
Exercise 4.3.2. 1
(3x2 x2
)dx
Sol.
Z 3 3
1 1 76
(3x2 2
)dx = (x3 + ) = :
1 x x 1 3
R2p
Exercise 4.3.3. 0
xdx
Sol.
Z 2 Z 2 2 p
p 1 2 3 4 2
xdx = x dx = x 2
2 = :
0 0 3 0 3
R1
Exercise 4.3.4. 1
(x2 2)2 dx
Sol.
Z 1 Z 1
2 2
(x 2) dx = (x4 4x2 + 4)dx
1 1
1
1 4 3 86
= ( x5 x + 4x) = :
5 3 1 15

R
Exercise 4.3.5. 0
4
tan2 xdx
 4. CHAPTER 4 37

Sol.
Z Z
4 4
2
tan xdx = (sec2 x 1)dx = (tan x x)j04 = 1 :
0 0 4
R 1
Exercise 4.3.6. 4
0 (1 sin x)(1+sin x)
dx
Sol.

Z Z Z
4 1 4 1 4 1
dx = dx = dx
0 (1 sin x)(1 + sin x) 0 1 sin2 x 0 cos2 x
Z
4
= sec2 xdx = tan xj04 = 1:
0

R1 p
Exercise 4.3.7. 0
(1 x2 ) xdx
Sol.
Z 1 Z 1 1
p2 1 5 2 3 2 7 8
(1 x ) xdx = (x 2 x 2 )dx = ( x 2 x2 ) = :
0 0 3 7 0 21

4.4. Exercises 4.4.

Exercise 4.4.1. Let f (x) = (1 + cos x)2 + sin2 x, x 2 [ ; ]. Find the area
between the graph of f and the x-axis.
Sol.
Note that f (x) 0; 8 x 2 [
; ]: The area is
Z Z
2 2
[(1 + cos x) + sin x]dx = (1 + 2 cos x + cos2 x + sin2 x)dx
Z
= (2 + 2 cos x)dx = (2x + 2 sin x) =4 :

Exercise 4.4.2. Let f (x) = x3 , x 2 [ 1; 1]. Find the area between the graph of
f and the y-axis.
Sol.
Solution 1:
Since the area between the graph of f and the y-axis is equal to
2 (1 1) the area between the graph of f and the x axis;
38 CONTENTS

and the area between the graph of f and the x-axis is equal to
Z 1 Z 0
3 1
x dx + (0 x3 )dx = :
0 1 2
( Note that f (x) 0; 8 x 2 [0; 1]; f (x) 0; 8 x 2 [ 1; 0]:) We have that the area
between the graph of f and the y-axis is equal to 23 :

Solution 2:
1
Consider x = g(y) = y 3 ; y 2 [ 1; 1]: The area between the graph of f and the
y-axis is equal to the area between the graph of g and the y-axis, so the area is
Z 1 Z 0
1 1 3 3 3
y 3 dy + (0 y 3 )dy = + = :
0 1 4 4 2
(Note that g(y) 0; 8 y 2 [0; 1]; g(y) 0; 8 y 2 [ 1; 0]:)

Exercise 4.4.3. Let f (x) = tan2 x, x 2 [0; 4 ]. Find the area between the graph
of f and the x-axis.
Sol. R
Note that f (x) 0; 8 x 2 [0; 4 ]: The area is 4
0
tan2 xdx = 1 4
: (see Exercise
4.3.5)

Exercise 4.4.4. Let f (x) = sec x tan x, x 2 [ 4

; 4 ]. Find the area between the
graph of f and the x-axis.
Sol.
Note that f (x) 0; 8 x 2 [0; 4 ]; f (x) 0; 8 x 2 [ 4 ; 0]: The area is
Z Z 0
4
sec x tan xdx + (0 sec x tan x)dx
0 4

0
p
= sec xj0 4
sec xj =2 2 2:
4

Exercise 4.4.5. Sketch the region bounded by the curves and …nd its area.
p 1
(a) y = x, y = x 4 .
(b) x = y 2 , y = x2 .
(c) y = 36x, y = x3 .
(d) y = sin( x), x = 0:5y.
Sol. p p
1 1
(a) Since x = x 4 at x = 0 or 1, and since x x 4 for x 2 [0; 1]; the area is
Z 1 1
1 p 4 5 2 3 2
(x 4 x)dx = ( x 4 x2 ) = :
0 5 3 0 15
 4. CHAPTER 4 39

p p p
(b) Consider x = y 2 , y = x: Since x = x2 at x = 0 or 1, and since x x2
for x 2 [0; 1]; the area is
Z 1 1
p 2 3 1 3 1
( x x2 )dx = ( x 2 x) = :
0 3 3 0 3

(c) Since 36x = x3 at x = 6; 0 or 6, and since 36x x3 for x 2 [0; 6]; 36x x3
for x 2 [ 6; 0]; the area is
Z 6 Z 0
3
(36x x )dx + (x3 36x)dx = 648:
0 6

(d) Consider x = 0:5y , y = 2x: Since sin( x) = 2x at x = 0 or 12 , and since

sin( x) 2x for x 2 [0; 12 ]: the area is
Z 1 1
2 1 2 1 1
(sin( x) 2x)dx = ( cos( x) x2 ) = :
0 0 4

4.5. Exercises 4.5.

Compute the following integrals.
R 2
Exercise 4.5.1. p3s 1 s3
ds
Sol.
Let u = 1 s3 ; du = 3s2 ds:
Z Z p
3s2 1 p
p ds = p du = 2 u+C = 2 1 s3 + C:
1 s3 u
R 1
Exercise 4.5.2. (1+t)2
dt
Sol.
Let u = 1 + t; du = dt:
Z Z
1 1 1 1
2
dt = du = +C = + C:
(1 + t) u2 u 1+t
R 1 2
Exercise 4.5.3. 3
x tan x3 dx
Sol.
Let u = x3 ; du = 3x2 dx:
Z Z
1 2 3 1 1 1
x tan x dx = tan udu = ln j sec uj + C = ln j sec x3 j + C:
3 9 9 9
40 CONTENTS

R4
Exercise 4.5.4. p 1 dx
2 x+1

Sol.
Let u = x + 1; du = dx:
Z 4 Z 5 p p
1 1 p 5
p dx = p du = 2 u 3
=2 5 2 3:
2 x+1 3 u
R p
Exercise 4.5.5. x2 x 1dx
Sol.
Let u = x 1; du = dx:
Z Z Z
2
p 2
p 5 3 1
x x 1dx = (u + 1) udu = (u 2 + 2u 2 + u 2 )du
2 7 4 5 2 3
= u2 + u2 + u2 + C
7 5 3
2 7 4 5 2 3
= (x 1) 2 + (x 1) 2 + (x 1) 2 + C:
7 5 3
R 2
Exercise 4.5.6. px dx
x+1

Sol.
Let u = x + 1; du = dx:
Z Z Z
x2 (u 1)2 3 1 1
p dx = p du = (u 2 2u 2 + u 2 )du
x+1 u
2 5 4 3 1
= u2 u 2 + 2u 2 + C
5 3
2 5 4 3 1
= (x + 1) 2 (x + 1) 2 + 2(x + 1) 2 + C:
5 3
R
Exercise 4.5.7. (9x + 4)(3x 1)9 dx
Sol.
Let u = 3x 1; du = 3dx:
Z Z Z
9 1 7
(9x + 4)(3x 1) dx = (3u + 7)u du = (u10 + u9 )du
9
3 3
1 11 7
= u + u10 + C
11 30
1 7
= (3x 1)11 + (3x 1)10 + C:
11 30
 4. CHAPTER 4 41

R sec2
p
x
Exercise 4.5.8. p
x
dx
Sol. p
Let u = x; du = 2p1 x dx:
Z p Z
sec2 x p
p dx = 2 sec2 udu = 2 tan u + C = 2 tan x + C:
x
R x
Exercise 4.5.9. csc x2
dx
Sol.
Let u = x2 ; du = 2xdx:
Z Z
x 1 1 1
2
dx = sin udu = cos u + C = cos x2 + C:
csc x 2 2 2
R
Exercise 4.5.10. (tan3 x)(sec3 x)dx
Sol.
Consider Z Z
3 3
(tan x)(sec x)dx = (sec2 x 1)(sec2 x)(tan x sec x)dx:

Let u = sec x; du = sec x tan xdx:

Z Z
2 2 1 1 3
(sec x 1)(sec x)(tan x sec x)dx = (u2 1)u2 du = u5 u +C
5 3
1 1
= sec5 x sec3 x + C:
5 3
R
Exercise 4.5.11. (sin3 x + cos3 x)dx
Sol.
Consider
Z Z Z
3 3 3
(sin x + cos x)dx = sin xdx + cos3 xdx
Z Z
2
= sin x(1 cos x)dx + cos x(1 sin2 x)dx:

Let u = cos x; du = sin xdx; v = sin x; dv = cos xdx:

Z Z
sin x(1 cos x)dx + cos x(1 sin2 x)dx
2

Z Z
1 1 3
= (1 u )du + (1 v 2 )dv = u3 u + C1 + v
2
v + C2
3 3
1 1 3
= cos3 x cos x + sin x sin x + C:
3 3
42 CONTENTS

R
Exercise 4.5.12. (sin3 x)(cos3 x)dx
Sol.
Let u = sin x; du = cos xdx:
Z Z
3 3
(sin x)(cos x)dx = (sin3 x)(1 sin2 x)(cos x)dx
Z
1 1 6
= (u3 u5 )du = u4 u +C
4 6
1 4 1 6
= sin x sin x + C:
4 6

Exercise 4.5.13. Suppose

Ra the following de…nite integrals both exist, and f (x) =
f ( x); 8 x. Prove that a f (x)dx = 0.
Sol.
Consider Z Z Z
a a 0
f (x)dx = f (x)dx + f (x)dx:
a 0 a
R0
For a
f (x)dx; let u = x; du = dx: Then we have
Z 0 Z 0 Z a
f (x)dx = f ( u)du = f ( u)du
a a 0
Z a Z a Z a
= f (u)du = f (u)du = f ( x)dx:
0 0 0

Thus
Z a Z a Z 0 Z a
f (x)dx = f (x)dx + f (x)dx = f (x)dx
a 0 a a
Z a Z a
= f (x)dx + f ( x)dx = 0:
0 0

Exercise
R 2x 4.5.14. Suppose f 0 (x) > 0; 8 x 2 R, and f ( 1) = 1; f (1) = 1.Set
F (x) = 0 f (3t)dt. Prove that
(a) F is twice di¤erentiable.
(b) F has a critical point on the interval ( 1; 1), and this critical point is a local
minimum point.
Sol.
 4. CHAPTER 4 43

(a) By the Fundamental Theorem of Calculus I, we have

Z 2x
0 d d(2x)
F (x) = f (3t)dt = 2f (6x):
d(2x) 0 dx
Then since f 0 (x) > 0; 8 x 2 R, we know that f 0 (x) exists; 8 x 2 R: So F is twice
di¤erentiable and F 00 (x) = 12f 0 (6x):

(b) Since f 0 (x) exists; 8 x 2 R; f (x) is continuous on R. Then since we have

f ( 1) = 1; f (1) = 1; by the intermediate value theorem (Thm 1.6.1), 9 c 2 ( 1; 1)
such that f (c) = 0: Therefore, 9 d = 16 c 2 ( 16 ; 61 ) ( 1; 1) such that f (6d) =
f (c) = 0: Then since F 0 (d) = 2f (6d) = 0; d is a critical point of F: Moreover, since
f 0 (x) > 0; 8 x 2 R; F 00 (d) = 12f 0 (6d) > 0: Then by the second derivative test (Thm
3.3.9), d is a local minimum point of F:
44 CONTENTS

5. Chapter 5
5.1. Exercises 5.1.
For the following problems, …nd the area of the region enclosed by the given
graphs.
2
Exercise 5.1.1. y = x3 ; y = x 3 :
Sol.
y = x3
Since the solutions of 2 are (0; 0) and (27; 9); the area of the enclosed
y = x3
region is
Z 27 27
2 x 3 5 1 2 243
(x 3 )dx = ( x 3 x) = :
0 3 5 6 0 10

Exercise 5.1.2. y 2 = x 4; y 2 = x2 :
Sol.
y2 = x 4
Since the solutions of x are (8; 2) and (8; 2); the area of the enclosed
y2 = 2
region is
Z 2 2
1 3
(y 2 + 4 2y 2 )dy = ( y + 4y) = 16:
2 3 2

Exercise 5.1.3. 4y = x; y = x2 ; x > 0:

Sol.
4y = x
Since the solutions of 1
are (0; 0) and ( 41 , 16 ); the area of the enclosed
y = x2
region is
Z 1 1
4 1 1 1 3 4 1
( x x )dx = ( x2
2
x) = :
0 4 8 3 0 384

Exercise 5.1.4. Find the number c such that the area of the region bounded by
these graphs y + x2 = c2 and y x2 = c2 is 576.
Sol.
y + x 2 = c2
Since the solutions of are (c; 0) and ( c; 0); the area of the
y x 2 = c2
enclosed region is
Z c c
2 8
[( x2 + c2 ) (x 2 2
c )]dx = ( x3 + 2c2 x) = c3 = 576:
c 3 c 3
So c = 6:
 5. CHAPTER 5 45

5.2. Exercises 5.2.

For the following problems, …nd the volume of the solid obtained by revolving the
region bounded by the graphs about the given line.
Exercise 5.2.1. y = x2 ; x = y 2 about the y-axis.
Sol.
y = 12 x
Since the solutions of are (0; 0) and (4; 2); the volume of the bounded
y2 = x
solid revolving about y-axis is
Z 2 2
2 2 2 4 3 1 5 64
[(2y) (y ) ]dx = [ ( y y )] = :
0 3 5 0 15

p
Exercise 5.2.2. y = x; y = x about y = 1:
Sol. p
y= x
Since the solutions of are (0; 0) and (1; 1); the volume of the bounded
y=x
solid revolving about y = 1 is
Z 1
1
[(x 1)2 (x 2 1)2 ]dx
0
1
1 3 2 4 3 1
= [ ( x3 x + x 2 )] = :
3 2 3 0 6

Exercise 5.2.3. y 2 = x; y = x2 about x = 1:

Sol.
y2 = x
Since the solutions of are (0; 0) and (1; 1); the volume of the bounded
y = x2
solid revolving about x = 1 is
Z 1
1
[(y 2 + 1)2 (y 2 + 1)2 ]dx
0
1
1 5 2 3 1 2 4 3 29
= [ ( y y + y + y 2 )] = :
5 3 2 3 0 30

Exercise 5.2.4. y = x2 4x + 3; y = x 1 about y = 3:

Sol.
46 CONTENTS

y = x2 4x + 3
Since the solutions of are (1; 0) and (4; 3); the volume of the
y=x 1
bounded solid revolving about y = 3 is
Z 4
[(x2 4x + 3 3)2 (x 1 3)2 ]dx
1
4
1 108
= [ ( x5 2x4 + 5x3 + 4x2 16x)] = :
5 1 5

Exercise 5.2.5. y = 4x + 6; y = x3 x2 + 2x 6 about the x-axis.

Sol.
y = 4x 6
Since the solutions of are (1; 0) and (4; 3); the volume
y = x3 x2 + 2x 6
of the bounded solid revolving about y = 2 is
Z 2
[(x3 x2 + 2x 6 2)2 (4x 6 2)2 ]dx
0
Z 0
+ [(4x 6 2)2 (x3 x2 + 2x 6 2)2 ]dx
1
2
1 1 6 4
= [ ( x7 x + x5 5x + x3 + 16x2 )]
4
7 3 3 0
0
1 1 4 3
+ [ ( x7 + x6 5
x + 5x 4
x 2
16x )]
7 3 3 1
766
= :
21

5.3. Exercises 5.3.

For the following problems, use the shell method to …nd the volume generated by
revolving the region bounded by the given graphs about the given line.
Exercise 5.3.1. y = 3; y = 4x x2 about x = 3:
Sol.
y=3
Since the solutions of are (1; 3) and (3; 3); the volume of the
y = 4x x2
bounded shell revolving about x = 3 is
Z 3
2 ( x + 3) [(4x x2 ) 3]dx
1
3
1 7 3 15 2 8
= [2 ( x4 x + x 9x)] = :
4 3 2 1 3
 5. CHAPTER 5 47

Exercise 5.3.2. (y 2)2 = x 1; x = 2 about the x-axis.

Sol.
2)2 = x 1
(y
Since the solutions of are (2; 3) and (2; 1); the volume of the
x=2
bounded shell revolving about x-axis is
Z 3
2 y[2 ((y 2)2 + 1)]dx
1
3
1 4 4 3 3 2 16
= [2 ( x + x x )] = :
4 3 2 1 3

Exercise 5.3.3. (y 1)2 = 1 x2 about the y-axis.

Sol.
Since (yp 1)2 = 1 x2 is a circle centered at (0; 1) with radius 1, consider
y = 1 + p1 x2
; then the volume of the bounded shell revolving about y-axis is
y=1 1 x2
Z 1 p p
2 x[(1 + 1 x2 ) (1 1 x2 )]dx:
0
2
Let u = 1 x ; du = 2xdx; then
Z 1 p p
2 x[(1 + 1 x2 ) (1 1 x2 )]dx
0
Z 1 p
= 4 x 1 x2 dx
0
Z 0 0
1 4 3 4
= ( 2 )u 2 du = u2 = :
1 3 1 3

Exercise 5.3.4. y = x3 ; y = 8x about the y-axis.

