1
2022
CALCULUS 1
LÊ XUÂN32ĐẠI
49
17
1
20
50
8
38
15
13
11
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6
31
24
26
SUMMARY
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48
9
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CALCULUS 11
CALCULUS
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47
30
43
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28
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45
37
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APPLIED MATHS
29
5
12
14
3
π
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16
41
42
4
4027
FACULTY OF APPLIED SCIENCE
39
Contents
1 Functions
I
Functions of single variable . . . . . . . . . . . . . . . . .
1
Four Ways to Represent a Function . . . . . . . .
2
Function and its graph . . . . . . . . . . . . . . . .
3
Piecewise defined functions . . . . . . . . . . . . .
4
The composite function . . . . . . . . . . . . . . .
5
One-to-one functions . . . . . . . . . . . . . . . .
6
Inverse functions . . . . . . . . . . . . . . . . . . .
II
Basic properties of functions . . . . . . . . . . . . . . . .
1
Periodic functions . . . . . . . . . . . . . . . . . .
2
Increasing and decreasing functions . . . . . . . .
3
Symmetry . . . . . . . . . . . . . . . . . . . . . . .
4
Boundedness . . . . . . . . . . . . . . . . . . . . .
III Elementary functions . . . . . . . . . . . . . . . . . . . .
1
Linear function . . . . . . . . . . . . . . . . . . . .
2
Polynomials . . . . . . . . . . . . . . . . . . . . . .
3
Power functions . . . . . . . . . . . . . . . . . . . .
4
Exponential functions . . . . . . . . . . . . . . . .
5
Logarithmic functions . . . . . . . . . . . . . . . .
6
Trigonometric functions . . . . . . . . . . . . . . .
7
Inverse trigonometric functions . . . . . . . . . .
8
Hyperbolic functions . . . . . . . . . . . . . . . . .
IV New Functions from Old Functions . . . . . . . . . . . .
1
Vertical and Horizontal Shifts . . . . . . . . . . . .
2
Vertical and Horizontal Stretching and Reflecting
V
Mathematical Models . . . . . . . . . . . . . . . . . . . .
1
Definition . . . . . . . . . . . . . . . . . . . . . . .
2
The modeling process . . . . . . . . . . . . . . . .
3
Linear Models . . . . . . . . . . . . . . . . . . . . .
4
Polynomial Functions . . . . . . . . . . . . . . . .
5
Trigonometric Functions . . . . . . . . . . . . . .
6
Exponential Functions . . . . . . . . . . . . . . . .
7
Logarithmic Functions . . . . . . . . . . . . . . . .
VI Sequences and Their Limits . . . . . . . . . . . . . . . . .
1
Sequences . . . . . . . . . . . . . . . . . . . . . . .
2
Limits of Sequences . . . . . . . . . . . . . . . . .
3
Bounded, monotonic sequences . . . . . . . . . .
VII Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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38
Contents
1
2
Essay Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Multiple-choice Questions . . . . . . . . . . . . . . . . . . . . . . . . .
2 The limit and continuity of a function
I
The limit of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Model real-world situations . . . . . . . . . . . . . . . . . . . . . . . .
2
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Calculating Limits using the limit laws . . . . . . . . . . . . . . . . .
4
One-sided limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
II
Limits involving infinity. Asymptotes . . . . . . . . . . . . . . . . . . . . . .
1
Infinite limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Vertical asymptote . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Limits at infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
Horizontal asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Slant asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
Infinite limits at infinity . . . . . . . . . . . . . . . . . . . . . . . . . .
III Infinitely Small (infinitesimals) and Infinitely Large (infinities) Quantities
1
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
0
2
Some basic limits of the form
. . . . . . . . . . . . . . . . . . . . .
0
∞
3
Some basic limits of the form
. . . . . . . . . . . . . . . . . . . . .
∞
4
Equivalent functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Some equivalent infinitesimals . . . . . . . . . . . . . . . . . . . . . .
6
Evaluating limits using equivalent infinitesimals . . . . . . . . . . .
7
Evaluating limits using equivalent infinites . . . . . . . . . . . . . . .
IV Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Continuity at a number . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Continuity from one-side . . . . . . . . . . . . . . . . . . . . . . . . .
