1. 8 Indefinite Integrals Case Study 8.1 Concepts of Indefinite Integrals
8.2 Indefinite Integration of Functions
8.3 Integration by Substitution
8.4 Integration by Parts
8.5 Applications of Indefinite Integrals
Chapter Summary

2. Case Study
Can you estimate the number of radioactive particles in the sample from your readings?
By measuring the level of radioactivity with a counter, it is estimated that the number of radioactive particles, y, in the sample is decreasing at a rate of 1000e–0.046t per hour, where t is expressed in hours.
I have already recorded the readings for the level of radioactivity.
According to what we learnt in Section 7.5 (Rates of Change), we have
In order to express y in terms of t, we need to find a function y of t such that its derivative is equal to –1000e–0.046t .
The process of finding a function from its derivative is called integration and will be discussed in this chapter.

3. 8.1 Concepts of Indefinite Integrals
A. Definition of Indefinite Integrals
In previous chapters, we learnt how to find the derivative of a given function.
Suppose we are given a function x2, by differentiation, we have
As 2x is the derivative of x2, we call x2 the primitive function (or antiderivative) of 2x.
Generally, for any differentiable function F(x), we have the following definition:
Definition 8.1
If , then F(x) is called a primitive function of f(x).
Although x2 is a primitive function of 2x, it is not the unique primitive function.
If we add an arbitrary constant C to x2 and differentiate it, we
have
Thus, x2 + C is also a primitive function of 2x for an arbitrary constant C.

4. 8.1 Concepts of Indefinite Integrals
A. Definition of Indefinite Integrals
In order to represent all the primitive functions of a function
f (x), we introduce the concept of indefinite integral as below:
Definition 8.2
If , then the indefinite integral of f(x), which is
denoted by , is given by
, where C is an arbitrary constant.
Note:
In the notation of , f (x) is called the integrand, and ‘ ’
is called the integral sign. The process of finding the primitive function is called integration.
2. C is called the constant of integration (or integration constant).

5. 8.1 Concepts of Indefinite Integrals
B. Basic Formulas of Indefinite Integrals
As integration is the reverse process of differentiation, the basic formulas for integrations can be derived from the differentiation formulas.
For example: Since ,
8.1 , for all real numbers n  1.
Note:
1. Formula 8.1 is also called the Power Rule for integration.
When n  0, the left hand side of the formula becomes
For convenience we usually express as

6. 8.1 Concepts of Indefinite Integrals
B. Basic Formulas of Indefinite Integrals
In addition to the Power Rule, we can use the similar way to derive the following integration formulas:
, where k is a constant.

7. 8.1 Concepts of Indefinite Integrals
C. Basic Properties of Indefinite Integrals
Theorem 8.1
If k is a non-constant, then
Proof:
By definition, C ¢, where C ¢ is an arbitrary constant.
On the other hand,
Since C ¢ and kC are arbitrary constants, the expressions kg(x)  C ¢ and kg(x)  kC represent the same family of primitive functions.

8. 8.1 Concepts of Indefinite Integrals
C. Basic Properties of Indefinite Integrals
Theorem 8.2
Proof:
Let and , where C1 and C2 are
arbitrary constants.
By definition,
On the other hand,
Since C1  C2 is an arbitrary constant, the expressions F(x)  G(x)  C and F(x)  G(x)  C1  C2 represent the same family of primitive functions.

9. Example 8.1T 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals
Example 8.1T
Find
Solution:
Use C to express the sum of all constants

10. Example 8.2T 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals
Example 8.2T
Find
Solution:
a3  b3  (a  b)(a2  ab  b2)
Cancel the common factor

11. Example 8.3T 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals
Example 8.3T
Find
Solution:

12. Example 8.4T 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals
Example 8.4T
Find
Solution:

13. Example 8.5T 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals
Example 8.5T
Find
Solution:

14. Example 8.6T 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals
Example 8.6T
Let y  ln x – ln (x  1).
Find
Hence find
Solution:
(a)
(b) By (a),

15. 8.2 Indefinite Integration of Functions
A. Integration of Functions Involving the Expression
(ax + b)
The integration formulas mentioned in Section 8.1 enable us to find the indefinite integrals of simple functions such as ex, sin x and cos x.
But how about e2x, sin 4x and cos (7x + 5)?
, where n  –1 and a  0.
Using the same method, we can obtain the following integration formulas:
Suppose a and b are real numbers with a  0.
8.12
8.13
8.14
8.15

16. Example 8.7T 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression
(ax + b)
Example 8.7T
Find
Solution:

17. Example 8.8T 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression
(ax + b)
Example 8.8T
Find
Solution:
Rationalize the denominator

18. Example 8.9T 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression
(ax + b)
Example 8.9T
Find
Solution:
Cancel the common factor

19. Example 8.10T 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression
(ax + b)
Example 8.10T
Find
Solution:

20. Example 8.11T 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression
(ax + b)
Example 8.11T
Find
Solution:

21. Example 8.12T 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression
(ax + b)
Example 8.12T
Find
Solution:
If y  ln 10, then ey  10 by definition,
i.e., eln 10  10.