Sol.
y = x3 p p p p
Since the solutions of are (2 2; 16 2); (0; 0) and ( 2 2; 16 2); the
y = 8x
volume of the bounded shell revolving about y-axis is
Z 2p2 Z 0
3
2 x[8x x ]dx + p
2 ( x)[x3 8x]dx
0 2 2
p
2 2 0
8 1 5 8 1 5
= [2 ( x3 x )] + [2 ( x3 x )] p
3 5 0 3 5 2 2
p
512 2
= :
15
48 CONTENTS

Exercise 5.3.5. y = 2; y = 4; y = 4x2 ; x = 0 about the y-axis.

Sol.
y=4 y=2
Since when x > 0; the solutions of 2 and are (1; 4) and ( p12 ; 2);
y = 4x y = 4x2
the volume of the bounded shell revolving about y-axis is
Z 1 Z 1
2 x[4 2]dx 2 x(4x2 2)dx
1
0 p
2

1 1 3
= [2 (x2 )] 0
[2 (x4 x2 )] p1
= :
2 2
 6. CHAPTER 6 49

6. Chapter 6
6.1. Exercises 6.1.
x
Exercise 6.1.1. Let f (x) = tan 2
+ x2 + 3, 4
x 4
. Show that f is an
0
1-1 function and …nd (f 1 ) (3).
Sol.
Since f is continuous on [ 4
; 4 ], is di¤erentiable on ( 4
; 4 ), and since
x
f 0 (x) = sec2 ( ) + 2x > 1+2 ( ) = 0; 8 x 2 ( ; );
2 2 2 4 4 4
by theorem 3.2.5, f is increasing on [ 4 ; 4 ]. Hence f is 1-1.
Then since f (0) = 3 and f 0 (0) = 2 , by theorem 6.1.4,

1 0 1 2
f (3) = = :
f 0 (0)

Rx p
Exercise 6.1.2. Let f (x) = 10
1 + t3 dt. Show that f is an 1-1 function and
0
…nd (f 1 ) (0).
Sol. p
Since 1 + t3 is continuous on [ 1; 1); by the fundamental theorem of calculus
I, f is continuous on [ 1; 1), is di¤erentiable on ( 1; 1); and
p
f 0 (x) = 1 + x3 > 0; 8 x 2 ( 1; 1):
Hence by theorem 3.2.5, f is increasingpon [ 1; 1). So f is 1-1.
Then since f (10) = 0 and f 0 (10) = 1001, by theorem 6.1.4,
1 0 1 1
f (0) = =p :
f 0 (10) 1001

6.2. Exercises 6.2.

6.3. Exercises 6.3.

Exercise 6.3.1. Find lim [ln(1 x2 ) ln(1 x)].
x!1

Sol.
x2
lim [ln(1 x2 ) ln(1 x)] = lim ln 11 x
= lim ln(1 + x) = ln 2:
x!1 x!1 x!1

Exercise 6.3.2. Use the de…nition of derivative to prove that lim ln(x+1)
x
.
x!0
50 CONTENTS

Sol.
Let f (x) = ln(x + 1). Then we have
ln(x + 1) ln(h + 1) ln(1)
lim = lim
x!0 x h!0 h
f (h) f (0)
= lim
h!0 h
1
= f 0 (0) = = 1:
0+1

Re 1
Exercise 6.3.3. 6 x ln x
dx =?

Sol.
Let u = ln x; du = x1 dx; then we have
Z e Z u=ln e
1 1
dx = du = ln jujj1ln 6 = ln 1 ln jln 6j = ln jln 6j :
6 x ln x u=ln 6 u

R 2 sin x cos x
Exercise 6.3.4. 1+cos2 x
dx =?

Sol. Let u = 1 + cos2 x; du = 2 sin x cos xdx; then we have

Z Z
2 sin x cos x 1
dx = du = ln juj + C = ln(1 + cos2 x) + C:
1 + cos2 x u

d
Exercise 6.3.5. dx
ln jtan 2xj = ?
d 1 d 1
Sol. dx
ln jtan 2xj = tan 2x dx
tan 2x = tan 2x
sec2 (2x) 2:

Exercise 6.3.6. Find the tangent line of the graph y = sin (ln x2 ) at the point
(1; 0).

dy d 1 2 cos(ln x2 )
Sol. Since dx
= dx
sin (ln x2 ) = cos (ln x2 ) x2
2x = x
; the slope of the
tangent line is
dy 2 cos (ln 12 )
= = 2:
dx (1;0) 1
Hence the tangent line is y = 2(x 1):
 6. CHAPTER 6 51

6.4. Exercises 6.4.

d ex
Exercise 6.4.1. dx
e =?
d ex x d x x
Sol. dx
e = ee dx
e = ee ex :

x
Exercise 6.4.2. Find y 0 if x y = ey .
Sol. Di¤erentiate both sides with respect to x; we have
x d x x y 1 xy 0
1 y0 = e y = ey :
dx y y2
x
y 2 ye y
So y 0 = x :
y 2 xe y

Exercise 6.4.3. Find the local extreme points of f (x) = ex + x.

Sol.
Since f 0 (x) = ex + 1; the critical point is 0. Then since f 00 (x) = ex ; f 00 (0) =
1 < 0: Hence by the second derivative test, 0 is a local maximum point.

R2 x
Example 1. 0
e dx = ?
Sol.
Z 2 Z 2 2
x 1 x 1 x 1 2 1
e dx = e d( x) = e = e + :
0 0 0

R p
Exercise 6.4.4. ex 1 + ex dx = ?
Sol.
Let u = 1 + ex ; du = ex dx; then we have
Z Z
x
p p 2 3 2 3
x
e 1 + e dx = udu = u 2 + C = (1 + ex ) 2 + C:
3 3
R
Exercise 6.4.5. etan x sec2 xdx = ?
Sol.
Let u = tan x; du = sec2 xdx; then we have
Z Z
tan x
e sec xdx = eu du = eu + C = etan x + C:
2

R2 1
Exercise 6.4.6. 1
x 2 e x dx = ?
52 CONTENTS

Sol. Z Z
2 2
2 1 1 1 1 2 1
x e dx =
x e x d( ) = ex = e 2 + e:
1 1 x 1

6.5. Exercises 6.5.

d x
Exercise 6.5.1. Find dx
a if a > 0.
Sol.
d x d x ln a d
a = e = ex ln a (x ln a) = ex ln a ln a = ax ln a:
dx dx dx

d 2
Exercise 6.5.2. dx
tan(4x ) = ?
Sol.
By exercise 6.5.1,
d 2 2 d x2
tan(4x ) = sec2 (4x ) (4 )
dx dx
2 2 d 2
= sec2 (4x ) 4x ln 4 x
dx
2 2
= sec2 (4x ) 4x ln 4 2x:

d sin x
Exercise 6.5.3. dx
x =?
Sol.
d sin x d sin x ln x d
x = e = esin x ln x (sin x ln x)
dx dx dx
1
= xsin x (cos x ln x + sin x ):
x

Exercise 6.5.4. d
dx
(cos x)x = ?
Sol.
d d x ln(cos x) d
(cos x)x = e = ex ln(cos x) [x ln(cos x)]
dx dx dx
1
= (cos x)x [ln(cos x) + x ( sin x)]:
cos x

Exercise 6.5.5. Find y 0 if y x = xy .

 6. CHAPTER 6 53

Sol.
Since
d x d x ln y d x 0
y = e = ex ln y (x ln y) = y x (ln y + y)
dx dx dx y
and
d y d y ln x d y
x = e = ey ln x (y ln x) = xy (y 0 ln x + );
dx dx dx x
we have
yxy
0 x
y x ln y
y = xy x :
y
xy ln x
R2
Exercise 6.5.6. 1
10x dx = ?
Sol. By exercise 6.5.1,
Z 2 2
1 100 10 90
10x dx = 10x = = :
1 ln 10 1 ln 10 ln 10 ln 10
R
Exercise 6.5.7. 3sin x cos xdx = ?
Sol.
Let u = sin x; du = cos xdx; then we have
Z Z
sin x 1 u 1 sin x
3 cos xdx = 3u du = 3 +C = 3 + C:
ln 3 ln 3

6.6. Exercises 6.6.

1
p 1
p
Exercise 6.6.1. Show that cos sin x = 1 x2 and sec (tan x) = 1 + x2 .
Find tan (sec 1 x).
Sol.
1 1
Since 2
sin x 2
; cos sin x 0: Then since
1 = cos2 sin 1 x + sin2 sin 1 x = cos2 sin 1
x + x2 ;
p
we have cos sin 1 x = 1 x2 :
Since 2 < tan 1 x < 2 ; sec (tan 1 x) > 0: Then since
sec2 tan 1 x = 1 + tan2 tan 1 x = 1 + x2 ;
p
we have sec (tan 1 x) = 1 + x2 :
If x 1; then 0 sec 1 x < 2 and tan (sec 1 x) 0: Then since
1 + tan2 sec 1
x = sec2 sec 1
x = x2 ;
p
we have tan (sec 1 x) = x2 1:
54 CONTENTS

1 1
If x 1; then 2
< sec x and tan (sec x) 0: Then since
1 + tan2 sec 1
x = sec2 sec 1
x = x2 ;
p
we have tan (sec 1 x) = x2 1:

p
Exercise 6.6.2. Let f (x) = 16 x2 + x sin 1 ( x4 ). Find f 0 (2).
Sol.
Since
1 x 1 x 1
f 0 (x) = (16 x2 )
( 2x) + sin 1 ( ) + p2
x
2 4 1 (4) 4
2

x x x
= p + sin 1 ( ) + p
16 x 2 4 16 x2
x
= sin 1 ( );
4
0 1 1
we have f (2) = sin 2 = 6 :

R1 sin 1 x
Exercise 6.6.3. 0
2 p
1 x2
dx =?
Sol.
1 p 1
Let u = sin x; du = 1 x2
dx; then we have
Z 1
1 Z u=sin 1 1
2
2 sin x 2 1 2 6
p dx = udu = u = :
0 1 x2 u=sin 1 0 2 0 72
R x+1
Exercise 6.6.4. x2 +1
dx =?
Sol. Z Z Z
x+1 x 1
dx = dx + dx
x2 + 1 2
x +1 x2 +1
Z 1
= 2
d(x2 + 1) + tan 1
x
x2 + 1
1
= ln x2 + 1 + tan 1
x + C:
2
R p
Exercise 6.6.5. p 1 dx = ? (Hint: Let u = x.)
x(x+1)

Sol. p
Let u = x; du = 2p1 x dx; then we have
Z Z
1 2 1 1
p
p dx = 2
du = 2 tan u + C = 2 tan x + C:
x(x + 1) u +1
 6. CHAPTER 6 55

R
Exercise 6.6.6. p 1 dx = ? (a > 0)
a2 x2

Sol.
Z Z Z
1 1 1 x x
p dx = q dx = q d( ) = sin 1 ( ) + C:
a2 x 2 a 1 x2
1 x2 a a
a2 a2

R 1
Exercise 6.6.7. a2 +x2
dx = ? (a > 0)
Sol. Z Z
1 1 x 1 x
dx = x2
d( ) = tan 1 ( ) + C:
a + x2
2
a(1 + a2
) a a a
R
Exercise 6.6.8. p1 dx = ? (a > 0)
jxj x2 a2

Sol. Z Z Z
1 1 1 x
p dx = q dx = q d( )
jxj x2 a2 a2 x
a
x2
a2
1 a x
a
x2
a2
1 a
1 x
= sec 1 ( ) + C:
a a
56 CONTENTS

7. Chapter 7
7.1. Exercises 7.1.
R x4
Exercise 7.1.1. x3 e dx = ?
Sol.
Let u = x4 ; du =
4x3 dx; then we have
Z Z
3 x4 1 u 1 u 1 x4
x e dx = e du = e +C = e + C:
4 4 4
R ln x
Exercise 7.1.2. x
dx =?
Sol. Z Z
ln x 1
dx = ln xd(ln x) = (ln x)2 + C:
x 2

Example 2.
R 2x+1
Exercise 7.1.3. x2 +x+1
dx =?
Sol.
Z Z
2x + 1 1
2
dx = d(x2 + x + 1) = ln x2 + x + 1 + C:
x +x+1 x2 +x+1
R
Exercise 7.1.4. sin5 x cos xdx = ?
Sol. Z Z
5 1 6
sin x cos xdx = sin5 xd(sin x) = sin x + C:
6
R tan x
Exercise 7.1.5. p dx =?
1 (lnjsec xj)2

Sol.
x tan x
Let u = ln jsec xj ; du = secsec x
dx = tan xdx; then we have
Z Z
tan x 1
p dx = p du = sin 1 u + C = sin 1 (ln jsec xj) + C:
1 (ln jsec xj) 2 1 u 2

R tan 1 x
Exercise 7.1.6. 1+x2
dx =?
Sol.
Z Z
tan 1 x 1 1 1 1
dx = tan xd(tan x) = (tan x)2 + C:
1 + x2 2
 7. CHAPTER 7 57

R p
sec( x)
Exercise 7.1.7. p
x
dx =?

Sol.
Z p Z
sec( x) p p p p
p dx = 2 sec( x)d( x) = 2 ln sec( x) + tan( x) + C:
x

R p
Exercise 7.1.8. x3 1 x2 dx = ?
Sol.
Let u = 1 x2 ; du = 2xdx; then we have
Z p Z
3 2
1 p
x 1 x dx = (1 u) udu
2
Z 1 3
u2 u2
= + du
2 2
1 3 1 5
= u2 + u2 + C
3 5
1 3 1 5
= (1 x2 ) 2 + (1 x2 ) 2 + C:
3 5

7.2. Exercises 7.2.

R
Example 3. tan 1 xdx = ?
Sol.
1 1
Let u = tan x; du = dv = dx; v = x; then we have
1+x2
dx;
Z Z
1 1 x
tan xdx = x tan x dx
1 + x2
Z 1
1
= x tan x 2
d(1 + x2 )
1 + x2
1
= x tan 1 x ln 1 + x2 + C:
2

R
Exercise 7.2.1. e2x sin 3xdx = ?
Sol.
Let u = e2x ; du = 2e2x dx; dv = sin 3xdx; v = 31 cos 3x; then we have
Z Z
2x 1 2x 2 2x
e sin 3xdx = e cos 3x + e cos 3xdx:
3 3
58 CONTENTS

Once again, let u = e2x ; du = 2e2x dx; dv = cos 3xdx; v = 13 sin 3x; then we have
Z Z
2x 1 2x 2 2x
e cos 3xdx = e sin 3x e sin 3xdx:
3 3
Hence we have
Z Z
1 2x 2
e2x sin 3xdx = e cos 3x + e2x cos 3xdx
3 3
Z
1 2x 2 2x 4
= e cos 3x + e sin 3x e2x sin 3xdx:
3 9 9
Thus Z
13 1 2x 2
e2x sin 3xdx = e cos 3x + e2x sin 3x + C;
9 3 9
that is,
Z
3 2x 2
e2x sin 3xdx = e cos 3x + e2x sin 3x + C:
13 13

R
Exercise 7.2.2. (ln x)2 dx = ?
Sol.
Let u = (ln x)2 ; du = 2 lnx x dx; dv = dx; v = x; then by example 7.2.6, we have
Z Z
2 2
(ln x) dx = x (ln x) 2 ln xdx = x (ln x)2 2x ln x + 2x + C:

R
Exercise 7.2.3. (ln x)3 dx = ?
Sol. 2
Let u = (ln x)3 ; du = 3(lnxx) dx; dv = dx; v = x; then by exercise 7.2.3, we have
Z Z
3 3
(ln x) dx = x (ln x) 3(ln x)2 dx

= x (ln x)3 3x (ln x)2 + 6x ln x 6x + C:

R
Exercise 7.2.4. x2 sin xdx = ?
Sol.
Let u = x2 ; du = 2xdx; dv = sin xdx; v = cos x; then we have
Z Z
2 2
x sin xdx = x cos x + 2x cos xdx:
 7. CHAPTER 7 59

Once again, let u = x; du = dx; dv = cos xdx; v = sin x; then we have

Z Z
2 2
x sin xdx = x cos x + 2 x cos xdx
Z
2
= x cos x + 2(x sin x sin xdx)

= x2 cos x + 2x sin x + 2 cos x + C:

R
Exercise 7.2.5. x3 ln xdx = ?
Sol.
Let u = ln x; du = x1 dx; dv = x3 dx; v = 14 x4 ; then we have
Z Z
3 1 4 1 3 1 1 4
x ln xdx = x ln x x dx = x4 ln x x + C:
4 4 4 16
R
Exercise 7.2.6. x2 (ln x)2 dx = ?
Sol.
Let u = (ln x)2 ; du = 2 lnx x dx; dv = x2 dx; v = 13 x3 ; then we have
Z Z
2 2 1 3 2 2 2
x (ln x) dx = x (ln x) x ln xdx:
3 3
Once again, let u = ln x; du = x1 dx; dv = x2 dx; v = 13 x3 ; then we have
Z Z
2 2 1 3 2 2
x (ln x) dx = x (ln x) x2 ln xdx
3 3
Z
1 3 2 2 1 3 1 2
= x (ln x) ( x ln x x dx)
3 3 3 3
1 3 2 3 2
= x (ln x)2 x ln x + x3 + C:
3 9 27
R xex
Exercise 7.2.7. (x+1)2
dx =?
Sol.
1 1
Let u = xex ; du = (xex + ex )dx; dv = (x+1)2 dx; v = x+1
; then we have
Z Z
xex xex xex + ex
= + dx
(x + 1)2 x+1 x+1
Z
xex
= + ex dx
x+1
xex
= + ex + C:
x+1
60 CONTENTS

7.3. Exercises 7.3.

R
Exercise 7.3.1. cos3 xdx = ?