3
Limit of composite function . . . . . . . . . . . . . . . . . . . . . . . .
V
Discontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Discontinuity at a number . . . . . . . . . . . . . . . . . . . . . . . .
2
Removable discontinuity . . . . . . . . . . . . . . . . . . . . . . . . .
3
Jump discontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
Infinite discontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . .
VI Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Essay Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Multiple-choice Questions . . . . . . . . . . . . . . . . . . . . . . . .
3 Derivatives and Differentials
I
Derivatives . . . . . . . . . . . . . . . . . . . . .
1
Tangents . . . . . . . . . . . . . . . . . .
2
Velocities . . . . . . . . . . . . . . . . . .
3
Definitions . . . . . . . . . . . . . . . . .
4
The derivative of elementary functions
5
Differentiation rules . . . . . . . . . . .
6
The chain rule . . . . . . . . . . . . . . .
II
Higher derivatives . . . . . . . . . . . . . . . . .
1
The second derivative . . . . . . . . . .
2
The n−th derivative . . . . . . . . . . .
22
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38
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Contents
3
Some basic formulas . . . . . . . . . . . . . . . . .
III Linear approximations and Differentials . . . . . . . . .
1
Linear approximations . . . . . . . . . . . . . . . .
2
The 1-st order differentials . . . . . . . . . . . . .
3
The 2-nd order differentials . . . . . . . . . . . . .
4
The n−th order differentials . . . . . . . . . . . . .
IV Rates of change and Related rates . . . . . . . . . . . . .
1
Rates of change in the natural and social sciences
2
Related rates . . . . . . . . . . . . . . . . . . . . . .
V
Indeterminate forms and L’ Hospital’s rule . . . . . . . .
0
1
Indeterminate form of type . . . . . . . . . . . .
0
∞
. . . . . . . . . . .
2
Indeterminate form of type
∞
3
Indeterminate products 0.∞ . . . . . . . . . . . .
4
Indeterminate form of type ∞ − ∞ . . . . . . . . .
5
Indeterminate powers 1∞ , 00 , ∞0 . . . . . . . . . .
VI Applications of Differentiation . . . . . . . . . . . . . . .
1
Global maximum, global minimum . . . . . . . .
2
Local extrema . . . . . . . . . . . . . . . . . . . . .
3
Critical number . . . . . . . . . . . . . . . . . . . .
4
The mean value theorem . . . . . . . . . . . . . .
5
Monotonicity . . . . . . . . . . . . . . . . . . . . .
6
How to find the local maximum and minimum .
7
How to find the global maximum and minimum
8
Concavity . . . . . . . . . . . . . . . . . . . . . . .
9
The second derivative test . . . . . . . . . . . . . .
VII Taylor - Maclaurin approximations . . . . . . . . . . . . .
1
Taylor- Maclaurin approximations . . . . . . . . .
2
Some basic Maclaurin approximations . . . . . .
VIII Curves defined by Parametric Equations . . . . . . . . .
1
Definition . . . . . . . . . . . . . . . . . . . . . . .
2
Graphing Devices . . . . . . . . . . . . . . . . . . .
3
Cycloid . . . . . . . . . . . . . . . . . . . . . . . . .
IX Calculus with Parametric Curves . . . . . . . . . . . . . .
1
Tangents . . . . . . . . . . . . . . . . . . . . . . . .
X
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Essay Questions . . . . . . . . . . . . . . . . . . . .
2
Multiple-choice Questions . . . . . . . . . . . . .
4 Integration
I
Anti-derivatives and indefinite integrals . . . . . . . .
1
Definition . . . . . . . . . . . . . . . . . . . . .
2
Some basic formulas of indefinite integrals . .
II Techniques of indefinite integration . . . . . . . . . .
1
The substitution rule . . . . . . . . . . . . . . .
2
Integration by Parts . . . . . . . . . . . . . . . .
III Integration of rational functions by partial fractions
1
Partial fractions . . . . . . . . . . . . . . . . . .
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100
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136
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151
151
152
153
154
154
156
159
159
Contents
IV
V
2
Integration of rational functions by partial fractions . . .
Integration of non-rational
functions
. . ¶. . . . µ. . . . . ¶. . ¶. . .