22. cos 2A  2 cos2 A  1 or cos 2A  1  2 sin2 A
8.2 Indefinite Integration of Functions
B. Integration of Trigonometric Functions
To find integrals where the integrand is the product or power of trigonometric functions, we can first use
double angle formulas and product-to-sum formulas to express the integrand in the sum of trigonometric functions.
cos 2A  2 cos2 A  1 or cos 2A  1  2 sin2 A
Product-to-sum Formulas
sin A cos B  [sin (A  B)  sin (A  B)]
cos A sin B  [sin (A  B)  sin (A  B)]
cos A cos B  [cos (A  B)  cos (A  B)]
sin A sin B   [cos (A  B)  cos (A  B)]

23. Example 8.13T 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions
Example 8.13T
Find
Solution:
sin 2A  2 sin A cos A

24. Example 8.14T 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions
Example 8.14T
Find
Solution:
sin 2A  2 sin A cos A
cos 2A  1  2 sin2 A

25. Example 8.15T 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions
Example 8.15T
Find
Solution:
Product-to-sum formula

26. Example 8.16T 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions
Example 8.16T
Find
Solution:
cos 2A  1  2 sin2 A
cot A 

27. 8.3 Integration by Substitution
A. Change of Variables
In Section 8.2, we learnt some basic formulas to find the indefinite integrals of functions.
However, not all functions can be integrated directly using these formulas.
In this case, we have to use the method of integration by substitution.
The following shows the basic principle of this method.
Let and u  g (x).
 f (u) g¢(x)
 f [g(x)] g¢(x)
By the definition of integration,

28. 8.3 Integration by Substitution
A. Change of Variables
For an integral , we can transform it into
a simpler integral , by the following steps.
Step 1: Separate the integrand into two parts: f [g(x)] and g¢(x)dx.
Step 2: Replace every occurrence of g(x) in the integrand by u.
Step 3: Replace the expression ‘g¢(x)dx’ by ‘du’.
Let us use this method to find the integral together.
Note that ,
so we let u  x2 + 1, such that  2x.

29. Example 8.17T 8.3 Integration by Substitution Solution:
A. Change of Variables
Example 8.17T
Find
Solution:
Let u  1 – x2. Then
Express the answer in terms of x

30. Example 8.18T 8.3 Integration by Substitution Solution:
A. Change of Variables
Example 8.18T
Find
Solution:
Let u  x2 – 7. Then
Rewrite x2 as (u  7)

31. 8.3 Integration by Substitution
A. Change of Variables
With the method of integration by substitution, we can find the integrals of trigonometric functions other than sine and cosine, as shown in the following example.

32. Example 8.19T 8.3 Integration by Substitution Solution:
A. Change of Variables
Example 8.19T
Find
Solution:
Alternative Solution:
Let u  csc x – cot x.
Let u  csc x – cot x.

33. 8.3 Integration by Substitution
A. Change of Variables
It is tedious to write u and du every time when finding the integrals by substitution, as shown in the previous examples.
After becoming familiar with the method of integration by substitution, the working steps can be simplified by omitting the use of the variable u.
Let us study the following example.

34. Example 8.20T 8.3 Integration by Substitution Solution:
A. Change of Variables
Example 8.20T
Find
Solution:

35. 8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric
Functions
Sometimes we need to handle indefinite integrals that involve
the products of powers of trigonometric functions, such as
or , where m and n are integers.
In the following discussion, we will see how to apply different strategies according to different values of m and n.
Strategies for finding integrals in the form
Case 1: m is an odd number.
Use sin x dx  –d(cos x) and express all the other sine terms
as cosine terms.
Case 2: n is an odd number.
Use cos x dx  d(sin x) and express all the other cosine terms
as sine terms.
Case 3: both m and n are even numbers.
Use the double-angle formula to reduce the powers of the
functions.