Sol.
Z Z
3
cos xdx = cos x(1 sin2 x)dx
Z
= (1 sin2 x)d(sin x)
1 3
= sin x sin x + C:
3

R
Exercise 7.3.2. sin4 xdx = ?

Sol.
Z Z Z
4 1
cos 2x 2 1 2 cos 2x + cos2 2x
sin xdx = ( ) dx = dx
2 4
Z
1 cos 2x 1 + cos 4x
= ( + )dx
4 2 8
x sin 2x x sin 4x
= + + + C:
4 4 8 32

R
Exercise 7.3.3. sin3 x cos2 xdx = ?

Sol.
Z Z
3 2
sin x cos xdx = sin x(1 cos2 x) cos2 xdx
Z
= (cos2 x cos4 x)d(cos x)
1 1
= cos3 x + cos5 x + C:
3 5

R
Exercise 7.3.4. sin4 x cos2 xdx = ?
 7. CHAPTER 7 61

Sol.
Z Z
4 2 1 cos 2x sin2 2x
sin x cos xdx = dx
2 4
Z Z
sin2 2x cos 2x sin2 2x
= dx dx
8 8
Z Z
1 cos 4x sin2 2x
= dx d(sin 2x)
16 16
x sin 4x sin3 2x
= + C:
16 64 48

R
Exercise 7.3.5. tan3 x sec3 xdx = ?

Sol.
Z Z
3 3
tan x sec xdx = tan x sec2 x 1 sec3 xdx
Z
= tan x sec5 x tan x sec3 xdx
Z
= sec4 x sec2 xd(sec x)
1 1
= sec5 x sec3 x + C:
5 3

R
Exercise 7.3.6. tan2 x sec4 xdx = ?

Sol.
Z Z
2 4
tan x sec xdx = tan2 x tan2 x + 1 sec2 xdx
Z
= tan4 x + tan2 xd(tan x)
1 1
= tan5 x + tan3 x + C:
5 3

R
Exercise 7.3.7. tan4 xdx = ?
62 CONTENTS

Sol.
Z Z
4
tan xdx = tan2 x sec2 x 1 dx
Z Z
2 2
= tan x sec xdx tan2 xdx
1 1
= tan5 x + tan3 x + C:
5 3

R
Exercise 7.3.8. tan2 x sec xdx = ?
Sol.
Let u = tan x; du = sec2 xdx; dv = tan x sec xdx; v = sec x; then we have
Z Z
2
tan x sec x = tan x sec x sec3 xdx
Z
= tan x sec x (1 + tan2 x) sec xdx
Z Z
= tan x sec x sec xdx tan2 x sec xdx
Z
= tan x sec x ln jsec x + tan xj tan2 x sec xdx:

Hence we have
Z
2 tan2 x sec x = tan x sec x ln jsec x + tan xj + C;

that is,
Z
1 1
tan2 x sec x = tan x sec x ln jsec x + tan xj + C:
2 2

R cos( ) cos( )
Exercise 7.3.9. sin 3x sin 5xdx = ? (Hint : sin sin = 2
)
Sol.
Z Z
cos 2x cos 8x sin 2x sin 8x
sin 3x sin 5xdx = dx = + C:
2 4 16

7.4. Exercises 7.4.

R p 9 x2
Exercise 7.4.1. x
dx = ?
 7. CHAPTER 7 63

Sol.
Let x = 3 sin u; dx = 3 cos udu; 2
u 2
: Then we have
Z p Z p
9 x2 9 9 sin2 u
dx = 3 cos udu
x 3 sin u
Z
3 cos2 u
= du
sin u
Z
3 3 sin2 u
= du
sin u
Z Z
= 3 csc udu 3 sin udu
= 3 ln jcsc u + cot uj + 3 cos u + C:
Then since r
x p x2
sin u = ; cos u = 1 sin2 u = 1 ;
3 9
r p
1 3 cos u 3 x2 9 x2
csc u = = ; cot u = = 1 = ;
sin u x sin u x 9 x
we have
Z p p r
9 x2 3 9 x2 x2
dx = 3 ln + +3 1 + C:
x x x 9

Rp
Exercise 7.4.2. 1 9x2 dx = ?
Sol.
Let x = 31 sin u; dx = 31 cos udu; 2 u 2 : Then we have
Z p Z p
1
2
1 9x dx = 1 sin2 u cos udu
3
Z
cos2 u
= du
3
Z
1 + cos 2u
= du
6
u sin 2u
= + + C:
6 12
Then since p p
sin u = 3x; cos u = 1 sin2 u = 1 9x2 ;
p
sin 2u = 2 sin u cos u = 6x 1 9x2 ;
64 CONTENTS

we have Z p p
1
sin (3x) x 1 9x2
1 9x2 dx = + + C:
6 2
R 3
Exercise 7.4.3. x2 (4 4x2 ) 2
dx = ?
Sol.
Let x = sin u; dx = cos udu; 2 u 2 : Then we have
Z Z
3 3
2 2
x 4 4x 2
dx = sin2 u(4 4 sin2 u) 2 cos udu
Z
sin2 u cos u
= du
8 cos3 u
Z
1
= tan2 udu
8
Z
1 1
= ( sec2 u )du
8 8
1 u
= tan u + C:
8 8
Then since p p
sin u = x; cos u = 1 sin2 u = 1 x2 ;
sin u x
tan u = =p ;
cos u 1 x2
we have Z
3 x sin 1 x
x2 4 4x2 2 dx = p + C:
8 1 x2 8
R p
Exercise 7.4.4. x 1 + x2 dx = ?
Sol.
Let x = tan u; dx = sec2 udu; 2 < u < 2 : Then we have
Z p Z p
2
x 1 + x dx = tan u 1 + tan2 u sec2 udu
Z
= tan u sec3 udu
Z
= sec2 ud(sec u)
1
= sec3 u + C:
3
Then since p p
tan u = x; sec u = 1 + tan2 u = 1 + x2 ;
 7. CHAPTER 7 65

we have Z p 1 3
x 1 + x2 dx = (1 + x2 ) 2 + C:
3

R
Exercise 7.4.5. p 1 dx =?
x 25x2 +49

Sol.
Let x = 57 tan u; dx = 75 sec2 udu; 2 < u < 2 : Then we have
Z Z
1 1 7
p dx = 7
p sec2 udu
2
x 25x + 49 tan u 49 tan2
u + 49 5
Z 5
sec u
= du
7 tan u
Z
1
= csc udu
7
1
= ln jcsc u + cot uj + C:
7
Then since
r
5 p 25x2
tan u = x; sec u = 1 + tan2 u = 1 + ;
7 49
q
2 p
1 7 sec u 1 + 25x
49 49 + 25x2
cot u = = ; csc u = = 5 = ;
tan u 5x tan u 7
x 5x
we have
Z p
1 1 49 + 25x2 7
p dx = ln + + C:
x 25x2 + 49 7 5x 5x

R
Exercise 7.4.6. p x dx =?
x2 1

Sol.
Let u = x2 1; du = 2xdx: Then we have
Z Z p
x 1 p
p dx = p du = u + C = x2 1 + C:
x2 1 2 u

R 1
Exercise 7.4.7. p
4x2 25
dx =?

Sol.
66 CONTENTS

Let x = 52 sec u; dx = 52 tan u sec udu; 0 u < 2 if x > 0; 2 < u if x < 0:

Then we have
Z Z
1 1 5
p dx = p tan u sec udu
4x2 25 25 sec2 u 25 2
Z
sec u
= du
2
1
= ln jsec u + tan uj + C:
2
Then since r
2 p 4x2
sec u = x; tan u = sec2 u 1 = 1;
5 25
we have r
Z
1 1 2 4x2
p dx = ln x + 1 + C:
4x2 25 2 5 25

R 3=p2 p
4x2 9
Exercise 7.4.8. 3=2 x2
dx =?
Sol.
Let x = 23 sec u; dx = 32 tan u sec udu; 0 u 4 since 32 x p32 : Then we have
Z 3=p2 p 2 Z p
4x 9 4 9 sec2 u 9 3
dx = 9 tan u sec udu
3=2 x2 0 4
sec 2u 2
Z
4 2 tan2 u
= du
0 sec u
Z
4 2 sec2 u 2
= du
0 sec u
Z Z
4 4
= 2 sec udu 2 cos udu
0 0

= [2 ln jsec u + tan uj 2 sin u]04

p p
= 2 ln 2 + 1 2:

R 1
Exercise 7.4.9. sec xdx = ?
Sol.
Let u = sec 1
x; du = jxjp1x2 1 dx; dv = dx; v = x; then we have
Z Z
1 1 x
sec xdx = x sec x p dx:
jxj x2 1
 7. CHAPTER 7 67

R R
If x > 0, then jxjpxx2 1 dx = px12 1 dx. Let x = sec u; dx = tan u sec udu; 0 u<
2
: Then we have
Z Z
1 1
p dx = p tan u sec udu
x2 1 sec2 u 1
Z
= sec udu
= ln jsec u + tan uj + C
p
= ln x + x2 1 + C:

So Z
1
p
sec xdx = x sec 1 x ln x + x2 1 + C:
R R
If x < 0, then jxjpxx2 1
dx = p 1
x2 1
dx. Let x = sec u; dx = tan u sec udu; 2
<
u : Then we have
Z Z
1 1
p dx = p tan u sec udu
x2 1 sec2 u 1
Z
= sec udu
= ln jsec u + tan uj + C
p
= ln x x2 1 + C:

So Z
1 1
p
sec xdx = x sec x ln x x2 1 + C:

7.5. Exercises 7.5.

R 1
Exercise 7.5.1. x2 5x+6
dx =?
Sol.
1
Since x2 5x + 6 = (x 3)(x 2), assume x2 5x+6 = xA 3 + xB 2 . Then A = 1 and
B = 1. So we have
Z Z
1 1 1
2
dx = ( )dx = ln jx 3j ln jx 2j + C:
x 5x + 6 x 3 x 2

R x2 +4x 33
Exercise 7.5.2. x2 +2x 3
dx =?
Sol.
68 CONTENTS

2
Since xx2+4x 33
+2x 3
= 1 + x22x+2x30 3 and x2 + 2x 3 = (x + 3)(x 1), assume 2x 30
x2 +2x 3
=
A
x+3
+ xB 1 . Then A = 9 and B = 7. So we have
Z 2 Z
x + 4x 33 9 7
2
dx = (1 + )dx
x + 2x 3 x+3 x 1
= x + 9 ln jx + 3j 7 ln jx 1j + C:

R 14
Exercise 7.5.3. x3 x
dx =?
Sol.
Since x3 x = x(x+1)(x 1),assume x314 x = Ax + x+1B
+ xC 1 . Then A = 14; B = 7
and C = 7. So we have
Z Z
14 14 7 7
dx = ( + + )dx
x3 x x x+1 x 1
= 14 ln jxj + 7 ln jx + 1j + 7 ln jx 1j + C:

R 11x+6
Exercise 7.5.4. (x 1)2
dx =?

Sol.
11x+6 A B
Assume (x 1)2
= x 1
+ (x 1)2
.
Then A = 11 and B = 17. So we have
Z Z
11x + 6 11 17
2 dx = ( + )dx
(x 1) x 1 (x 1)2
17
= 11 ln jx 1j + C:
x 1

R x2 +x 3
Exercise 7.5.5. (x 1)2 (x 2)
dx =?

Sol.
x2 +x 3 A
Assume (x 1)2 (x 2)
= x 1
+ (x B1)2 + xC 2 . Then A = 2; B = 1 and C = 3. So we
have
Z Z
x2 + x 3 2 1 3
dx = ( + + )dx
(x 1)2 (x 2) x 1 (x 1) 2
x 2
1
= 2 ln jx 1j + 3 ln jx 2j + C:
x 1

R x2 2
Exercise 7.5.6. x(x2 +2)
dx =?
 7. CHAPTER 7 69

Sol.
x2 2 A Bx+C
Assume x(x2 +2)
= x
+ x2 +2
.
Then A = 1; B = 1 and C = 0. So we have
Z Z
x2 2 1 x
2
dx = ( + 2 )dx
x (x + 2) x x +2
1
= ln jxj + ln x2 + 2 + C:
2
R 1
Exercise 7.5.7. (x2 +a2 )2
dx = ? (a > 0)
Sol.
Let x = a tan u; dx = a sec2 udu; 2 < u < 2 : Then we have
Z Z
1 1 2
2 dx = 2 2 a sec udu
2
(x + a ) 2 2
(a tan u + a ) 2
Z
cos2 u
= du
a3
Z
1 + cos 2u
= du
2a3
u sin 2u
= + + C:
2a3 4a3
Then since
r
x p x2 1 a
tan u = ; sec u = 1 + tan2 u = 1 + 2 ; cos u = =p ;
a a sec u a + x2
2

tan u x 2ax
sin u = =p ; sin 2u = 2 sin u cos u = 2 ;
sec u a2 + x 2 a + x2
we have Z
1 tan 1 ( xa ) x
2 dx = 3
+ 2 2 + C:
(x2 + a2 ) 2a 2a (a + x2 )
R x5
Exercise 7.5.8. (x2 +4)2
dx =?
Sol.
5 8x3 16x 8x3 16x
Since (x2x+4)2 = x+ (x 2 +4)2 ,assume (x2 +4)2 = Ax+B
x2 +4
+ (xCx+D
2 +4)2 . Then A = 8; B =
0; C = 16 and D = 0. So we have
Z Z
x5 8x 16x
2 2
dx = (x + 2 )dx
(x + 4) x + 4 (x + 4)2
2

1 2 8
= x 4 ln x2 + 4 2
+ C:
2 x +4
70 CONTENTS

R x4 +13x2 5x+17
Exercise 7.5.9. (x2 +1)2 (x 3)
dx =?
4 2
Sol. Assume x (x+13x 5x+17
2 +1)2 (x 3) = xA 3 + Bx+C
x2 +1
+ (xDx+E
2 +1)2 . Then A = 2; B = 1; C =
3; D = 1 and E = 2. So by exercise 7.5.7, we have
Z 4
x + 13x2 5x + 17
dx
(x2 + 1)2 (x 3)
Z
2 x 3 x 2
= ( 2 2
+ 2 2
)dx
x 3 x + 1 x + 1 (x + 1) (x + 1)2
2

1
= 2 ln jx 3j ln x2 + 1 3 tan 1 x
2
1 x
2
tan 1 x 2
+ C:
2(x + 1) x +1
R 5x+2
Exercise 7.5.10. x3 8
dx =?
Sol.
Since x3 8 = (x 2)(x2 + 2x + 4), assume 5x+2 x3 8
= xA 2 + x2Bx+C
+2x+4
. Then A =
1; B = 1; and C = 1. So we have
Z Z
5x + 2 1 x+1
dx = ( + )dx
x3 8 x 2 x2 + 2x + 4
Z Z
1 x+1
= dx + dx:
x 2 (x + 1)2 + 3
Let u = x + 1; du = dx; then by exercise 6.6.7 we have
Z Z Z
5x + 2 1 (u 1) + 1
3
dx = dx + du
x 8 x 2 u2 + 3
Z Z
u 2
= ln jx 2j 2
du + 2
du
u +3 u +3
1 2 u
= ln jx 2j ln u2 + 3 + p tan 1 ( p ) + C
2 3 3
1 2 x+1
= ln jx 2j ln (x + 1)2 + 3 + p tan 1 ( p ) + C:
2 3 3
 8. CHAPTER 8 71

8. Chapter 8
8.1. Exercises 8.1.
Exercise 8.1.1. Find the least upper bound ( sup S ) and the greatest lower bound
( inf S ) of the set S.
(a) S = [0; 1]
(b) S = fx j x4 81g
(c) S = fx j x3 8g
(d) S = fx j ln x < 1g
(e) S = fx j x2 + x + 2 0g
Sol.
(a) sup S = 1; inf S = 0:

(b) Since S = fx j x4 81g = fx j 3 x 3g = [ 3; 3]; sup S = 3; inf S =

3:

(c) Since S = fx j x3 8g = fx j x 2g = [2; 1); sup S does not exist, inf S =

2:

(d) Since S = fx j ln x < 1g = fx j 0 < x < eg = (0; e); sup S = e; inf S =

0:

(e) Since x2 + x + 2 = (x + 21 )2 + 3
4
> 0; 8 x 2 R; S = R: So sup S and inf S do
not exist.

Exercise 8.1.2. Suppose M is an upper bound of a set S of real numbers. Show

that if M 2 S, then sup S = M .
Sol.
Since M is an upper bound of S; sup S M: On the other hand, since M 2
S; M sup S: So sup S = M:

Exercise 8.1.3. Suppose S is a nonempty bounded set of real numbers and T is

a nonempty subset of S.
(a) Show that T is bounded.
(b) Show that inf S inf T sup T sup S.
Sol.
(a) Since S is bounded, let M be an upper bound of S and m be a lower bound
of S: Then 8 x 2 T S; m x M: So T is also bounded.
72 CONTENTS

(b) It is obviously that inf T sup T . Then since T S and since sup S is an
upper bound of S; sup S is also an upper bound of T: So sup T sup S: Similarly, since
inf S is a lower bound of S; inf S is also a lower bound of T: So inf S inf T: Hence
we have inf S inf T sup T sup S:

Exercise 8.1.4. Let S be a nonempty set of real numbers and T = fjxj j x 2 Sg.
Show that S is bounded if and only if T is bounded above.

Sol.
If S is bounded, then let M be an upper bound of S and m be a lower bound of
S: Then 8 x 2 S; m x M; that is, 8 jxj 2 T; jxj jmj + jM j : So jmj + jM j is
an upper bound of T; that is, T is bounded above.
On the other hand, if T is bounded above, then let K be an upper bound of T: Then
8 jxj 2 T; jxj K; that is, 8 x 2 S; K x K: Hence S is bounded.