¶ µ
Z µ µ
ax + b p n
ax + b p 1 ax + b p 2
,
,...,
1
Type 1: R x,
dx
cx + d
cx + d
cx + d
Z
dx
. . . . . . . . . . . . . . . . . . . .
2
Type 2: p
ax 2 + bx + c
Trigonometric ZIntegrals . . . . . . . . . . . . . . . . . . . . . . . .
1
Type 1: R(sin x, cos x)d x . . . . . . . . . . . . . . . . . . .
VI
Definite integrals . . . . . . . . . . . . . . . . . . .
1
Area under a curve . . . . . . . . . . . . . .
2
Principle of Mathematical Induction . . .
3
Geometric meaning . . . . . . . . . . . . .
4
Applications of integral in Construction . .
5
Properties of the definite integrals . . . . .
VII Techniques of definite integration . . . . . . . . .
1
The Fundamental Theorem of Calculus . .
2
Newton-Leibniz’s formula . . . . . . . . . .
3
Integration by parts for definite integrals .
4
The substitution rule for definite integrals
5
Integral of Symmetric functions . . . . . .
VIII Improper integral of Type 1: Infinite intervals . .
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164
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172
+∞
1
f (x)d x
Definition of an improper integral of type 1
. . . . . . . 172
a
IX
2
Geometric meaning . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Newton-Leibniz’s Formula . . . . . . . . . . . . . . . . . . . . . .
4
A comparison test for improper integrals of type 1 . . . . . . . .
Improper integral of Type 2: Infinity discontinuous integrands . . . . .
Z b
1
Definition of an improper integral of type 2
f (x)d x on [a, b)
a
X
XI
2
Geometric meaning . . . . . . . . . . . . . . . . . . . . . . . . .
3
Newton-Leibniz’s formula . . . . . . . . . . . . . . . . . . . . . .
4
A comparison test for improper integrals of type 2 . . . . . . .
Application of integration . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Area between the graph of a function y = f (x) and the x−axis .
2
Area between curves . . . . . . . . . . . . . . . . . . . . . . . . .
3
Volume problem . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
The volume of a solid of revolution . . . . . . . . . . . . . . . .
5
The volume by cylindrical shells . . . . . . . . . . . . . . . . . .
6
Arc length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
Area of a surface of Revolution . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Essay Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Multiple-choice Questions . . . . . . . . . . . . . . . . . . . . .
5 Ordinary Differential Equations
I
Ordinary differential equations . . . . .
1
Introduction . . . . . . . . . . . .
2
Ordinary Differential Equations
3
Direction fields . . . . . . . . . .
44
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173
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177
. . . 177
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217
217
217
217
218
Contents
II
Separable Differential Equations . . . . . . . . . . . . . . . . . . . .
1
Definition of separable equations . . . . . . . . . . . . . . .
2
Another form of separable equations . . . . . . . . . . . . .
3
A cool problem . . . . . . . . . . . . . . . . . . . . . . . . . .
4
Radiocarbon Dating . . . . . . . . . . . . . . . . . . . . . . .
5
Mixing Problem . . . . . . . . . . . . . . . . . . . . . . . . . .
III Linear Differential Equations . . . . . . . . . . . . . . . . . . . . . .
1
Definition of linear equation . . . . . . . . . . . . . . . . . .
2
The solution of linear equation . . . . . . . . . . . . . . . . .
3
Electric Circuit . . . . . . . . . . . . . . . . . . . . . . . . . .
4
Hormone Level . . . . . . . . . . . . . . . . . . . . . . . . . .
IV Bernoulli Differential Equations . . . . . . . . . . . . . . . . . . . .
1
Definition of Bernoulli Differential Equations . . . . . . . .
2
Solving Bernoulli Differential Equations
. . . . . . . . . . .
³y´
. . . . . . . . . . . . . .
V
Homogeneous Equations of the form f
x
³y´
. .
1
Definition of homogeneous equations of the form f
x
2
Another form of homogeneous equations . . . . . . . . . .
VI The second order differential equations with constant coefficients
1
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
VII Initial and boundary value problems . . . . . . . . . . . . . . . . .
VIII Homogeneous Differential Equation with constant coefficients . .
1
Linear Independence and the Wronskian . . . . . . . . . . .
2
Characteristic equation . . . . . . . . . . . . . . . . . . . . .