36. Example 8.21T 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric
Functions
Example 8.21T
Find
Solution:

37. Example 8.22T 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric
Functions
Example 8.22T
Find
Solution:

38. Example 8.23T 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric
Functions
Example 8.23T
Find
Solution:

39. 8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric
Functions
Similarly, integrals in the form may be
found by using the method of integration by substitution.
Strategies for finding integrals in the form
Case 1: m is an odd number.
Use tan x sec x dx  d(sec x) and then express all other
tangent terms as secant terms.
Case 2: n is an even number.
Use sec2x dx  d(tan x) and then express all other secant
terms as tangent terms.

40. Example 8.24T 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric
Functions
Example 8.24T
Find
Solution:

41. Example 8.25T 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric
Functions
Example 8.25T
Find
Solution:

42. 8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric
Functions
In the above examples, the case that m is even while n is odd is not considered. This is because there is no standard technique and the method varies from case to case.
For example, to find (m  0 and n  1), we may follow the
method in Example 8.19.
In some other cases, such as (m  2 and n  1), we may
need to use the technique ‘integration by parts’, which will be discussed later in this chapter.

43. cos2q  1  sin2q , sec2q  1  tan2q , tan2q  sec2q  1
8.3 Integration by Substitution
C. Integration by Trigonometric Substitution
If an indefinite integral involves radicals in the form
, or , we can use the method of integration by substitution to eliminate the radicals.
The following three trigonometric identities are very useful for the elimination:
cos2q  1  sin2q , sec2q  1  tan2q , tan2q  sec2q  1
For example, if we substitute x  a sin q into the expression ,
we have
Then we can express the integrand in terms of q.
After finding the indefinite integral in terms of q (say, 3q + C), the final answer should be expressed in terms of the original variable, say, x.

44. 8.3 Integration by Substitution
C. Integration by Trigonometric Substitution
In order to express q in terms of x, let us first introduce the following notations:
Inverse of Trigonometric Functions
Let x be a real number.
1. sin–1x is defined as the angle q such that sin q  x (where –1  x  1)
and
2. cos–1x is defined as the angle q such that cos q  x (where –1  x  1)
and 0  q  p.
3. tan–1x is defined as the angle q such that tan q  x and

45. Example 8.26T 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution
Example 8.26T
Find
Solution:
Let x  sinq. Then dx  cosq dq.
Since sin q  x,

46. Example 8.27T 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution
Example 8.27T
Find
Solution:
Let x  sinq. Then dx  cosq dq.
Since sin q  x,

47. Example 8.28T 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution
Example 8.28T
Find
Solution:
Let x  3tanq. Then dx  3sec2q dq.
Since

48. Example 8.29T 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution
Example 8.29T
Find
Solution: q 1
x + 2
__________
(x  2)2  1
Let x + 2  secq.
Then dx  secq tanq dq.
Since secq  x + 2,

49. 8.4 Integration by Parts Some indefinite integrals such as , and
cannot be found by using the techniques we have
learnt so far.
To evaluate them, we need to introduce another method called integration by parts.
Theorem Integration by Parts
If u(x) and v(x) are two differentiable functions, then
In other words,
Proof:
Suppose u and v are two differentiable functions.
Since (uv)  uv¢ + vu¢, by definition,

50. 8.4 Integration by Parts
From Theorem 8.3, we can see that the problem of finding
can be transformed into the problem of finding instead.
If the integral is much simpler than , then the original integral
can be found easily.
If we want to apply the technique of integration by parts to an integral,
such as , we need to transform the integral into the form
first, such as or

51. 8.4 Integration by Parts
Example 8.30T
Find
Solution:

52. 8.4 Integration by Parts
Example 8.31T
Find
Solution:

53. 8.4 Integration by Parts
In some cases, there may be more than one choice for u and v.
For example, we can transform into or
However, if we choose the former, then
As a result, we get an integrand x2 cos x which is more complicated than the original one x sin x.
Thus we should try sin x dx  d(–cos x) instead.

54. 8.4 Integration by Parts
Example 8.32T
Find
Solution:

55. Example 8.33T 8.4 Integration by Parts Solution: Show that
Hence find
Solution:
(b)

56. 8.4 Integration by Parts
Example 8.34T
Find
Solution:

57. 8.4 Integration by Parts
Example 8.35T
Find
Solution:
Therefore,

58. 8.5 Applications of Indefinite Integrals
A. Geometrical Applications
In previous chapters, we learnt that of a curve y  F(x) is the slope function of the curve.
Since integration is the reverse process of differentiation,
if we let  f(x), then by the definition of integration, we have
where C is an arbitrary constant.
Thus we can see that the equation of a family of curves y  F(x) + C
can be found by integration, providing that the slope function of the curve is known.