Exercise 8.1.5. Suppose the sequences fan g1 1

n=1 and fbn gn=1 are bounded. Show
that fcn = an bn g1
n=1 is also bounded.

Sol.
Since fan g1 1 1 1
n=1 and fbn gn=1 are bounded, by exercise 8.1.4, fjan jgn=1 and fjbn jgn=1
are bounded above. Then let M1 be an upper bound of fjan jg1 n=1 and M2 be an
1
upper bound of fjbn jgn=1 : Then since 8 n 2 N; jan j M1 and jbn j M2 ; we
have jcn j = jan bn j M1 M2 : Hence fjcn jg1 n=1 is bounded above. Then by exercise
8.1.4 once again, fcn g1
n=1 is bounded.

8.2. Exercises 8.2.

Exercise 8.2.1. For each of the following sequence, …nd an upper bound and a
lower bound, and determine whether the sequence is increasing or decreasing.
23
(a) an = 10n , n 2 N.
(b) an = (1:000001)n , n 2 N.
(c) an = p4n4n2 +1 , n 2 N.
2n
(d) an = ln( n+2 ), n 2 N.
Sol.
23 23
(a) Since 8 n 2 N; an = 10n > 10n+1
= an+1 ; fan g1
n=1 is decreasing. Then since
8 n 2 N; 1023 = a1 an > 0; 1023 is an upper bound and 0 is a lower bound.
 8. CHAPTER 8 73

(b) Since 8 n 2 N; an = (1:000001)n < (1:000001)n+1 = an+1 ; fan g1

n=1 is increas-
ing. Then since 8 n 2 N; we have
an = (1:000001)n
= (1 + 0:000001)n
= C0n + C1n 0:000001 + + Cnn (0:000001)n
1 + n 0:000001;
fan g1
n=1 is not bounded above. Finally, since 8 n 2 N; an > 0; 0 is a lower bound.

(c) Since 8 n 2 N;
4n 4 4 4(n + 1)
an = p =q <q =p = an+1 ;
2
4n + 1 1 1 4(n + 1)2 + 1
4+ n2
4 + (n+1)2

fan g1
n=1 is increasing. Then since 8 n 2 N; 0 < an =
p 4n
4n2 +1
< p4n
4n2
= 4n
2n
= 2; 2 is
an upper bound and 0 is a lower bound.

(d) Since 8 n 2 N;
2n 2 2 2(n + 1)
an = ln( ) = ln( 2 ) < ln( 2 ) = ln( ) = an+1 ;
n+2 1+ n 1 + n+1 (n + 1) + 2
fan g1 2n 2n
n=1 is increasing. Then since 8 n 2 N; 0 < an = ln( n+2 ) < ln( n ) = ln 2; ln 2 is
an upper bound and 0 is a lower bound.

pn 1
Exercise 8.2.2. Let p 2 N. Show that n! n=1
is decreasing for n p.
Sol.
p
Since 8 n p; n+1
< 1; hence we have
pn pn p pn+1
> = = an+1 ;
n! n! n + 1 (n + 1)!
pn 1
that is, n! n=1
is decreasing for n p.

8.3. Exercises 8.3.

a2n +2
Exercise 8.3.1. Let a1 = 2, an+1 = 2an
, n 1.
p
(a) Show that an 2, 8 n 1.
(b) Show that fan g1
n=1 converges.
p
(c) Show that lim an = 2.
n!1
74 CONTENTS

p
(d) Show that the greatest lower bound of fan g1
n=1 is 2.
Sol. p
(a) a1 = 2 > 2: Then by the arithmetic-geometric mean inequality, 8 n 1; we
have
r
a2n + 2 a2n 2 an 1 an 1 p
an+1 = = + = + 2 = 2:
2an 2an 2an 2 an 2 an
p
an 2 p1 1 p1 ;
(b) Since by (a), 8 n 1; 2 2
= 2
and an 2
we have
a2n + 2 a2 2 an 1 1 1
an an+1 = an = n = p p = 0:
2an 2an 2 an 2 2
Hence fan g1 1
n=1 is nonincreasng and bounded below. Then by theorem 8.3.8, fan gn=1
converges to its greatst lower bound.

(c) Since by (b), fan g1

n=1 converges to its greatst lower bound, let n!1
lim an = L:
p 1
Then since 2 is a lower bound of fan gn=1 and L is the greatst lower bound, L
p
2 > 0: Thus we have
a2n + 2 L2 + 2
L = lim an = lim an+1 = lim = ;
n!1 n!1 n!1 2an 2L
p
that is, L = 2:

(d) This follows immediately from (c).

n+( 1)n
Exercise 8.3.2. lim n
=?
n!1

Sol.
Since
n 1 n + ( 1)n n+1
;
n n n
and since
1
n 1 1 n
lim = lim = 1;
n!1 n n!1 1
1
n+1 1+ n
lim = lim = 1;
n!1 n n!1 1
n+( 1)n
by the pinching theorem, lim n
= 1:
n!1

1
Exercise 8.3.3. lim 2 =?
n!1 n
 8. CHAPTER 8 75

Sol.
1 1
Since lim = 0; by the product rule, lim 2 = 0:
n!1 n n!1 n

p
Exercise 8.3.4. lim ( 1)n n=?
n!1

Sol. p p
Assume lim ( 1)n n converges, that is, lim ( 1)n n = L: Then by remark
n!1 p p n!1 p 1
8.3.11, jLj = lim j( 1)n nj = lim n: However, since f ngn=1 is not bounded
n!1 p n!1
above, by theorem 8.3.8, lim n diverges. This leads to a contradiction.
n!1

Exercise 8.3.5. lim sin( 2n ) = ?

n!1

Sol.
Since lim = 0; by theorem 8.3.14,
n!1 2n

lim sin( ) = sin( lim ) = sin 0 = 0:

n!1 2n n!1 2n

2n
Exercise 8.3.6. lim ln( n+2 )=?
n!1

Sol.
Since
2n 2
lim = lim 2 = 2;
n!1 n +2 n!1 1+ n
by theorem 8.3.14,
2n 2n
lim ln( ) = ln( lim ) = ln 2:
n!1 n+2 n!1 n + 2

3n
Exercise 8.3.7. lim =?
n!1 n!

Sol.
3
Since 8 n 4; n
1;
3n 3 3 3 3 3 3 3 27
0< = = ;
n! 1 2 3 n 1 2 n 2n
27 3n
and since lim = 0; by the pinching theorem, lim = 0:
n!1 2n n!1 n!

1
Exercise 8.3.8. lim 2 n = ?
n!1
76 CONTENTS

Sol.
1 1
Since 2 n = e n ln 2 ; by theorem 8.3.14,
1
1 1 lim ln 2
lim 2 n = lim e n ln 2 = en!1 n = e0 = 1:
n!1 n!1

Exercise 8.3.9. Let fan g1

n=1 be a sequence of real numbers, bn = a2n 1 , n 1,
and cn = a2n , n 1. Show that lim an = L if and only if lim bn = lim cn = L.
n!1 n!1 n!1

Sol.
If lim an = L; then 8 > 0; 9 n0 2 N such that if n n0 ; then jan Lj < : So
n!1
if n n0 ; then 2n 1 n0 ; hence

jbn Lj = ja2n 1 Lj < ;

and if n n0 ; then 2n n0 ; hence
jcn Lj = ja2n Lj < :
Therefore, lim bn = lim cn = L.
n!1 n!1
On the other hand, if lim bn = lim cn = L; then 8 > 0; 9 n1 2 N such that if
n!1 n!1
n n1 ; then jbn Lj < ; and 9 n2 2 N such that if n n2 ; then jcn Lj < : So if
n 2n1 + 2n2 + 1 and n is odd, then
n+1 2n1 + 2n2 + 2
n1
2 2
and
jan Lj = b n+1 L < ;
2

and if n 2n1 + 2n2 + 1 and n is even, then

n 2n1 + 2n2 + 1
n2
2 2
and
jan Lj = c n2 L < :
Hence if n 2n1 + 2n2 + 1; jan Lj < : So lim an = L.
n!1

8.4. Exercises 8.4.

Exercise 8.4.1. lim sin 2x
sin 3x
=?
x!0

Sol.
sin 2x 00 2 cos 2x 2
lim = lim = :
x!0 sin 3x x!0 3 cos 3x 3

Exercise 8.4.2. lim 1 cos x

x2
=?
x!0
 8. CHAPTER 8 77

Sol.
1 cos x 00 sin x 1
lim 2
= lim = :
x!0 x x!0 2x 2

Exercise 8.4.3. lim tan x x

x sin x
=?
x!0

Sol.
tan x x 00 sec2 x 1 00 2 tan x sec2 x
lim = lim = lim = lim 2 sec3 x = 2:
x!0 x sin x x!0 1 cos x x!0 sin x x!0

Exercise 8.4.4. lim 33tan 4x

sin 4x
12 tan x
12 sin x
=?
x!0

Sol.
3 tan 4x 12 tan x 00 12 sec2 4x 12 sec2 x
lim = lim
x!0 3 sin 4x 12 sin x x!0 12 cos 4x 12 cos x
2 2
0
0 96 tan 4x sec 4x 24 tan x sec x
= lim
x!0 48 sin 4x + 12 sin x
0
0 384 sec 4x + 768 tan2 4x sec2 4x 24 sec4 x 48 tan2 x sec2 x
4
= lim
x!0 192 cos 4x + 12 cos x
384 24
= = 2:
192 + 12

x
Exercise 8.4.5. lim e x2 1 = ?
x!0

Sol.
ex 1 00 ex
lim = lim diverges.
x!0 x2 x!0 2x

p
3
tan x 1
Exercise 8.4.6. lim 2 sin 2x 1
=?
x! 4

Sol.
p 1 2 1 2
3
tan x 1 00 3
tan 3 x sec2 x 13 2 1
lim 2 = lim = 3 1 1 = :
x! 4 2 sin x 1 x! 4 4 sin x cos x 4 2p p
2
3

x +1) 2(ex 1)
Exercise 8.4.7. lim x(e x3
=?
x!0
78 CONTENTS

Sol.
x (ex + 1) 2 (ex 1) 0
0
ex + xex + 1 2ex
lim = lim
x!0 x3 x!0 3x2
0
0 e + e + xex
x x
2ex
= lim
x!0 6x
ex 1
= lim = :
x!0 6 6

1 cos(x2 )
Exercise 8.4.8. lim 2 2 =?
x!0 x sin(x )

Sol.
1 cos (x2 ) 00 2x sin (x2 )
lim = lim
x!0 x2 sin (x2 ) x!0 2x sin (x2 ) + 2x3 cos (x2 )

sin (x2 ) 0
0 2x cos (x2 )
= lim = lim
x!0 sin (x2 ) + x2 cos (x2 ) x!0 2x cos (x2 ) + 2x cos (x2 ) 2x3 sin (x2 )
cos (x2 ) 1
= lim 2 2 2
= :
x!0 2 cos (x ) x sin (x ) 2

1 (2x) 1
Exercise 8.4.9. lim sin x3
2 sin x
=?
x!0

Sol.
1 1 p 2 p 2
sin (2x) 2 sin x 0
0 1 4x2 1 x2
lim = lim
x!0 x3 x!0 3x2
8x 2x
0 3 3
0 (1 4x2 ) 2 (1 x2 ) 2
= lim
x!0 6x
4 1
= lim 3 3 = 1:
x!0 3(1 4x2 ) 2 3(1 x2 ) 2

x x
Exercise 8.4.10. lim 3 x22 = ?
x!0

Sol.
3x 2x 0
0 ln 3 3x ln 2 2x
lim = lim diverges.
x!0 x2 x!0 2x

8.5. Exercises 8.5.

2x3 x2 +3x+1
Exercise 8.5.1. lim 3 2 =?
x!1 3x +2x x 1
 8. CHAPTER 8 79

Sol.
2x3 x2 + 3x + 1 1
1
6x2 2x + 3
lim = lim
x!1 3x3 + 2x2 x 1 x!1 9x2 + 4x 1
1
1 12x 2 11
12 2
= lim = lim = :
x!1 18x + 4 x!1 18 3

ln x
Exercise 8.5.2. lim 0:1 =?
x!1 x

Sol.
1
ln x 1
1 x 1
lim = lim = lim = 0:
x!1 x0:1 x!1 0:1 x 0:9 x!1 0:1 x0:1

x100
Exercise 8.5.3. lim x =?
x!1 e

Sol.
x100 1
1 100x99 1
1 100 99x98 1
1 1
1 100!
lim = lim = lim = = lim = 0:
x!1 ex x!1 ex x!1 ex x!1 ex

tan x
Exercise 8.5.4. lim lnjcos xj
=?
x! 2

Sol.
tan x 1 1 sec2 x 1
lim = lim sin x = lim diverges.
x! 2 ln jcos xj x! 2 x! 2 sin x cos x
cos

Exercise 8.5.5. lim+ x (ln x)2 = ?

x!0

Sol.
01 (ln x)2 1
2 ln x 1
lim+ x (ln x)2 = lim+ 1
1
= lim+ 1
x
x!0 x!0 x!0
x x2
2
2 ln x 1
1
x
= lim+ 1 = lim+ 1 = lim+ 2x = 0:
x!0 x!0 x!0
x x2

2
Exercise 8.5.6. lim+ xx = ?
x!0

Sol.
2 2
Since xx = ex ln x and since
1
2 01 ln x 1
1 x x2
lim+ x ln x = lim+ 1 = lim+ 2 = lim+ = 0;
x!0 x!0
x2
x!0
x3
x!0 2
80 CONTENTS

we have
2 2
lim+ xx = lim+ ex ln x
= e0 = 1:
x!0 x!0

Exercise 8.5.7. lim+ (sin x)tan x = ?

x!0

Sol.
Since (sin x)tan x = etan x ln(sin x) and since
cos x
ln(sin x) 1
01
lim tan x ln(sin x) = lim+ = lim+ sin x2 = lim+
1
sin x cos x = 0;
x!0+ x!0 cot x x!0 csc x x!0
we have
lim+ (sin x)tan x = lim+ etan x ln(sin x) = e0 = 1:
x!0 x!0

p
Exercise 8.5.8. lim n
n=?
n!1

Sol. p
1 1
Since n n = n n = e n ln n and since
1
ln n 1
lim 1
= lim n = 0;
n!1 n n!1 1

by theorem 8.3.14,
p 1 ( lim ln n
)
lim n
n = lim e n ln n = e n!1 n
= e0 = 1:
n!1 n!1

Exercise 8.5.9. lim (1 + na )n = ?

n!1

Sol.
a
Since (1 + na )n = en ln(1+ n ) and since
1 a
a 10 ln(1 + na ) 00 a
1+ n
( n2
) a
lim n ln(1 + ) = lim 1 = lim 1 = lim a = a;
n!1 n n!1
n
n!1
n2
n!1 1 +
n

by theorem 8.3.14,
a n a a
( lim n ln(1+ n ))
lim (1 + ) = lim en ln(1+ n ) = e n!1 = ea :
n!1 n n!1

Exercise 8.5.10. Suppose lim f (x) = lim f 0 (x) = lim f 00 (x) = lim f 000 (x) = 0 and
x!0 x!0 x!0 x!0
2 000 x2 f 0 (x)
lim x ff00 (x)(x) = 2. Find lim .
x!0 x!0 f (x)
 8. CHAPTER 8 81

Sol.
x2 f 0 (x) 00 2xf 0 (x) + x2 f 00 (x) x2 f 00 (x) x2 f 00 (x)
lim = lim = lim 2x + lim = lim
x!0 f (x) x!0 f 0 (x) x!0 x!0 f 0 (x) x!0 f 0 (x)
0
0 2xf 00 (x) + x2 f 000 (x) x2 f 000 (x)
= lim = lim 2x + lim = 2:
x!0 f 00 (x) x!0 x!0 f 00 (x)

Exercise 8.5.11. Suppose f is twice di¤erentiable at the point c: Find lim f (c+h) 2f (c)+f (c h)
h2
:
h!0

Sol.
Since f is twice di¤erentiable at c; f 0 (x) exists on (c ; c+ ) for some > 0; hence
f (c + h) and f 0 (c h) exist when h approaches to 0: So we have
0

f (c + h) 2f (c) + f (c h) 0
0 f 0 (c + h) f 0 (c h)
lim = lim :
h!0 h2 h!0 2h
( Note: We can’t apply L’Hospital rule twice since f 00 (c + h) and f 00 (c h) may not
exist.)
Then since
00 f 0 (c + h) f 0 (c)
f (c) = lim ;
h!0 h
let s = h; we have
f 0 (c) f 0 (c h) f 0 (c) f 0 (c + s) f 0 (c + s) f 0 (c)
lim = lim = lim = f 00 (c):
h!0 h s!0 s s!0 s
Thus we have
f 0 (c + h) f 0 (c h) f 0 (c + h) f 0 (c) + f 0 (c) f 0 (c h)
lim = lim
h!0 2h h!0 2h
0 0
f (c + h) f (c) f 0 (c) f 0 (c h) f 00 (c) f 00 (c)
= lim + lim = + = f 00 (c):
h!0 2h h!0 2h 2 2

8.6. Exercises 8.6.

R1
Exercise 8.6.1. 0 x12 dx = ?
Sol.
Z 1 Z 1 1
1 1 1 1
dx = lim+ dx = lim+ = lim+ ( 1 + ) diverges.
0 x2 a!0 a x2 a!0 x a a!0 a

R1 1
Exercise 8.6.2. 0 1+x2
dx =?
82 CONTENTS

Sol.
Z 1 Z b
1 1 1
b
dx = lim dx = lim tan x
0 1 + x2 b!1 0 1 + x2 b!1 0
1 1
= lim (tan b tan 0) = :
b!1 2

R5 1
Exercise 8.6.3. 0 5 x
dx =?