3
Reduction of order . . . . . . . . . . . . . . . . . . . . . . . .
IX Non-homogeneous Equation with constant coefficients . . . . . .
1
Method of Undetermined Coefficients . . . . . . . . . . . .
2
Principle of Superposition . . . . . . . . . . . . . . . . . . . .
3
Mechanical Vibration . . . . . . . . . . . . . . . . . . . . . .
X
Homogeneous linear systems with constant coefficients . . . . . .
1
Elimination method . . . . . . . . . . . . . . . . . . . . . . .
XI Non-homogeneous linear systems with constant coefficients . . .
1
Elimination method . . . . . . . . . . . . . . . . . . . . . . .
2
Interconnected fluid tanks . . . . . . . . . . . . . . . . . . .
XII Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Essay Questions . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Multiple-choice Questions . . . . . . . . . . . . . . . . . . .
55
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219
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220
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227
230
230
231
231
231
233
234
237
237
238
241
244
244
247
247
250
251
251
256
Contents
66
Chapter 1
Functions
Learning Objectives
Recall some elementary functions.
Study some applications of functions.
Study sequences and their limits.
I
Functions of single variable
1
Four Ways to Represent a Function
Functions arise whenever one quantity depends on another. Consider the following 4
situations:
1. The area A of a circle depends on the radius r of the circle: A = πr 2 .
2. The human population of the world P depends on the time t . The table gives estimates of the world population P (t ) at the time t
7
Chapter 1. Functions
3. The cost C of mailing an envelope depends on its weight w. Although there is no
simple formula that connects w and C , the post office has a rule for determining C
when w is known.
4. The vertical acceleration a of the ground as measured by a seismograph during an
earthquake is a function of the elapsed time t
There are 4 possible ways to represent a function:
1. verbally (by a description in words)
2. numerically (by a table of values)
3. visually (by a graph)
4. algebraically (by an explicit formula)
2
Function and its graph
Definition 1
A function f is a rule that assigns to each element x in a set X ⊂ R exactly one
element y, called f (x) in a set E ⊂ R. Denoted by:
f : X −→ E
y = f (x), x − independent, y − dependent
The set X = {x ∈ R : f (x)is defined} is called the domain of the function f and is
denoted by D( f ). The set f (X ) = {y = f (x) ∈ R : x ∈ X } is called the range of the
function f and is denoted by E ( f ).
88
I. Functions of single variable
f (x) is the value of f at x and is read " f of x".
Definition 2
The set consists of all points (x, f (x)), x ∈ X in the coordinate plane Ox y is called
the graph of the function f .
Example 1
Find the domain and range of function f (x) =
p
x + 2.
SOLUTION
1. The domain of f consists of all values of x such that x +2 Ê 0 ⇔ x Ê −2, so the domain
is the interval [−2, +∞)
p
2. The range of f consists of all values of y such that y = x + 2 Ê 0, so the range is the
interval [0, +∞)
3
Piecewise defined functions
Definition 3
The functions which are defined by different formulas in different parts of their
domain, are called piecewise defined functions.
Example 2
A function f is defined by
½
f (x) =
4
1 − x, if x É −1
x 2 , if x > −1
The composite function
Suppose that y = f (u), where u is a function of x : u = g (x). We compute this by substitution
y = f (g (x))
99
Chapter 1. Functions
The procedure is called composition because the new function is composed of the 2 given
functions f and g .
Definition 4
Given 2 functions f and g , the composite function f ◦ g (read: f circle g ) is defined
by
( f ◦ g )(x) = f (g (x))
(1.1)
Example 3
If f (x) = x 2 , and g (x) = x 3 − 7, find the composite function f ◦ g and g ◦ f .
SOLUTION We have
( f ◦ g )(x) = f (g (x)) = f (x 3 − 7) = (x 3 − 7)2
(g ◦ f )(x) = g ( f (x)) = g (x 2 ) = (x 2 )3 − 7 = x 6 − 7
5
One-to-one functions
Definition 5
A function f is called a one-to-one function if it never takes on the same value twice;
that is,
f (x 1 ) ̸= f (x 2 ) whenever x 1 ̸= x 2
(1.2)
Example 4
1. The function f (x) = x 3 is one-to-one because if x 1 ̸= x 2 then x 13 ̸= x 23 .
2. The function g (x) = x 2 is not one-to-one because, for instance g (1) = 1 =
g (−1), and so 1 and −1 have the same output.