59. Example 8.36T 8.5 Applications of Indefinite Integrals Solution:
A. Geometrical Applications
Example 8.36T
The equation of the slope of a curve at the point (x, y) is given by
. If the curve passes through (2, 2), find the equation of
the curve.
Solution:
where C is an arbitrary constant
When x  2, y  2, we have
∴ The equation of the curve is

60. Example 8.37T 8.5 Applications of Indefinite Integrals Solution:
A. Geometrical Applications
Example 8.37T
It is given that at each point (x, y) on a certain curve, If
the curve passes through and , find the equation of that curve.
Solution:
Since and lie on the curve, we have
∴ The equation of the curve is

61. 8.5 Applications of Indefinite Integrals
B. Applications in Physics
In Section 7.5 of Book 1, we learnt that for a particle moving along a straight line, its velocity v and acceleration a at time t are given by
where s is the displacement of the particle at time t.
Since integration is the reverse process of differentiation, we have

62. Example 8.38T 8.5 Applications of Indefinite Integrals Solution:
B. Applications in Physics
Example 8.38T
A particle moves along a straight line such that its velocity v at time t is given by Find the displacement s of the particle at time t, given that s  6 when t  2.
Solution:
When t  2, s  6, we have

63. Example 8.39T 8.5 Applications of Indefinite Integrals Solution:
B. Applications in Physics
Example 8.39T
A particle moves along a straight line so that its acceleration a at time t is
given by a  for t > 0. When t  1, the velocity of the particle is 8 and
its displacement is 24. Find the displacement of the particle at time t.
Solution:
When t  1, v  8, we have
When t  1, s  24, we have

64. Chapter Summary 8.1 Concepts of Indefinite Integrals
1. If , then the indefinite integral of f(x)
is defined by , where C is an arbitrary constant. 2. 3. 4. 5.

65. Chapter Summary 8.1 Concepts of Indefinite Integrals 6. 7. 8. 9. 10.
11.
12.
13.

66. Chapter Summary 8.2 Indefinite Integration of Functions
Let a and b be real numbers with a  0. 1. 2. 3. 4. 5.

67. Chapter Summary 8.3 Integration by Substitution
1. Let u  g(x) be a differentiable function. Then,
2. If the integrated involves terms like , and , we can simplify the integrand by substituting x  a sin q, x  a tan q
or x  a sec q respectively.

68. Chapter Summary 8.4 Integration by Parts
If u(x) and v(x) are two differentiable functions, then

69. Chapter Summary 8.5 Applications of Indefinite Integrals
If the slope of a curve at point (x, y) is f(x), then the equation of the family of curves is given by
where F ¢(x)  f(x).
2. Let s, v and a be the displacement, velocity and acceleration of a particle moving along a straight line respectively, then

70. Follow-up 8.1 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals
Follow-up 8.1
Find
Solution:

71. Follow-up 8.2 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals
Follow-up 8.2
Find
Solution:
Cancel the common factor

72. Follow-up 8.3 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals
Follow-up 8.3
Find
Solution:
Express the answer with positive indices.

73. Follow-up 8.4 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals
Follow-up 8.4
Find
Solution:

74. Follow-up 8.5 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals
Follow-up 8.5
Find
Solution:
cot A 

75. Follow-up 8.6 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals
Follow-up 8.6
Let y  x ln x.
Find in terms of ln x.
Hence find
Solution:
(a)
Chain Rule
(b) By (a),

76. Follow-up 8.7 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression
(ax + b)
Follow-up 8.7
Find
Solution:

77. Follow-up 8.8 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression
(ax + b)
Follow-up 8.8
Find
Solution:

78. Follow-up 8.9 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression
(ax + b)
Follow-up 8.9
Find
Solution:
Cancel the common factor

79. Follow-up 8.10 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression
(ax + b)
Follow-up 8.10
Find
Solution:

80. Follow-up 8.11 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression
(ax + b)
Follow-up 8.11
Find
Solution:

81. Follow-up 8.12 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression
(ax + b)
Follow-up 8.12
Find
Solution:
If y  ln 3, then ey  3 by definition,
i.e., eln 3  3.