Sol.
Z 5 Z b b
1 1
dx = lim dx = lim ln j5 xj
0 5 x b!5 5 x
0 b!5 0
= lim ( ln j5 bj + ln 5) diverges.
b!5

R1
Exercise 8.6.4. 1
e x dx = ?

Sol.
Z 1 Z b
x
b 1
e dx = lim e x dx = lim e x
= lim ( e b
+ e 1) = :
1 b!1 1 b!1 1 b!1 e

R2 1
Exercise 8.6.5. 0
p
3
x 1
dx =?

Sol.
Z 2 Z 1 Z 2
1 1 1
p
3
dx = p
3
dx + p
3
dx
0 x 1 0 x 1 1 x 1
Z b Z 2
1 1
= lim p
3
dx + lim+ p3
dx
b!1 0 x 1 a!1 a x 1
b 2
3 2 3 2
= lim (x 1) 3 + lim+ (x 1) 3
b!1 2 0 a!1 2 a
3 2 3 3 3 2
= lim [ (b 1) 3 ] + lim+ [ (a 1) 3 ] = 0:
b!1 2 2 a!1 2 2

R5 x
Exercise 8.6.6. 0 (x2 1)2
dx =?

Sol.
 8. CHAPTER 8 83

x x
Assume (x2 1)2
= (x 1)2 (x+1)2
= xA 1 + B
(x 1)2
+ C
x+1
+ D
(x+1)2
: Then A = 0; B =
1 1
4
; C = 0; and D = 4
: Hence we have
Z 5
x
dx
0 (x
2 1)2
Z 5
1 1
= ( 2 )dx
0 4 (x 1) 4 (x + 1)2
Z b Z 5
1 1 1 1
= lim ( 2 2 )dx + lim+ ( )dx
b!1 0 4 (x 1) 4 (x + 1) a!1 a 4 (x 1)2 4 (x + 1)2
b 5
11 1 1
= lim ( + ) + lim+ ( + )
b!1 4 (x 1) 4 (x + 1) 0 a!1 4 (x 1) 4 (x + 1) a
1 1 1 1 1 1 1 1
= lim [ + ] + lim+ [ + + ]
b!1 4 (b 1) 4 (b + 1) 4 4 a!1 16 24 4 (a 1) 4 (a + 1)
diverges.

R1 3x
Exercise 8.6.7. Does 1 x3 +1
dx converge or diverge?
Sol. R1
3x 3x 3 3
Since 8 x 2 [1; 1); 0 = ; and since by example 8.6.7, dx =
R1 3
x +1 x 3 x 2
R1 1 x2
3 1 x12 dx converges, by the comparison test, 1 x33x+1 dx converges.

R1 3x
Exercise 8.6.8. Does 2 x2 1
dx converge or diverge?
Sol. R1
Since 8 x 2 [2; 1); x23x 1 3x
= x3 0; and since by example 8.6.7, 3
dx diverges,
R1 x2
3x
2 x
by the comparison test, 2 x2 1
dx diverges.
84 CONTENTS

9. Chapter 9
9.1. Exercises 9.1.

Exercise 9.1.1. Determine the following series converges or diverges, and …nd
its value if it converges.

P
1
(a) 1.
k=1
P1
6
(b) (k+2)(k+3)
.
k=1
P
1
2k+1 +( 3)k
(c) 5k+2
.
k=1
P
1
k2 2k+3
(d) 2k2 +k+1
.
k=1
P
1
(k 1)(k+1)
(e) ln k2
.
k=2
Sol.
P
n P
1
(a) Since sn = 1 = n and lim sn = lim n = 1 diverges, by de…nition, 1
k=1 n!1 n!1 k=1
diverges.

(b) Since

X
n
6 X n
1 1
sn = = 6
k=1
(k + 2) (k + 3) k=1 k+2 k+3
1 1 1 1 1 1
= 6 + + +
1+2 1+3 2+2 2+3 n+2 n+3
1 1
= 6
3 n+3

1 1
P
1
6
and lim sn = lim 6 3 n+3
= 2, by de…nition, (k+2)(k+3)
= 2 converges.
n!1 n!1 k=1

(c) First, we have

k k
2k+1 + ( 3)k 2k+1 ( 3)k 2 2 1 3
k+2
= k+2
+ k+2 = +
5 5 5 25 5 25 5
 9. CHAPTER 9 85

P
1
2 k
2
2
P
1
3 k
3
3
for all k 2 N. Then since = 5
2 = and = 5
= converge,
k=1
5 1 5
3
k=1
5 1 ( 53 ) 8

by the sum rule and constant multiple, we get

X1
2k+1 + ( 3)k 2 X
1
2
k
1 X
1
3
k
= +
k=1
5k+2 25 k=1 5 25 k=1 5
2 2 1 3 23
= + =
25 3 25 8 600
converges.

k2 2k+3 1
P
1
k2 2k+3
(d) Since lim 2 = 6= 0, 2k2 +k+1
diverges by theorem 9.1.6.
k!1 2k +k+1 2
k=1

(e) Since
X
n
(k 1) (k + 1)
sn = ln
k=2
k2
(2 1) (2 + 1) (3 1) (3 + 1) (n 1) (n + 1)
= ln + ln + + ln
2 2 3 3 n n
1 3 2 4 3 5 (n 1) (n + 1) 1 n+1
= ln = ln ;
2 2 3 3 4 4 n n 2 n
and
1 n+1 1
lim sn = lim ln = ln ;
n!1 n!1 2 n 2
P
1
(k 1)(k+1) 1
by de…nition, ln k2
= ln 2
converges.
k=2

9.2. Exercises 9.2.

Exercise 9.2.1. Determine the following series converges or diverges.
P
1
2k+3
(a) k2 +3k+2
.
k=1
P1
ln k
(b) k
.
k=1
P
1
(c) p1 .
k
k=1
P
1
1
(d) p
k k
.
k=1
P
1
k 1
(e) k3
.
k=1
86 CONTENTS

P
1
ln(k+1)
(f) (k+1)3
.
k=1
P
1
k+1
(g) k 2k
.
k=1
P1
(h) p 1 .
k=1 k(k+1)
P1 p k
k
(i) k
.
k=1
P1
3k
(j) p
3 5
k +1
.
k=1
Sol.
(a) Suppose f (x) = x22x+3
+3x+2
, then f (x) is continuous, positive and decreasing on
[1; 1). Since
Z 1 Z b
2x + 3 b
f (x)dx = lim 2
dx = lim ln x2 + 3x + 2 1
1 b!1 1 x + 3x + 2 b!1

= lim ln b2 + 3b + 2 ln j6j = 1;
b!1

R1 P
1 P
1
2k+3
1
f (x)dx diverges. Hence f (k) = k2 +3k+2
diverges by the integral test
k=1 k=1
(Theorem 9.2.3).

(b) Suppose f (x) = lnxx , f 0 (x) = 1 xln2 x < 0, 8 x 3, then f (x) is continuous,
positive and decreasing on [3; 1). Since
Z 1 Z b
ln x
f (x)dx = lim dx = lim ln jln xj jb3
3 b!1 3 x b!1

= lim (ln jln bj ln jln 3j) = 1;

b!1

R1 P
1 P
1
ln k
3
f (x)dx diverges. Hence by the integral test (Theorem 9.2.3), f (k) = k
k=3 k=3
diverges. Therefore
X
1
ln k ln 1 ln 2 X ln k ln 2 X ln k
1 1
= + + = +
k=1
k 1 2 k=3
k 2 k=3
k

also diverges.
P
1 P
1 P
1
(c) Since p1 = 1
1 , where p = 1
< 1, p1 diverges by p-series (Example
k k2 2 k
k=1 k=1 k=1
9.2.5).
 9. CHAPTER 9 87

P
1
1
P
1
1 3
P
1
1
(d) Since k k
p = 3 , where p = 2
> 1, p
k k
converges by p-series (Ex-
k=1 k=1 k2 k=1
ample 9.2.5).

(e) First, we have

k 2 (k 1) = k 3 k2 k3; 8 k 1;
implies
k 1 1
0 ; 8 k 1:
k2k3
P
1
1
P
1
k 1
Then since k2
converges (p-series, p = 2), k3
converges by the basic compar-
k=1 k=1
ison test (Theorem 9.2.8).

(f) First, we have

ln (k + 1) k + 1; 8 k 1;
implies
ln (k + 1) (k + 1) 1
0< 3 3 = ; 8 k 1:
(k + 1) (k + 1) (k + 1)2
P
1
1
P
1
1
P
1
ln(k+1)
Then since (k+1)2
= n2
converges, (k+1)3
converges by the basic comparison
k=1 n=2 k=1
test (Theorem 9.2.8).

(g) First, we have

k+1 k + k = 2k; 8 k 1;
implies
k+1 2k 1
0< = ; 8 k 1:
k 2k k 2k 2k 1
P
1
1
P
1
k+1
THen since 2k 1 converges (geometric series), k 2k
converges by the basic com-
k=1 k=1
parison test (Theorem 9.2.8).
P
1
1
(h) Since k
diverges (p-series, p = 1) and since
k=1
1
p r
k k (k + 1) k2 + k
lim = lim = lim = 1;
k!1 p 1 k!1 k k!1 k2
k(k+1)

P
1
1
by the limit comparison test (Theorem 9.2.11), p also diverges.
k=1 k(k+1)
88 CONTENTS

P
1
1
(i) Since k
diverges (p-series, p = 1) and since
k=1
p
k
k p
k
k
lim 1 = lim k = 1;
k!1 k!1
k
P
1 p
k
k
by the limit comparison test (Theorem 9.2.11), k
also diverges.
k=1
P
1
1
(j) Since 2 diverges (p-series, p = 32 ) and since
k=1 k 3

p 3k 5
3 5
k +1 3k 3 3
lim 1 = lim p = lim q = 3;
k!1 k!1 3 k 5 + 1 k!1 3 1
k3
2
1+ k5

P
1
3k
by the limit comparison test (Theorem 9.2.11), p
3 5
k +1
also diverges.
k=1

9.3. Exercises 9.3.

Exercise 9.3.1. Determine the following series converges or diverges.
P
1
1
(a) kk
.
k=1
P1
1
(b) (ln k)k
.
k=1
P
1
k
(c) 2k
.
k=1
P
1
k100
(d) ek
.
k=1
P
1 p p k
(e) k+1 k .
k=1
P1
1
(f) k!
.
k=1
P
1
3k
(g) k!
.
k=1
P1
k4
(h) k!
.
k=1
P
1
1 3 (2k 1)
(i) (2k)!
.
k=1
P1
3k k!
(j) kk
.
k=1
Sol.
 9. CHAPTER 9 89

q P
1
k 1 1 1
(a) Since lim kk
= lim = 0 < 1, by the root test (Theorem 9.3.1), kk
k!1 k!1 k k=1
converges.
q
1 1
(b) Since lim k
(ln k)k
= lim = 0 < 1, by the root test (Theorem 9.3.1),
k!1 k!1 ln k
P
1
1
converges.
(ln k)k
k=1
q p
k P
1
(c) Since lim k 2kk = lim k
= 1
< 1, by the root test (Theorem 9.3.1), k
2k
k!1 k!1 2 2
k=1
converges.
q 100
k k100 k k 1
(d) Since lim ek
= lim e
= e
< 1; by the root test (Theorem 9.3.1),
k!1 k!1
P
1
k100
ek
converges.
k=1

(e) Since

r
k
p p k p p
lim k+1 k = lim k+1 k
k!1 k!1
p p p p
k+1
k+1+ k k
= lim p p
k!1 k+1+ k
(k + 1) k 1
= lim p p = lim p p = 0 < 1;
k!1 k + 1 + k k!1 k + 1 + k

P
1 p p k
by the root test (Theorem 9.3.1), k+1 k converges.
k=1
1
(k+1)! 1
P
1
1
(f) Since lim 1 = lim = 0 < 1, by the ratio test (Theorem 9.3.4),
k!1 k! k!1 k+1 k=1
k!

converges.
3k+1 P
1
(k+1)! 3 3k
(g) Since lim 3k
= lim = 0 < 1, by the ratio test (Theorem 9.3.4),
k!1 k! k!1 k+1 k=1
k!

converges.
(k+1)4
(k+1)! (k+1)4
(h) Since lim k4
= lim k 4 (k+1) = 0 < 1. by the ratio test (Theorem 9.3.4),
k!1 k! k!1
P
1
k4
k!
converges.
k=1
90 CONTENTS

(i) Since
13 (2k 1) (2k+1)
(2(k+1))! 2k + 1
lim 1 3 (2k 1)
= lim
k!1 k!1 (2k + 1) (2k + 2)
(2k)!
1
= lim = 0 < 1;
k!1 2k + 2
P
1
13 (2k 1)
by the ratio test (Theorem 9.3.4), (2k)!
converges.
k=1

(j) First, we have

3k+1 (k+1)! k
(k+1)k+1 3 (k + 1) k k 3 kk k
lim = lim = lim = lim 3 ;
k!1 3k k!
kk
k!1 (k + 1)k+1 k!1 (k + 1)k k!1 k+1
x x x
Then let y = x+1 , ln y = x ln x+1
. By L’Hopital rule, we have
x x+1 1(x+1) 1 x
ln x+1 x (x+1)2 x
lim ln y = lim 1 = lim 1 = lim = 1:
x!1 x!1 x!1 x!1 x + 1
x x2

Therefore, by theorem 8.3.14

x
x lim ln y 1 1
lim = lim y = ex!1 =e = :
x!1 x+1 x!1 e
k k 3
P
1
3k k!
Then by the ratio test (Theorem 9.3.4), lim 3 k+1
= e
> 1 implies kk
k!1 k=1
diverges.

9.4. Exercises 9.4.

Exercise 9.4.1. Do the following series converge absolutely, converge condition-
ally or diverge?
P
1
sin n
(a) n3
.
n=1
P1
1+cos n
(b) n!
.
n=1
P1
( 1)n ln n
(c) n
.
n=1
P
1
( 1)n nn
(d) n!
.
n=1
P1
( 1)n
(e) n ln n
.
n=2
 9. CHAPTER 9 91

P
1
( 1)n
(f) n(ln n)2
:
n=2
Sol.
sin n 1
(a) First, since 0 jsin nj 1; 8 n 2 N; we have 0 n3 n3
8 n 2 N. THen
P1
1
since n3
converges (p-series, p = 2), by the basic comparison test (Theorem 9.2.8),
n=1
P
1
sin n
P
1
sin n
n3
converges. Hence n3
converges absolutely.
n=1 n=1

(b) First, since

0 j1 + cos n j j1j + jcos n j 1 + 1 = 2; 8 n 2 N;

2
1+cos n 2 1
0 n!
8 n 2 N. Then since lim (n+1)!
n!
; 2 = lim n+1 = 0 < 1, by
n!1 n! n!1
P 2
1
the ratio test (Theorem 9.3.4), n!
converges. As a result, by the basic compari-
n=1
P
1
1+cos n
P
1
1+cos n
son test (Theorem 9.2.8), n!
also converges. Hence n!
converges
n=1 n=1
absolutely.

( 1)n ln n ln n
P
1
ln n
(c) First, since n
= n
; 8n 3; and since by exercise 9.2.1.(a), n
n=3
P
1
( 1)n ln n
diverges, n
also diverges.
n=1
1
Then since lnnn n=3 is positive, decreasing and lim lnnn = 0, by the alternating
n!1
P1
( 1)n ln n P1
( 1)n ln n
test (Theorem 9.4.5), n
converges, that is, n
converges. Hence
n=3 n=1
P1
( 1)n ln n
n
converges conditionally.
n=1

(d) First, since

n! 1 2 3 n 1 2 n 1 1
0 n
= = 1 1=
n n n n n n n n n n
1 n!
and lim 0 = lim = 0, by the pinching theorem, lim n = 0. Then since
n!1 n!1 n n!1 n

nn 1
lim= lim n! = 1
n!1 n! n!1 n
n

nn
P
1
( 1)n nn
does not exist, lim 6= 0. Thus by theorem 9.1.6, diverges.
n!1 n! n=1
n!
92 CONTENTS

(e) First, let f (x) = x ln1 x , then f (x) is continuous, positive and decreasing on
[3; 1). Let u = ln x, du = x1 dx, then we have
Z 1 Z b Z ln b
1 1
f (x)dx = lim dx = lim du
3 b!1 3 x ln x b!1 ln 3 u

= lim ln juj jln b

ln 3 = lim (ln jln bj ln jln 3j) = 1;
b!1 b!1

P
1 P
1
1
diverges. Hence f (n) = n ln n
diverges by the integral test (Theorem 9.2.3).
n=3 n=3
P
1
1
Thus n ln n
also diverges.
n=2
1 1
Then since n ln n n=3
positive, decreasing and lim n ln1 n = 0, by the alternating
n!1
P1
( 1)n P
1
( 1)n P
1
( 1)n
test (Theorem 9.4.5), n ln n
converges, that is, n ln n
converges. Hence n ln n
n=3 n=2 n=2
converges conditionally.

(f) First, let f (x) = x(ln1x)2 , then f (x) is continuous, positive and decreasing on
[3; 1). Let u = ln x, du = x1 dx, then we have
Z 1 Z b Z ln b
1 1
f (x)dx = lim 2 dx = lim 2
du
3 b!1 3 x (ln x) b!1 ln 3 u
!
ln b
1 1 1 1
= lim = lim + = ;
b!1 u ln 3 b!1 ln b ln 3 ln 3

P
1 P
1
1
converges. Hence f (n) = n(ln n)2
converges by the integral test (Theorem
n=3 n=3
P
1
( 1)n 1
P1
1
P1
( 1)n
9.2.3). Thus n(ln n)2
= 2 ln 2
+ n(ln n)2 also converges. Hence n ln n
converges
n=2 n=3 n=2
absolutely.