10
10
I. Functions of single variable
6
Inverse functions
Definition 6
Let f be a one-to-one function with domain D and range E . Then its inverse function
f −1 (read: f inverse) has domain E and range D and is defined by
f −1 (y) = x ⇔ f (x) = y
(1.3)
Remark 1 (Cancellation equations)
f −1 ( f (x)) = x, ∀x ∈ D
(1.4)
f ( f −1 (y)) = y, ∀y ∈ E
(1.5)
The graph of f −1 (x) is obtained by reflecting the graph of f (x) about the line y = x.
11
11
Chapter 1. Functions
Remark 2
How to find the inverse function of a one-to-one function f
1. Write y = f (x).
2. Solve this equation for x in terms of y (if possible).
3. To express f −1 as a function of x, interchange x and y. The resulting equation
is y = f −1 (x).
Example 5
Find the inverse function of f (x) = x 3 + 2.
SOLUTION.
1. Write y = x 3 + 2
2. Solve this equation for x in terms of y : x 3 = y − 2 ⇒ x =
p
3
y −2
of x, interchange x and y. The resulting equation is
3. To express f −1 as a function
p
3
−1
y = f (x). Therefore, y = x − 2 = f −1 (x)
II
1
Basic properties of functions
Periodic functions
Definition 7
The function f is called periodic of period T > 0 if for all x ∈ X , such that x −T, x +T ∈
X and
f (x + T ) = f (x − T ) = f (x)
(1.6)
Example 6
Function sine f (x) = sin x is periodic function and has period T = 2π. This means
that, for all values of x
f (x + 2π) = sin(x + 2π) = f (x − 2π) = sin(x − 2π) = f (x) = sin x.
2
Increasing and decreasing functions
Definition 8
A function f is called
1. increasing on an interval X if f (x 1 ) < f (x 2 ) whenever x 1 < x 2 in X .
2. decreasing on an interval X if f (x 1 ) > f (x 2 ) whenever x 1 < x 2 in X .
12
12
II. Basic properties of functions
Example 7
1. Function f (x) = x is increasing on the R.
2. Function g (x) = x 2 is decreasing on the interval (−∞, 0) and increasing on the
interval (0, +∞).
3. Function h(x) = c = const, according to the definition is not decreasing and
not increasing.
3
Symmetry
Definition 9
1. Function f is called an even function on the X ⊂ R if for all x ∈ X such that
−x ∈ X and
f (−x) = f (x)
(1.7)
2. Function f is called an odd function on the X ⊂ R if for all x ∈ X such that
−x ∈ X and
f (−x) = − f (x)
(1.8)
Example 8
1. Function f (x) = x is odd on R because for each x ∈ R then −x ∈ R and f (−x) =
−x = − f (x).
2. Function g (x) = x 2 is even on R because for each x ∈ R then −x ∈ R and
g (−x) = (−x)2 = x 2 = g (x).
4
Boundedness
Definition 10
Let function f : X → Y be defined on a set D ⊂ X . Function f is called
1. bounded from above if there is a number M ∈ R such that for all x ∈ D from
the domain D one has f (x) É M ;
2. bounded from below if there is a number m ∈ R such that for all x ∈ D from
the domain D one has f (x) Ê m;
3. bounded if there is a number C > 0 such that for all x ∈ D from the domain D
one has | f (x)| É C ;
4. unbounded if for all C > 0, exists x 0 ∈ D such that | f (x 0 )| > C .
13
13
Chapter 1. Functions
Example 9
1
is bounded from above on the set D = R because
1 + x2
0 < f (x) É 1, ∀x ∈ R.
1. Function f (x) =
1
is bounded on the open interval [1, +∞) because 0 < g (x) É
x
1, ∀x ∈ [1, +∞) but g (x) is unbounded on the interval (0, +∞) because for all
1
C > 0, exists x 0 =
∈ (0, +∞) such that |g (x 0 )| = |C + 1| > C .