82. Follow-up 8.13 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions
Follow-up 8.13
Find
Solution:
cos 2A  1  2 sin2 A

83. Follow-up 8.14 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions
Follow-up 8.14
Find
Solution:
cos 2A  2 cos2 A  1
cos 2A  2 cos2 A  1

84. Follow-up 8.15 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions
Follow-up 8.15
Find
Solution:

85. Follow-up 8.16 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions
Follow-up 8.16
Find
Solution:

86. Follow-up 8.17 8.3 Integration by Substitution Solution:
A. Change of Variables
Follow-up 8.17
Find
Solution:
Let u  1 + x3. Then
Express the answer in terms of x

87. Follow-up 8.18 8.3 Integration by Substitution Solution:
A. Change of Variables
Follow-up 8.18
Find
Solution:
Let u  x – 4. Then
Rewrite x as (u + 4)

88. Follow-up 8.19 8.3 Integration by Substitution Solution:
A. Change of Variables
Follow-up 8.19
Find
Solution:
Note that
Let u  sin x. Then

89. Follow-up 8.20 8.3 Integration by Substitution Solution:
A. Change of Variables
Follow-up 8.20
Find the following integrals.
(a)
(b)
Solution:
(a)
(b)

90. Follow-up 8.21 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric
Functions
Follow-up 8.21
Find
Solution:

91. Follow-up 8.22 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric
Functions
Follow-up 8.22
Find
Solution:

92. Follow-up 8.23 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric
Functions
Follow-up 8.23
Find
Solution:

93. Follow-up 8.24 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric
Functions
Follow-up 8.24
Find
Solution:

94. Follow-up 8.25 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric
Functions
Follow-up 8.25
Find
Solution:

95. Follow-up 8.26 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution
Follow-up 8.26
Find
Solution:
Let x  3 sinq. Then dx  3 cosq dq.

96. Follow-up 8.27 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution
Follow-up 8.27
Find
Solution:
Let Then
Since sin q  2x,

97. Follow-up 8.28 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution
Follow-up 8.28
Find
Solution:
Let Then
Since

98. Follow-up 8.29 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution
Follow-up 8.29
Find
Solution:
Let x  8 secq .
Then dx  8 tanq secq dq. q 8 x
_______
 x2  64

99. 8.4 Integration by Parts
Follow-up 8.30
Find
Solution:

100. 8.4 Integration by Parts
Follow-up 8.31
Find
Solution:

101. 8.4 Integration by Parts
Follow-up 8.32
Find
Solution:

102. Follow-up 8.33 8.4 Integration by Parts Solution: Find Hence find
(a) Let y  2x.
(b)
ln y  x ln 2
Differentiate both sides
with respect to x,

103. Follow-up 8.34
Find
Solution:

104. 8.4 Integration by Parts
Follow-up 8.35
Find
Solution:
Therefore,

105. Follow-up 8.36 8.5 Applications of Indefinite Integrals Solution:
A. Geometrical Applications
Follow-up 8.36
The equation of the slope of a curve at the point (x, y) is given by
 16 sin x cos x. If the curve passes through , find its equation.
Solution:
where C is an arbitrary constant.
When y  7, we have
∴ The equation of the curve is y  –4 cos 2x – 5.

106. Follow-up 8.37 8.5 Applications of Indefinite Integrals Solution:
A. Geometrical Applications
Follow-up 8.37
It is given that at each point (x, y) on a certain curve,  18e–3x.
If  1 and y  1 when x  0, find the equation of that curve.
Solution:
When x  0, y  1, we have
∴ The equation of the curve is
y  2e–3x + 7x – 1.
When x = 0,  1, we have

107. Follow-up 8.38 8.5 Applications of Indefinite Integrals Solution:
B. Applications in Physics
Follow-up 8.38
A particle moves along a straight line such that its acceleration a at time t is given by Find the velocity v of the particle at time t if v  7 when t  4.
Solution:
When t  4, v  7, we have

108. Follow-up 8.39 8.5 Applications of Indefinite Integrals Solution:
B. Applications in Physics
Follow-up 8.39
A particle is moving along a straight line such that its acceleration
a cm s–2 after t seconds is given by a  6(t – 1). The displacement of the particle is 13 cm after 2 seconds and 45 cm after 4 seconds. Find the displacement of the particle when its velocity is at its minimum.
Solution:
(2) – (1),
Substituting C1  6 into (1),
When t  2, s  13, we have
When t  4, s  45, we have
∴ The velocity is minimum when t = 1.
When t  1,
∴ The displacement is 9 cm.