P
1 P
1
Exercise 9.4.2. Show that if an converges absolutely, then a2n converges.
n=1 n=1

Sol. 1
P P
1
Since an converges absolutely, that is, jan j converges, by Theorem 9.1.6,
n=1 n=1
lim jan j = 0. Then for = 1 > 0; 9 N 2 N such that if n N then jjan j 0j =
n!1
jan j < 1. So we have

0 a2n = jan j jan j < 1 jan j = jan j ; 8n N:

 9. CHAPTER 9 93

P
1 P
1
Then since jan j converges, by the basic comparison test (Theorem 9.2.8), a2n
n=N n=N
P
1
converges. Thus a2n also converges.
n=1

P
1 P
1
Exercise 9.4.3. Give an example that an converges, but a2n diverges.
n=1 n=1

Sol. n o1
( 1)n 1
Consider an = p , then a2n = . Since p1
is positive, decreasing and
n n n
n=1
P
1 P
1
( 1)n
lim p1 = 0, by the alternating test (Theorem 9.4.5), an = p converges.
n!1 n n
n=1 n=1
P
1 P
1
1
However, a2n = n
diverges (p-series, p = 2).
n=1 n=1

9.5. Exercises 9.5.

Exercise 9.5.1. Show that the Taylor series of the following functions at 0 con-
verge to f (x) for all x 2 R.
(a) f (x) = sin(2x):
(b) f (x) = e x :
(c) f (x) = cos x:
Sol.
(a) Since
f 0 (x) = 2 cos(2x); f 00 (x) = 4 sin(2x); f (3) (x) = 8 cos(2x); f (4) (x) = 16 sin(2x);
..
.
f (4m+1) (x) = 24m+1 cos(2x); f (4m+2) (x) = 24m+2 sin(2x);
f (4m+3) (x) = 24m+3 cos(2x); f (4m+4) (x) = 24m+4 sin(2x);
we have
2 0 23 3 0 4 f (n) (0) n
x + x2
f (x) = 0 + x x + + x + Rn (x):
1! 2! 3! 4! n!
Then by Lagrange’s estimate, since f (n+1) (t) 2n+1 ,
jxjn+1 2n+1 jxjn+1
jRn (x)j max f (n+1) (t) :
t2J (n + 1)! (n + 1)!
Fixed x; let k 2 N such that k + 1 > 2 jxj : Then since for n k;
2n+1 jxjn+1 2k jxjk 2 jxj 2 jxj 2k jxjk 2 jxj
0 jRn (x)j = ;
(n + 1)! k! k+1 n+1 k! n+1
94 CONTENTS

and since
2k jxjk 2 jxj
lim ( ) = 0;
n!1 k! n+1
by the pinching theorem, lim jRn (x)j = 0; that is, lim Rn (x) = 0: Thus we have
n!1 n!1

X
1
( 1)n 22n+1 2n+1
sin(2x) = x :
n=0
(2n + 1)!

(b) Since
f 0 (x) = e x ; f 00 (x) = e x ; ; f (n) (x) = ( 1)n e x ;
we have
1 1 1 3 ( 1)n n
f (x) = 1 x + x2 x + + x + Rn (x):
1! 2! 3! n!
Then since if x > 0; then 8 t 2 [0; x]; f (n+1) (t) = e t 1; and if x < 0; then
8 t 2 [x; 0]; f (n+1) (t) = e t e x ; by Lagrange’s estimate,
( jxjn+1
n+1
jxj (n+1)!
; x > 0;
0 jRn (x)j max f (n+1) (t) e x jxjn+1
t2J (n + 1)! ; x < 0: (n+1)!

Fixed x; let k 2 N such that k + 1 > jxj : Then since for n k;

jxjn+1 jxjk jxj jxj jxjk jxj
= ;
(n + 1)! k! k + 1 n+1 k! n + 1
k
and since lim ( jxjk! jxj
n+1
) = 0; by the pinching theorem,
n!1

jxjn+1 e x jxjn+1
lim = 0 = lim :
n!1 (n + 1)! n!1 (n + 1)!

So lim jRn (x)j = 0; that is, lim Rn (x) = 0: Thus we have

n!1 n!1

X1
( 1)n n
x
e = x :
n=0
n!

(c) Since
f 0 (x) = sin x; f 00 (x) = cos x; f (3) (x) = sin x; f (4) (x) = cos x;
..
.

f (4m+1) (x) = sin x; f (4m+2) (x) = cos x;

f (4m+3) (x) = sin x; f (4m+4) (x) = cos x;
 9. CHAPTER 9 95

we have
0 1 2 0 3 1 4 f (n) (0) n
f (x) = 1 x x + x + x + + x + Rn (x):
1! 2! 3! 4! n!
Then by Lagrange’s estimate, since f (n+1) (t) 1,

(n+1) jxjn+1 jxjn+1

0 jRn (x)j max f (t) :
t2J (n + 1)! (n + 1)!
jxjn+1
Then since lim = 0; by the pinching theorem, lim jRn (x)j = 0; that is,
n!1 (n+1)! n!1
lim Rn (x) = 0: Thus we have
n!1
X1
( 1)n 2n
cos x = x :
n=0
(2n)!

Exercise 9.5.2. Show that the Taylor series of f (x) = ln (1 + x) at 0 converges

P
1
( 1)n+1
to f (x) for x = 1. Find n
.
n=1

Sol.
Since
1 1 2
f 0 (x) = ; f 00 (x) = 2
; f (3) (x) = ;
1+x (1 + x) (1 + x)3
..
.
(n) ( 1)n 1 (n 1)!
f (x) = ;
(1 + x)n
we have
1 1 2 2! 3 3! 4 ( 1)n 1 (n 1)! n
f (x) = 0 + x x + x x + + x + Rn (x):
1! 2! 3! 4! n!
Then by Lagrange’s estimate, since for t 2 [0; 1];
n!
f (n+1) (t) = n!;
(1 + t)n+1
we have
(n+1) j1jn+1 n! 1
0 jRn (1)j max f (t) = :
t2[0;1] (n + 1)! (n + 1)! n+1
1
Then since lim n+1 = 0; by the pinching theorem, lim jRn (1)j = 0; that is, lim Rn (1) =
n!1 n!1 n!1
0: Thus we have
X1
( 1)n+1
ln 2 = f (1) = :
n=1
n
96 CONTENTS

9.6. Exercises 9.6.

Find the intervals of convergence of the following series.
P
1
1
Exercise 9.6.1. 2n2 +n 1
xn .
n=1

Sol.
First, since
1
2(n+1)2 +(n+1) 1
xn+1 2n2 + n 1
lim 1
= lim x = jxj ;
n!1
2n2 +n 1
xn n!1 2n2 + 5n + 2
by the ratio test (Theorem 9.3.4) and Theorem 9.6.3, the radius of convergence is 1.
P
1
1
Then if x = 1, since n2
converges (p-series, p = 2) and
n=1
1
2n2 +n 1 n2 1
lim 1 = lim = > 0;
n!1 n!1 2n2 + n 1 2
n2
P
1
1
by the limit comparison test (Theorem 9.2.11), 2n2 +n 1
converges. Then if x = 1,
n=1
P
1
( 1)n P
1
1
P
1
( 1)n
since 2n2 +n 1
= 2n2 +n 1
converges, 2n2 +n 1
converges absolutely, which
n=1 n=1 n=1
P
1
( 1)n P
1
1
implies that 2n2 +n 1
also converges. So the interval of convergence of 2n2 +n 1
xn
n=1 n=1
is [ 1; 1].

P
1
Exercise 9.6.2. nxn .
n=1

Sol.
First, since
j(n + 1) xn+1 j n+1
lim n
= lim x = jxj ;
n!1 jnx j n!1 n
by the ratio test (Theorem 9.3.4) and Theorem 9.6.3, the radius of convergence is 1.
P
1
Then if x = 1, since lim n 6= 0, by Theorem 9.1.6, n diverges. Then if x = 1,
n!1 n=1
n P
1
since lim n 6= 0 implies lim ( 1) n 6= 0, by Theorem 9.1.6, ( 1)n n diverges. So
n!1 n!1 n=1
P
1
n
the interval of convergence of nx is ( 1; 1).
n=1

P
1
( 1)n n
Exercise 9.6.3. n!
x .
n=1
 9. CHAPTER 9 97

Sol.
First, since
( 1)n+1 n+1
(n+1)!
x 1
lim n
= lim x = 0;
n!1 ( 1)
xn n!1 n+1
n!

by the ratio test (Theorem 9.3.4) and Theorem 9.6.3, the radius of convergence is 1.
P
1
( 1)n n
So the interval of convergence of n!
x is ( 1; 1) = R.
n=1

P
1
n
Exercise 9.6.4. ln n
xn .
n=1

Sol.
First, we have
n+1
ln(n+1)
xn+1 n+1 ln n
lim n
= lim x = jxj ;
n!1
ln n
xn n!1 n ln (n + 1)
n+1
where lim n
= 1 and
n!1
1
ln n x x+1
lim = lim 1 = lim =1
n!1 ln (n + 1) x!1 x!1 x
x+1

by L’Hopital’s rule. So by the ratio test (Theorem 9.3.4) and Theorem 9.6.3, the
radius of convergence is 1.
Then if x = 1, since by L’Hopital’s rule,
x 1
lim = lim 1 = lim x = 1;
x!1 ln x x!1 x!1
x

n
P
1
n
that is, lim = 1, by Theorem 9.1.6, diverges. Then if x = 1, since
n!1 ln n n=1
ln n
P
1
lim n
6= 0 implies lim ( 1)n n
6= 0, by Theorem 9.1.6, ( 1)n n
diverges. So
n!1 ln n n!1 ln n
n=1
ln n
P
1
n
the interval of convergence of ln n
xn is ( 1; 1).
n=1

P
1 n
3
Exercise 9.6.5. n
xn .
n=1

Sol.
First, since
3n+1 n+1
n+1
x 3n
lim 3n n
= lim x = 3 jxj ;
n!1
n
x n!1 n+1
98 CONTENTS

by the ratio test (Theorem 9.3.4) and Theorem 9.6.3, the radius of convergence is 13 .
P1 n
1 n
P1
Then if x = 31 , 3
n 3
= 1
n
diverges (p-series, p = 1). Then if x = 31 ,
n=1 n=1
P
1 n
3 1 n
P
1
( 1)n
n 3
= n
converges by alternating test (Theorem 9.4.5). So the interval
n=1 n=1
P
1 n
3 1 1
of convergence of n
xn is [ ; ).
3 3
n=1

P
1
ln n
Exercise 9.6.6. 2n
xn .
n=1

Sol.
First, since
ln(n+1) n+1
2n+1
x ln (n + 1) 1
lim ln n n
= lim x = jxj ;
n!1
2n
x n!1 2 ln n 2
by the ratio test (Theorem 9.3.4) and Theorem 9.6.3, the radius of convergence is 2.
P
1
ln n n
P
1
Then if x = 2, since lim ln n 6= 0, by Theorem 9.1.6, 2n 2 = ln n diverges.
n!1 n=1 n=1
P
1
Then if x = 2, since lim ( 1)n ln n 6= 0, by Theorem 9.1.6, ln n
2n
( 2)n =
n!1 n=1
P
1
n P
1
ln n n
( 1) ln n diverges. So the interval of convergence of 2n
x is ( 2; 2).
n=1 n=1

P
1
(3n)!
Exercise 9.6.7. (n!)2
xn .
n=1

Sol.
First, since
(3(n+1))! n+1 (3n+3)!
((n+1)!)2
x (n+1)! (n+1)!
lim = lim (3n)!
x
n!1 (3n)! n n!1
(n!)2
x n! n!

(3n + 1) (3n + 2) (3n + 3)

= lim x = 1;
n!1 (n + 1) (n + 1)
by the ratio test (Theorem 9.3.4) and Theorem 9.6.3, the radius of convergence is 0.
P1
ln n n
So 2n
x only converges at x = 0.
n=1
 10. CHAPTER 10 99

10. Chapter 10
10.1. Exercises 10.1.
Exercise 10.1.1. f (x; y) = sin2 xy cos x2 + y 2 : Compute fx and fy .
Sol.
fx = (2 sin xy y) cos x2 + sin2 xy ( sin x2 2x):
fy = (2 sin xy x) cos x2 + 2y:

Exercise 10.1.2. f (x; y) = esin xy tan 1

x2 + y 2 : Compute fx and fy .
Sol.
fx = (esin xy cos xy y) tan 1
x2 + esin xy ( 1+(x1 2 )2 2x):
fy = (esin xy cos xy x) tan 1 2
x + 2y:

Exercise 10.1.3. f (x; y) = 1 x2 y 2 : Compute fx and fy .

Sol.
fx = 2x: fy = 2y:

Ry
Exercise 10.1.4. f (x; y) = x
sin t3 dt: Compute fx and fy .
Sol.
By the fundamental theorem of calculus I, fx = sin x3 : fy = sin y 3 :

4 y2 sin z 3
Exercise 10.1.5. f (x; y; z) = ex . Compute fx , fy and fz .
Sol.
4 2 3
fx = ex y sin z (4x3 )y 2 sin z 3 :
4 2 3
fy = ex y sin z x4 (2y) sin z 3 :
4 2 3
fz = ex y sin z x4 y 2 (cos z 3 3z 2 ):

Exercise 10.1.6. f (x; y; z) = f1 (x)f2 (y)f3 (z): Describe fx , fy and fz in terms of

f1 , f2 and f3 .
Sol.
0
fx = f1 (x)f2 (y)f3 (z):
0
fy = f1 (x)f2 (y)f3 (z):
0
fz = f1 (x)f2 (y)f3 (z):
100 CONTENTS

10.2. Exercises 10.2.

(
p2xy ; (x; y) 6= (0; 0)
2 2
x +y
Exercise 10.2.1. f (x; y) = . Prove that f (x; y) is
0; (x; y) = (0; 0)
continuous at 0.
Sol. p
Since x2 + y 2 2 x2 y 2 = 2 jxyj ; we have
(x2 + y 2 ) 2 jxyj 2xy 2 jxyj (x2 + y 2 )
p p p p p :
x2 + y 2 x2 + y 2 x2 + y 2 x2 + y 2 x2 + y 2
Then since
1 (x2 + y 2 )
0 = lim [ (x2 + y 2 )] 2 = lim p
(x;y)!(0;0) (x;y)!(0;0) x2 + y 2
2xy (x2 + y 2 )
lim p lim p
(x;y)!(0;0) x2 + y 2 (x;y)!(0;0) x2 + y 2
1
= lim (x2 + y 2 ) 2 = 0:
(x;y)!(0;0)

by the pinching theorem, lim p2xy = 0 = f (0; 0): So f (x; y) is continuous at

(x;y)!(0;0) x2 +y 2
0.

x
Exercise 10.2.2. f (x; y) = p : Prove that lim f (x; y) does not exist.
x2 +y 2 (x;y)!(0;0)

Sol.
lim f (h; 0) = lim phh2 +0 = 1; but lim f (0; h) = lim p0+h
0
2 = 0:
h!0 h!0 h!0 h!0

xy 3
Exercise 10.2.3. f (x; y) = x2 +y 6
: Prove that lim f (x; y) does not exist.
(x;y)!(0;0)

Sol.
6
lim f (h; 0) = lim h20+0 = 0; but lim f (h3 ; h) = lim h6h+h6 = 12 :
h!0 h!0 h!0 h!0

3x2 +7y 2
Exercise 10.2.4. f (x; y) = x+y 2
: Prove that lim f (x; y) does not exist.
(x;y)!(0;0)

Sol.
2 2
lim f (h; 0) = lim 3hh = 0; but lim f (0; h) = lim 7h
h2
= 7:
h!0 h!0 h!0 h!0

Exercise 10.2.5. f (x; y) = ln(ex + ey ): Prove that fxx fyy 2

fxy = 0:
 10. CHAPTER 10 101

Sol.
x y
fx = exe+ey : fy = exe+ey :
x x y ) ex ex x y
fxx = e (e(e+e
x +ey )2 = (exe+ee y )2 :
ey (ex +ey ) ey ey x y
fyy = (ex +ey )2
= (exe+ee y )2 :
ex ey
fxy = (ex +ey )2
:
x ey )2 ( ex ey )2
fxx fyy fxy 2
= (e(ex +e y )4 (ex +ey )4
= 0:

1 1 1
Exercise 10.2.6. f = f (x; y; z); where f
= x
+ y
+ z1 : Compute fx ; fy and fz :
Sol.
fx
( f1 )x = f2
=) fx = f 2 ( f1 )x = f 2 1
x2
:
fy
( f1 )y = f2
=) fy = f 2 ( f1 )y = f 2 1
y2
:
fz
( f1 )z = f2
=) fz = f 2 ( f1 )z = f 2 1
z2
:

0
Exercise 10.2.7. Suppoe f1 (x) and f20 (y) are continuous with respect to x and y
respectively, and let f (x; y) = f1 (x)f2 (y): Prove that fxy = fyx :
Sol.
0 0 0
Since f1 (x) and f2 (y) are continuous, fx (x; y) = f1 (x)f2 (y) is continuous, and
0 0 0
since f1 (x) and f2 (y) are continuous, fy (x; y) = f1 (x)f2 (y) is continuous. Then since
0 0 0 0
fxy (x; y) = f1 (x)f2 (y) and fyx (x; y) = f1 (x)f2 (y) are both continuous, by Remark
10.2.7, fxy = fyx :
102 CONTENTS