C +1
2. Function g (x) =
III
1
Elementary functions
Linear function
Definition 11
Linear function has a form
y = mx + b,
(1.9)
where m is the slope of the line and b is the y−intercept.
Proposition 3.0
The slope, m, of the line through (x 1 , y 1 ) and (x 2 , y 2 ) is given by the following equation,
if x 1 ̸= x 2
m=
y2 − y1
x2 − x1
(1.10)
The slope of a line can be interpreted as the rate of change in the y−coordinates for each
1-unit increase in the x−coordinates.
Example 10
Find the slope and the equation of the line through the points (4, 3) and (2, 5).
SOLUTION. The slope is
m=
y2 − y1 5 − 3
=
= −1.
x2 − x1 2 − 4
14
14
III. Elementary functions
The equation of the line that has slope −1 and passes through point (4, 3) is
y − 3 = −1(x − 4) ⇔ y = −x + 7
2
Polynomials
Definition 12
A function P is called a polynomial if
P (x) = a n x n + a n−1 x n−1 + . . . + a 2 x 2 + a 1 x + a 0
where n is a non-negative integer and the numbers a 0 , a 1 , a 2 , . . . , a n are constants
called the coefficients of the polynomial.
If the leading coefficient a n ̸= 0, then the degree of the polynomial is n
Example 11
1. P (x) = mx + b is a polynomial of degree 1 - linear function
2. P (x) = ax 2 + bx + c, (a ̸= 0) is a polynomial of degree 2 - quadratic function
3. P (x) = ax 3 + bx 2 + c x + d , (a ̸= 0) is a polynomial of degree 3 - cubic function
3
Power functions
Definition 13
Power function has a form
y = x α , α ∈ R.
(1.11)
(Read: x to the α power)
The domain and range of power functions depend on the parameter α.
Example 12
Case α = 2 ⇒ y = x 2 - square function
1. Domain: D = R.
2. Range: E = [0, +∞).
3. Function is increasing on the interval (0, ∞) and decreasing on the interval
(−∞, 0).
4. Function is even, the graph is symmetric with respect to the y−axis.
15
15
Chapter 1. Functions
Example 13
Case α = −1 ⇒ y =
1
- reciprocal function
x
1. Domain: D = R \ {0}.
2. Range: E = (−∞, 0) ∪ (0, ∞).
3. Function is decreasing on the interval (−∞, 0) and (0, +∞)
4. Function is odd, the graph is symmetric about the origin O(0, 0).
Example 14
Case α =
p
1
⇒ y = x -square root function
2
1. Domain: D = [0, +∞).
2. Range: E = [0, +∞).
3. Function is increasing on the interval (0, +∞)
4. Function does not have symmetry.
16
16
III. Elementary functions
4
Exponential functions
Definition 14
Exponential function has a form
y = a x , a > 0, a ̸= 1.
(1.12)
(Read: base a to the x power)
Laws of exponents
a x .a y = a x+y
a x .b x = (ab)x
ax
= a x−y
y
a
a x ³ a ´x
=
bx
b
(a x ) y = a x y
a −x =
1
ax
Example 15
Function y = a x , (a > 1)
1. Domain: D = R
2. Range: E = (0, ∞)
3. Function is increasing on the interval (−∞, +∞)
4. The graph always passes through the point at (0, 1)
17
17
Chapter 1. Functions
Example 16
Function y = a x , (0 < a < 1)
1. Domain: D = R.
2. Range: E = (0, +∞).
3. Function is decreasing on the interval (−∞, +∞)
4. The graph always passes through the point at (0, 1)
5
Logarithmic functions
Definition 15
Logarithmic function has a form
y = loga x, a > 0, a ̸= 1
(1.13)
(Read: logarithm of x with base a)
Logarithmic function y = loga x is the inverse function of exponential function y = a x , this
means that, if y = loga x then x = a y > 0. Therefore, the domain of logarithmic function is:
D = {x ∈ R | x > 0}.
Properties of logarithmic functions
µ ¶
1
loga (x.y) = loga x + loga y
loga
= − loga x
x
µ ¶
x
α
loga
= loga x − loga y
loga β x α = loga x
y
β
18
18
III. Elementary functions
Example 17
Function y = loga x, (a > 1)
1. Domain: D = (0, +∞).
2. Range: E = R.
3. Function is increasing on the interval (0, +∞).
4. The graph of logarithmic function y = loga x is the reflection of the graph of
exponential function y = a x about the line y = x.