11. Chapter 11
11.1. Exercises 11.1.
11.2. Exercises 11.2.
For Problem 1 to 7,
(a) Find 5f .
(b) Evaluate 5f at the given point P .
(c) Find the directional derivative of f at P in the direction of the given vector ~u.
p p p
Exercise 11.2.1. f (x; y; z) = x2 + yz, P = (1; 0; 0), ~u = ( 22 ; 22 ; 0):
Sol.
(a) 5f = ( @f ; @f ; @f ) = ( p
@x @y @z
x
; pz ; py ):
x2 +yz 2 x2 +yz 2 x2 +yz
(b) 5f (P ) = (1; 0; 0): p
(c) jj~ujj = 1; f~u 0 (P ) = 5f (P ) ~u = ( 22 ; 0; 0):

p
3
Exercise 11.2.2. f (x; y; z) = sin(x + y + z), P = ( 3 ; 3 ; 3 ), ~u = ( 12 ; 2
; 0):
Sol.
(a) 5f = ( @f ; @f ; @f ) = (cos(x + y + z); cos(x + y + z); cos(x + y + z)):
@x @y @z
(b) 5f (P ) = ( 1; 1; 1): p
(c) jj~ujj = 1; f~u 0 (P ) = 5f (P ) ~u = ( 21 ; 23 ; 0):

p
Exercise 11.2.3. f (x; y; z) = x2 + y 2 + z 2 , P = ( 31 ; 23 ; 2
3
), ~u = ( 23 ; 31 ; 23 ):
Sol.
(a) 5f = ( @f ; @f ; @f ) = ( p
@x @y @z
x
; p y
;p z
):
x2 +y 2 +z 2 x2 +y 2 +z 2 x2 +y 2 +z 2
(b) 5f (P ) = ( 31 ; 23 ; 32 ):
(c) jj~ujj = 1; f~u 0 (P ) = 5f (P ) ~u = ( 29 ; 29 ; 4
9
):

p p p
Exercise 11.2.4. f (x; y; z) = arctan(x2 +y 2 +z 2 ), P = ( 3
; 3
; 3
), ~u = (0; 1; 0):
Sol.
(a) 5f = ( @f ; @f ; @f ) = ( (x2 +y22x
@x @y @z
; 2y
; 2z
+z 2 )2 +1 (x2 +y 2 +z 2 )2 +1 (x2 +y 2 +z 2 )2 +1
):
p p p
(b) 5f (P ) = ( 92 +1 ; 9 9
2 +1 ; 2 +1 ):
p
9
(c) jj~ujj = 1; f~u 0 (P ) = 5f (P ) ~u = (0; 2 +1 ; 0):

xyz
Exercise 11.2.5. f (x; y; z) = p , P = (1; 1; 1), ~u = ( p13 ; p1 ; p1 ):
3 3
x2 +y 2 +z 2

Sol.
 11. CHAPTER 11 103

yz(y 2 +z 2 ) xz(x2 +z 2 ) xy(x2 +y 2 )

(a) 5f = ( @f ; @f ; @f ) = (
@x @y @z 3 ; 3 ; 3 ):
(x2 +y 2 +z 2 ) 2 (x2 +y 2 +z 2 ) 2 (x2 +y 2 +z 2 ) 2
(b) 5f (P ) = ( 3p2 3 ; 3p2 3 ; 3p2 3 ):
(c) jj~ujj = 1; f~u 0 (P ) = 5f (P ) ~u = ( 29 ; 2 2
; ):
9 9

2 +y 2 +z 2
Exercise 11.2.6. f (x; y; z) = ex , P = (1; 2; 3), ~u = ( 45 ; 3
5
; 0):
Sol.
2 2 2 2 2 2 2 2 2
(a) 5f = ( @f ; @f ; @f ) = (2xex +y +z ; 2yex +y +z ; 2zex +y +z ):
@x @y @z
(b) 5f (P ) = (2e14 ; 4e14 ; 6e14 ):
14 14
(c) jj~ujj = 1; f~u 0 (P ) = 5f (P ) ~u = ( 8e5 ; 12e5 ; 0):

Exercise 11.2.7. f (x; y) = ex cos y + ey sin x, P = (0; 0), ~u = ( 1; 0):

Sol.
(a) 5f = ( @f ; @f ) = (ex cos y + ey cos x; ex sin y + ey sin x):
@x @y
(b) 5f (P ) = (2; 0):
(c) jj~ujj = 1; f~u 0 (P ) = 5f (P ) ~u = ( 2; 0):

Exercise 11.2.8. Let h(x; y) = ex cos y. Find the tangent line of the curve
h(x; y) = 2 at P = (ln 2; 0):
Sol.
First, we have 5h = (ex cos y; ey sin x) ; and 5h(P ) = (2; 0):
Then since h(x; y) = 2 is a level curve of z = h(x; y); that is, h(x; y) is a constant
on the curve, if ~u is a tangent vector of the curve at P; then the direcional derivative
h~u 0 (P ) = 0: So
h~u0 (P ) = 5h(P ) ~u = (2; 0) ~u = 0;
that is, the vector 5h(P ) = (2; 0) is normal to the tangent line of h(x; y) = 2 at P:
Therefore, the tangent line is
2(x ln 2) = 0:

p
Exercise 11.2.9. Let h(x; y; z) = sin x2 + y 2 + z 2 . Find the tangent plane of
the surface h(x; y; z) = 0 at P = ( 23 ; 23 ; 3 )
Sol.
First, we have
p p p
x cos( x2 + y 2 + z 2 ) y cos( x2 + y 2 + z 2 ) z cos( x2 + y 2 + z 2 )
5h = ( p ; p ; p )
x2 + y 2 + z 2 x2 + y 2 + z 2 x2 + y 2 + z 2
2 2 1
and 5h(P ) = ( 3
; 3
; 3
):
104 CONTENTS

Then since h(x; y; z) = 0 is a level surface of w = h(x; y; z); that is, h(x; y; z) is
a constant on the surface, if ~u is a vector at P lies on the tangent plane, then the
direcional derivative h~u 0 (P ) = 0: So
2 2 1
h~u0 (P ) = 5h(P ) ~u = ( ; ; ) ~u = 0;
3 3 3
that is, the vector 5h(P ) = ( 32 ; 23 ; 13 ) is normal to the tangent plane of h(x; y; z) =
0 at P: Therefore, the tangent plane is
2 2 2 2 1
(x ) (y ) (z ) = 0:
3 3 3 3 3 3

11.3. Exercises 11.3.

2 +t2
Exercise 11.3.1. u = sin2 x + cos2 y, x = es , y = st. Find @u
@s
and @u
@t
.
Sol.
@u @u @x @u @y 2 2
@s
= @x @s
+ @y @s
= 4s sin x cos xes +t 2t cos y sin y:
@u @u @x @u @y 2 2
@s
= @x @t
+ @y @t
= 4t sin x cos xes +t 2s cos y sin y:

@u @u
Exercise 11.3.2. u = f (x)g(y), x = f1 (s)f2 (t), y = g1 (s)g2 (t). Find @s
and @t
.
Sol.
@u @u @x @u @y
@s
= @x @s
+ @y @s
= f 0 (x)g(y)f1 0 (s)f2 (t) + f (x)g 0 (y)g10 (s)g2 (t):
@u @u @x @u @y
@s
= @x @t
+ @y @t
= f 0 (x)g(y)f1 (s)f2 0 (t) + f (x)g 0 (y)g1 (s)g20 (t):

@u @u
Exercise 11.3.3. u = arctan xy, x = sin st, y = s2 + t2 . Find @s
and @t
.
Sol.
@u @u @x @u @y ty cos st
@s
= @x @s
+ @y @s
= x2 y 2 +1
+ x22sx
y 2 +1
:
@u @u @x @u @y sy cos st 2tx
@s
= @x @t
+ @y @t
= x2 y 2 +1
+ x2 y2 +1 :

dy dx
Exercise 11.3.4. Let ecos x sin y + x2 + y 2 = 0. Find dx
and dy
.
Sol.
Let u = ecos x sin y + x2 + y 2 :
@u
dy @x sin x sin yecos x sin y + 2x
= @u
= :
dx @y
cos x cos yecos x sin y + 2y
@u
dx @y cos x cos yecos x sin y + 2y
= @u
= :
dy @x
sin x sin yecos x sin y + 2x
 11. CHAPTER 11 105

x y dy dx
Exercise 11.3.5. Let p = 0. Find dx
and dy
.
x2 +y 2 +1

Sol.
x y
Let u = p :
x2 +y 2 +1
p 2x
x2 +y 2 +1 (x y) p
@u 2 x2 +y 2 +1
dy @x x2 +y 2 +1 y 2 + xy + 1
= @u
= p = :
dx @y
x2 +y 2 +1 (x y) p
2
2y
x2 +y 2 +1
x2 + xy + 1
x2 +y 2 +1
@u
dx @y x2 + xy + 1
= @u
= :
dy @x
y 2 + xy + 1
dy dx
Exercise 11.3.6. Let sin x2 tan y 2 = 0. Find dx
and dy
.
Sol.
Let u = sin x2 tan y 2 :
@u
dy cos x2 tan y 2
dx
= @u@x
= 2x 2y sin x2 sec y 2
= x
y
cot x2 sin y 2 :
@y
@u
dx y
dy
= @y
@u = x
tan x2 csc y 2 :
@x

dy dx
Exercise 11.3.7. Let f (x)g(y) = 0. Find dx
and dy
.
Sol.
Let u = f (x)g(y):
@u
dy f 0 (x)g(y)
dx
= @x
@u = f (x)g 0 (y)
:
@y
@u
dx @y f (x)g 0 (y)
dy
= @u = f 0 (x)g(y)
:
@x

11.4. Exercises 11.4 and 11.5.

For Problem 1 to 5,
(a) Find all local extreme values and saddle points of f on R2 .
(b) Find the absolute extreme values of f on the indicated domain D.
Exercise 11.4.1. f (x; y) = sin x cos y, D = R2
Sol.
(a) Since
rf (x; y) = (cos x cos y; sin x sin y);
cos x cos y = 0;
and since if and only if
sin x sin y = 0;
x = 2 + k1 ; k1 2 Z; x = k3 ; k3 2 Z;
or
y = k2 ; k2 2 Z; y = 2 + k4 ; k4 2 Z;
106 CONTENTS

the critical points are ( 2 + k1 ; k2 ); (k3 ; 2

+ k4 ); k1 ; k2 ; k3 ; k4 2 Z:
Then since
A = fxx (x; y) = sin x cos y;
B = fxy (x; y) = cos x sin y;
C = fyy (x; y) = sin x cos y;
D = AC B 2 = sin2 x cos2 y cos2 x sin2 y;
we have
A D Result
( 2 + 2n ; 2m ) 1 1 local max
( 2 + (2n + 1) ; 2m ) 1 1 local min
:
( 2 + 2n ; (2m + 1) ) 1 1 local min
( 2 + (2n + 1) ; (2m + 1) ) 1 1 local max
(k3 ; 2 + k4 ) 0 1 saddle
(b) Since 1 sin x 1; 1 cos y 1; 8 x; y 2 R;
1 f (x) = sin x cos y 1; 8 (x; y) 2 D = R2 :
So these local maximum points are also absolute maximum points, and these local
minimum points are also absolute minimum points.

Exercise 11.4.2. f (x; y) = x2 + 6xy + 8y 2 , D = [ 10; 0] [ 1; 8]

Sol.
(a) Since
rf (x; y) = (2x + 6y; 6x + 16y);
and since
2x + 6y = 0;
) x = 0; y = 0;
6x + 16y = 0;
the only critical point is (0; 0):
Then since
A = fxx (x; y) = 2; B = fxy (x; y) = 6; C = fyy (x; y) = 16;
D = AC B 2 = 4 < 0;
(0; 0) is a saddle point.
(b)
Line Equation Maximum Minimum
2
x=0 f (0; y) = 8y f (0; 8) = 512 f (0; 0) = 0
x = 10 f ( 10; y) = 100 60y + 8y f ( 10; 1) = 168 f ( 10; 15
2
4
) = 25
2
y= 1 f (x; 1) = x2 6x + 8 f ( 10; 1) = 168 f (3; 1) = 1
y=8 f (x; 8) = x2 + 48x + 512 f (0; 8) = 512 f ( 10; 8) = 132
 11. CHAPTER 11 107

So the absolute maximum point is (0; 8) and the absolute minimum point is ( 10; 15
4
):

Exercise 11.4.3. f (x; y) = arctan(x2 + y 2 ), D = [ p1 ; p1 ] [ p1 ; p1 ]

2 2 2 2

Sol.
(a) Since
2x 2y
rf (x; y) = ( ; 2 );
(x2 + y ) + 1 (x + y 2 )2 + 1
2 2

and since (
2x
(x2 +y 2 )2 +1
= 0;
2y ) x = 0; y = 0;
(x2 +y 2 )2 +1
= 0;
the only critical point is (0; 0):
Then since
2((x2 + y 2 )2 + 1) 8x2 (x2 + y 2 )
A = fxx (x; y) = ;
((x2 + y 2 )2 + 1)2
8xy(x2 + y 2 )
B = fxy (x; y) = ;
((x2 + y 2 )2 + 1)2
2((x2 + y 2 )2 + 1) 8y 2 (x2 + y 2 )
C = fyy (x; y) = ;
((x2 + y 2 )2 + 1)2
we have
A B C D result
(0; 0) 2 0 2 4 local min
(b) Since f (0; 0) = 0; and since
Line Equation Maximum Minimum
x = p2 f (x; y) = arctan(y + 2 ) f ( p12 ; p12 ) =
1 2 1
4
f( p1 ; 0)
2
= arctan( 12 )
y = p12 f (x; y) = arctan(x2 + 12 ) f ( p12 ; p12 ) = 4
f (0 p1 )
2
= arctan( 12 )

the absolute maximum point are ( p1 ; p1 ) and the absolute minimum point is
2 2
(0; 0):

p
Exercise 11.4.4. f (x; y) = j1 x2 y 2 j, D = fx2 + y 2 16g
Sol. p
(a) For the region x2 + y 2 > 1; f (x; y) = x2 + y 2 1; so
x y
rf (x; y) = ( p ;p ):
x2 + y2 1 x + y2
2 1
108 CONTENTS

Then since 8
x
< p = 0;
x2 +y 2 1
p y ) x = 0; y = 0;
: = 0;
x2 +y 2 1
2 2
but (0; 0) does not satisfy x + y > 1; there are no critical points in the region.
For the region x2 + y 2 = 1; rf does not exist 8 (x; y) in the region. Hence
all the points are critical points. Then since f (x; y) 0; 8 (x; y) 2 R2 ; and since
f (x; y) = 0; 8 (x; y) in the region, all these points are local and absolute minimum
points. p
For the region x2 + y 2 < 1; f (x; y) = 1 x2 y 2 ; so
x y
rf (x; y) = ( p ; p ):
1 x2 y 2 1 x2 y 2
Then since 8
< p x 2 2 = 0;
1 x y
p y ) x = 0; y = 0;
: = 0;
1 x2 y 2

(0; 0) is a critical point. Then since

p x2
1 x2 y2 p
1 x2 y 2
A = fxx (x; y) = ;
1 x2 y2
p xy
1 x2 y 2
B = fxy (x; y) = ;
1 x2 y 2
p y2
1 x2 y 2 p
1 x2 y 2
C = fyy (x; y) = ;
1 x2 y2
we have
A B C D result
(0; 0) 1 0 1 1 local min
p
(b) Since f (0; 0) = 1 and since f (x; y) = 15; 8 (x; y) on the circle x2 + y 2 = 16;
the absolute maximum points are the points on the circle x2 + y 2 = 16 and the
absolute minimum points are the points on the circle x2 + y 2 = 1:

Exercise 11.4.5. f (x; y) = jxj + jyj, D = [ 6; 5] [ 5; 6]

Sol.
(a) Since rf does not exist on x = 0 or y = 0; the points on the x-axis or y-axis are
critical points. Then since f (x; y) 0; 8 (x; y) 2 R2 ; and since f (x; y) = 0; 8 (x; y) on
the x-axis or y-axis, all these points are local and absolute minimum points.
Then since for xy 6= 0; @f @x
= 1 6= 0; @f
@y
= 1 6= 0; there are no other critical
points.
 11. CHAPTER 11 109

(b)
Line Equation Maximum Minimum
x = 6 f ( 6; y) = jyj + 6 f ( 6; 6) = 12 f ( 6; 0) = 6
x=5 f (5; y) = jyj + 5 f (5; 6) = 11 f (5; 0) = 5
y = 5 f (x; 5) = jxj + 5 f ( 6; 5) = 11 f (0; 5) = 5
y=6 f (x; 6) = jxj + 6 f ( 6; 6) = 12 f (0; 6) = 6
So the absolute maximum point is ( 6; 6) and the absolute minimum point is (0; 0):

Exercise 11.4.6. Under the condition x + y + z = 1 and x; y; z 0, …nd the

maximum of x2 + y 2 + z 2 and xyz.
Sol.
Write z = 1 x y; then we have
x2 + y 2 + z 2 = x2 + y 2 + (1 x y)2 ;
xyz = xy(1 x y);
and the region is bounded by the lines x = 0; y = 0; and 1 x y = 0:
Let f (x; y) = x2 + y 2 + (1 x y)2 : Then since
rf (x; y) = (2x 2(1 x y); 2y 2(1 x y));
and since
2x 2(1 x y) = 0; 1 1
)x= ; y= ;
2y 2(1 x y) = 0; 3 3
( 13 ; 13 ) is the only critical point. Then since
A = fxx (x; y) = 4; B = fxy (x; y) = 2; C = fyy (x; y) = 4;
D = AC B 2 = 12 > 0;
( 13 ; 13 ) is a local minimum point and f ( 13 ; 31 ) = 13 . Then since
Line Equation Maximum Minimum
x=0 f (x; y) = 2(y 12 )2 + 1
2
f (0; 1) = f (0; 0) = 1 f (0; 21 ) = 12
y=0 f (x; y) = 2(x 12 )2 + 1
2
f (1; 0) = f (0; 0) = 1 f ( 21 ; 0) = 12
y = 1 x f (x; y) = 2(x 12 )2 + 1
2
f (0; 1) = f (1; 0) = 1 f ( 21 ; 12 ) = 21
So the maximum of x2 + y 2 + z 2 is 1 at the points (1; 0; 0),(0; 1; 0); (0; 0; 1):
Now let g(x; y) = xy(1 x y) = xy x2 y xy 2 : Then since
rg(x; y) = (y 2xy y2; x x2 2xy);
and since
y 2xy y 2 = 0; 1 1
2 ) x = ; y = ; or x = 0; y = 0;
x x 2xy = 0; 3 3
110 CONTENTS

( 13 ; 13 ) and (0; 0) are the critical points. Then since

A = gxx (x; y) = 2y; B = gxy (x; y) = 1 2x 2y; C = gyy (x; y) = 2x;
we have
A D result
( 13 ; 13 ) 2
3
1
3
local max :
(0; 0) 0 1 saddle
Then since g( 13 ; 13 ) = 1
27
; and since

Line Equation Maximum Minimum

x=0 g(x; y) = 0 0 0
y=0 g(x; y) = 0 0 0
y = 1 x g(x; y) = 0 0 0
1
the maximum of xyz is 27
at the point ( 13 ; 31 ; 13 ):

Exercise 11.4.7. Find the volume of the largest rectangular box in the …rst octant
with 3 faces in the coordinate planes, and one vertex on the plane 3x + 2y + 4z = 9:
Sol.
It means that we need to …nd the maximum of xyz for (x; y; z) satis…es 3x + 2y +
4z = 9 and x; y; z 0. Write z = 9 3x4 2y , then we have xyz = xy( 9 3x4 2y ) and the
region is bounded by the lines x = 0; y = 0; and 9 3x 2y = 0:
Let f (x; y) = xy( 9 3x4 2y ) = 14 (9xy 3x2 y 2xy 2 ): Then since
1 1
rf (x; y) = ( (9y 6xy 2y 2 ); (9x 3x2 4xy));
4 4
and since
1
4
(9y 6xy 2y 2 ) = 0; 3
1 ) x = 1; y = ; or x = 0; y = 0;
4
(9x 3x2 4xy) = 0; 2
(1; 23 ) and (0; 0) are the critical points. Then since
3 1
A = fxx (x; y) = y; B = fxy (x; y) = (9 6x 4y);
2 4
C = fyy (x; y) = x;
we have
A D result
(1; 23 ) 1 27
16
local max :
81
(0; 0) 0 16
saddle
 11. CHAPTER 11 111

Then since f (1; 32 ) = 98 ; and since

Line Equation Maximum Minimum
x=0 f (x; y) = 0 0 0
y=0 f (x; y) = 0 0 0
9 3x 2y f (x; y) = 0 0 0
9
the maximum of xyz is 8
at the point (1; 23 ; 34 ): Thus the volume of the largest rec-
tangular box is 98 .
112 CONTENTS

12. Chapter 12
12.1. Exercises 12.1.
R3R5
Exercise 12.1.1. Compute 2 4
( xy + xy )dydx = ?
Sol.
Z 3 Z 5 Z 3 y=5
x y 1
( + )dydx = [(x ln y + y 2 ) ]dx
2 4 y x 2 2x y=4
Z 3 3
5 9 x2 5 9 9 5
= (x ln + )dx = ( ln + ln x) = ln :
2 4 2x 2 4 2 2 2 2
R4R3
Exercise 12.1.2. Compute 0 1
p xy dydx = ?
x2 +y 2

Sol. Let u = x2 + y 2 ; du = 2ydy:

Z 4Z 3 Z 4 Z x2 +9
xy x
p dydx = p dudx
0 1 x2 + y 2 0 x2 +1 2 u
Z 4 y=x2 +9 Z 4 p p
1
= [(xu 2 ) ]dx = (x x2 + 9 x x2 + 1)dx:
0 y=x2 +1 0
2
v = x + 9; dv = 2xdx
Let ;
w = x2 + 1; dw = 2xdx
Z 4 p p Z 25 Z 17
1 1 1 1
(x x2 +9 x x2 + 1)dx = v 2 dv w 2 dw
0 9 2 1 2
25 17 3
1 3 1 3 17 2
= v2 w2 = 33 :
3 9 3 1 3
R2R2p
Exercise 12.1.3. Compute 0 0
3x + 4ydydx = ?
Sol.
Z 2 Z 2 Z 2 y=2
1 1 3
(3x + 4y) dydx =2 [ (3x + 4y) 2 ]dx
0 0 0 6 y=0
Z 2
1 3 3 2 5 2 5 2
= [(3x + 8) 2 (3x) 2 ]dx = [(3x + 8) 2 (3x) 2 ]
0 6 45 0 0
2 5 5 5
= (14 2 82 6 2 ):
45
R1R1
Exercise 12.1.4. Compute 0 x
sin x2 dxdy = ?
 12. CHAPTER 12 113

Sol.
Z 1 Z 1 Z 1 Z y
2
sin x dydx = sin x2 dydx
0 x 0 0
Z 1 Z 1
2 y=x
= sin x y=0 dx = x sin x2 dx:
0 0

Let u = x2 ; du = 2xdx:
Z 1 Z 1 1
2 1 1 1 1
x sin x dx = sin udu = cos u = cos 1:
0 0 2 2 0 2 2

R1R1 2 +y 2
Exercise 12.1.5. Compute 0 0
xyex dydx = ?
Sol.
Z 1 Z 1 Z 1 y=1
x2 +y 2 x 2 2
xye dydx = [( ex +y ) ]dx
0 0 0 2 y=0
Z 1 Z 1 1 1
x x2 +1 x x2 1 2 1 x2
= e dx e dx = ex +1 e
0 2 0 2 4 0 4 0
1 2
= (e 2e + 1):
4

R R
Exercise 12.1.6. Compute 2
0
4
0
sec y cos(x + y)dydx = ?
Sol.
Z Z Z Z
2 4 2 4
sec y cos(x + y)dydx = sec y(cos x cos y sin x sin y)dydx
0 0 0 0
Z Z Z
2 4 2 y=
= (cos x sin x tan y)dydx = f[y cos x sin x( ln jcos xj)]jy=04 gdx
0 0 0
Z
2 1 1 2
= [ cos x + sin x(ln p )]dx = sin x cos x(ln p )
0 4 2 4 2 0
1
= + ln p :
4 2

RR
Exercise 12.1.7. Compute D
ydxdy for

D = f(x; y) j x2 + y 2 4; x 0; y 0g:
114 CONTENTS

Sol.
p
Z Z Z 2 Z p
4 x2 Z 2 y= 4 x2
1 2
ydxdy = ydydx = y dx
D 0 0 0 2 y=0
Z 2 2
1 2 1 3 8
= (4 x )dx = (2x x) = :
0 2 6 0 3

12.2. Exercises 12.2.

a a 2 a3
Exercise 12.2.1. Prove that the determinant b b2 b3 = abc(a b)(b c)(c
c c2 c3
a):

Sol.
a a 2 a3 1 a a2
b b2 b3 = abc 1 b b2
c c2 c3 1 c c2
1 a a2
= abc 0 b a b2 a2 = abc[(b a)(c2 a2 ) (c a)(b2 a2 )]
0 c a c 2 a2
= abc(b a)(c a)[(c + a) (b + a)] = abc(a b)(b c)(c a):

x y z
Exercise 12.2.2. u = x+y+z
;v = x+y+z
;w = x+y+z
: Compute the Jacobian de-
tenminant.

Sol.
y+z y z
1
J = x x+z z
(x + y + z)2 x y x+y
0 y z
1
= 0 x+z z = 0:
(x + y + z)2 0 y x+y

x y z
Exercise 12.2.3. u = p ;v = p ;w = p : Compute the
x2 +y 2 +z 2 x2 +y 2 +z 2 x2 +y 2 +z 2
Jacobian detenminant.
 12. CHAPTER 12 115

Sol.

y2 + z2 xy xz
1 2 2
J = 3 xy x + z yz
(x2 + y 2 + z 2 ) 2 xz yz x2 + y 2
1
= 3 [(y 2 + z 2 )(x2 + z 2 )(x2 + y 2 ) 2x2 y 2 z 2
(x2 + + y2 z2) 2
x z (x2 + z 2 )
2 2
y 2 z 2 (y 2 + z 2 ) x2 y 2 (x2 + y 2 )]
1
= 3 (0) = 0:
(x2 + y 2 + z 2 ) 2

1 1 1
Exercise 12.2.4. u = yz
;v = xz
;w = xy
: Compute the Jacobian detenminant.

Sol.
z y
0 (xz)2 (xy)2
z x
J = (yz)2
0 (xy)2
y x
(yz)2 (xz)2
0
xyz xyz 2
= + = :
x4 y 4 z 4 x4 y 4 z 4 (xyz)3

Exercise 12.2.5. u = sin(yz); v = cos(xz); w = tan(xy): Compute the Jacobian

detenminant.

Sol.

0 sin(xz) z sec2 (xy) y

J = cos(yz) z 0 sec2 (xy) x
cos(yz) y sin(xz) x 0
= 2 sin(xz) cos(yz) sec2 (xy)xyz:

12.3. Exercises 12.3.

R 3 R p9 2
Exercise 12.3.1. Compute p x xydydx = ?
3 9 x2

Sol.
116 CONTENTS

Z 3 Z p
9 x2 Z 2 Z 3

p
xydydx = (r2 sin cos )rdrd
3 9 x2 0 0
Z 2 r=3 Z 2
1 81
= [( r4 sin cos ) ]d = sin cos d
0 4 r=0 0 4
Z 2 2
81 81 2
= sin d(sin ) = sin = 0:
0 4 8 0

R 1 R px 2
Exercise 12.3.2. Compute p x (x2 + y 2 )dydx = ?
0 x x2

Sol.
Since the region is bounded by y 2 = x x2 ; 0 x 1; which is inside the circle
(x 21 )2 + y 2 = 14 ; if we let u = x 21 ; v = y; then the region is inside the circle
u2 + v 2 = 1 and
@x @y
@u @u
1 0
J= @x @y = = 1:
@v @v
0 1

Then let u = r cos ; v = r sin ; we have

Z 1 Z p
x x2 Z 1 Z p1 u2
2 2
4 4 1
p
(x + y )dydx = p1 ((u + )2 + v 2 ) 1dvdu
0 x x2 1
4 4
u2 2
Z 2 Z 1 Z 2 r= 21
2
2 1 1 1 1
= (r + r cos + )rdrd = [( r4 + r2 cos + r2 ) ]d
0 0 4 0 4 2 8 r=0
Z 2 2
3 1 3 1 3
= ( + cos )d = ( sin ) = :
0 64 8 64 8 0 32

R2Rxp
Exercise 12.3.3. Compute 0 0
x2 + y 2 dydx = ?

Sol.
Since under the polar coordinates, the line x = 2 can be written by r cos = 2; that
is, r = 2 sec ; and since the y-axis is = 0 and the line y = x is = 4 , the region
bounded the lines x = 2; y = x and the y-axis is
n o
(r cos ; r sin ) j 0 r 2 sec ; 0 :
4
 12. CHAPTER 12 117

Then by the solution of Exercise 7.3.8, we have

Z 2Z xp Z Z 2 sec
4
2 2
x + y dydx = r2 drd
0 0 0 0
Z r=2 sec Z
4 1 3 4 8
= r d = sec3 d
0 3 r=0 0 3
8 1 1 4
= ( tan x sec x + ln jsec x + tan xj)
3 2 2 0
8 1 1 p
= ( p + ln 2 + 1 ):
3 2 2

12.4. Exercises 12.4.

RRR (x2 +y2 +z2 )3=2
Exercise 12.4.1. Compute e dxdydz, where

= (x; y; z) j x2 + y 2 + z 2 1
.
Sol.
Since
= f( sin cos ; sin sin ; cos ) j 0 1; 0 2 ;0 g;
we have
Z Z p
1 z2 Z p1 y2 z2
RRR (x2 +y2 +z2 )3=2 1 3=2
e(x +y +z )
2 2 2
e dxdydz = p p dxdydz
1 1 z2 1 y2 z2
Z Z 2 Z 1 3=2
Z Z 2 Z 1
e( )
2 3
2 2
= sin d d d = e sin d d d
0 0 0 0 0 0
Z Z ! Z Z
2 1 2
1 3 1 1
= e sin d d = e sin sin d d
0 0 3 0 0 0 3 3
Z ! Z
2
1 2
= sin (e 1) d = sin (e 1) d
0 3 0 0 3
2 4
= cos (e 1) = cos (e 1) :
3 0 3

2
Exercise 12.4.2. Find the volumn of the solid
p bounded inside the sphere x +
y 2 + z 2 = 1 and bounded below by the cone z = x2 + y 2 .
118 CONTENTS

Sol.
Since under the spherical coordinates,
p the sphere x2 + y 2 + z 2 = 1 can be written
by = 1 and the cone z = x2 + y 2 can be written by cos = sin ; that is,
tan = 1; that is, = 4 ; the region is
n o
( sin cos ; sin sin ; cos ) j 0 1; 0 2 ;0
4
So the volume is
Z Z 2 Z 1 Z Z 2 1
4
2
4 1 3
sin d d d = sin d d
0 0 0 0 0 3 0
Z Z 2 Z
4 1 4 2
= sin d d = sin d
0 0 3 0 3
p
2 4 2 2
= cos = + :
3 0 3 3

R 2 R p4 2 R 2+p4 x2 y 2 1
Exercise 12.4.3. p x p (x2 + y 2 + z 2 ) 2 dzdydx = ?
2 4 x2 2 4 x2 y 2

Sol.
Since the region is = (x; y; z) j x2 + y 2 + (z 2)2 4 and under the spheri-
cal coordinates,
4 = x2 + y 2 + (z 2)2 = x2 + y 2 + z 2 4z + 4 = 2
4 cos + 4
() 2 = 4 cos () = 4 cos ;
we can consider the region by
n o
( sin cos ; sin sin ; cos ) j 0 4 cos ; 0 2 ; 0 :
2
So
Z Z p Z p
2 4 x2 2+ 4 x2 y 2 1

p p x2 + y 2 + z 2 2
dzdydx
2 4 x2 2 4 x2 y 2
Z Z 2 Z 4 cos Z Z 2 Z 4 cos
2 1 2
2 2 3
= 2
sin d d d = sin d d d
0 0 0 0 0 0
Z Z 2 =4 cos Z Z 2
2 1 4
2
= sin d d = 64 cos3 sin d d
0 0 4 =0 0 0
Z Z
2 2
= 128 cos3 sin d = 128 cos3 d(cos )
0 0
4 =2
= 32 cos =0
= 32 :
 12. CHAPTER 12 119

12.5. Exercises 12.5.

R 1 R p1 x2 R3
Exercise 12.5.1. p xdzdydx = ?
0 0 x2 +y 2

Sol.

Z 1 Z p
1 x2 Z 3 Z Z 1 Z 3
2

p xdzdydx = p r cos rdzdrd

0 0 x2 +y 2 0 0 x2 +y 2
Z Z 1 Z Z 1
2 z=3 2
2
= zr cos z=r
drd = 3r2 cos r3 cos drd
0 0 0 0
Z ! Z
1
2
3 1 4 2 1
= r cos r cos d = cos cos d
0 4 0 0 4
Z
3 2 3 3
= cos d = sin j02 = :
4 0 4 4

R 2 R p4 2 R p9 x2 y 2
Exercise 12.5.2. p x zdzdydx = ?
2 4 x2 0

Sol.

Z 2 Z p
4 x2 Z p9 x2 y 2 Z 2 Z 2 Z p
9 r2

p
zdzdydx = zrdzdrd
2 4 x2 0 0 0 0
p
Z 2 Z 2 z= 9 r2 Z 2 Z 2
1 2 9r r3
= z r drd = drd
0 0 2 z=0 0 0 2
Z ! Z
2 2 2
9 2 1 4
= r r d = 7d = 7 j20 = 14 :
0 4 8 0 0

Exercise 12.5.3. Find the volumn of the solid bounded below by z = x2 + y 2 and
2
bounded inside the ellipsoid x2 + y 2 + z4 = 3.

Sol.
Since under the cylindrical coordinates, z = x2 + y 2 can be written by z = r2 ; and
2 2
x2 + y 2 + z4 = 3 can be written by r2 + z4 = 3; and since these two graphs intersect
120 CONTENTS

z =p2;
when the volume is
r = 2;
Z 2 Z p2 Z p12 4r2 Z 2 Z p
2 p
rdzdrd = ( 12 4r2 r2 )rdrd
0 0 r2 0 0
p
Z 2 Z p
2 p Z 2 2
3 1 2 3 1 4
= r 12 4r2 r drd = (12 4r ) 2 r d
0 0 0 12 4 0
Z 2
1 3 4 1 3
= ( (12 8) 2 ) ( (12) 2 )d
0 12 4 12
Z p
2
8 24 3 p 5
= ( 1+ )d = 2 (2 3 ):
0 12 12 3