Example 18
Function y = loga x, (0 < a < 1)
1. Domain: D = (0, +∞).
2. Range: E = R.
3. Function is decreasing on the interval (0, +∞).
4. The graph of logarithmic function y = loga x is the reflection of the graph of
exponential function y = a x about the line y = x.
19
19
Chapter 1. Functions
6
Trigonometric functions
Definition 16
Function sine y = sin x
1. Domain: D = R
2. Range: E = [−1, 1]
3. Function is periodic of period 2π : sin(x) = sin(x + 2π) = sin(x − 2π)
³ π π´
4. Function is increasing on the interval − , , and decreasing on the interval
2 2
¶
µ
π 3π
,
2 2
5. Function is odd, the graph is symmetric about the origin O(0, 0).
Definition 17
Function cosine y = cos x
1. Domain: D = R
2. Range: E = [−1, 1]
3. Function is periodic of period 2π : cos(x) = cos(x + 2π) = cos(x − 2π)
4. Function is increasing on the interval (−π, 0) , and decreasing on the interval
(0, π) ,
5. Function is even, the graph is symmetric with respect to the y−axis.
Definition 18
Function tangent y = tan x =
sin x
cos x
20
20
III. Elementary functions
1. Domain: D = R \
nπ
2
o
+ kπ, k ∈ Z
2. Range: E = R
3. Function is periodic of period π : tan(x) = tan(x + π) = tan(x − π)
³ π π´
4. Function is increasing on the interval − , .
2 2
5. Function is odd, the graph is symmetric about the origin O(0, 0).
Definition 19
Function cotangent y = cot x =
cos x
sin x
1. Domain: D = R \ {kπ, k ∈ Z}
2. Range: E = R
3. Function is periodic of period π : cot(x) = cot(x + π) = cot(x − π)
4. Function is decreasing on the interval (0, π).
5. Function is odd, the graph is symmetric about the origin O(0, 0).
21
21
Chapter 1. Functions
Some basic formulas
³ π´
cos ± = 0.
2
sin2 x + cos2 x = 1
sin 2x = 2 sin x cos x
tan x =
sin 3x = 3 sin x − 4 sin3 x
sin x
cos x
tan(π − x) = tan(−x) = − tan x
1 − cos 2x
sin x =
2
2
sin
tan(π + x) = tan(x)
³π´
tan 0 = 0, tan
is undefined
2
π
= 1; sin(kπ) = 0
2
cot x =
cos 2x = 2 cos2 x − 1 = 1 − 2 sin2 x
cot(π − x) = cot(−x) = − cot x
cos2 x =
1 + cos 2x
2
cot(π + x) = cot x
³π´
cot
= 0, cot 0 is undefined
2
cos 0 = 1; cos π = −1
7
cos x
sin x
cos 2x = cos2 x − sin2 x
Inverse trigonometric functions
Definition 20
Function arcsine y = arcsin x = sin−1 x
y = arcsin x ⇐⇒ x = sin y
π
π
−1 É x É 1,
− ÉyÉ
2
2
22
22
(1.14)
III. Elementary functions
Definition 21
Function arccosine y = arccos x = cos−1 x
y = arccos x ⇐⇒ x = cos y
−1 É x É 1
0Éy Éπ
(1.15)
Definition 22
Function arctangent y = arctan x = tan−1 x
y = arctan x
⇐⇒ x = tan y
π
π
−∞ < x < ∞
− <y<
2
2
23
23
(1.16)
Chapter 1. Functions
8
Hyperbolic functions
Definition 23
Function sinh x =
e x − e −x
is called hyperbolic sine.
2
1. Domain: D = R
2. Range: E = R
3. Function is increasing on the R
4. Function is odd, the graph is symmetric about the origin O(0, 0).
Definition 24
Function cosh x =
e x + e −x
is called hyperbolic cosine.
2
1. Domain: D = R
2. Range: E = [1, +∞)
3. Function is increasing on (0, +∞) and decreasing on (−∞, 0)
4. Function is even, the graph is symmetric with respect to the y−axis.
24
24
IV. New Functions from Old Functions
Definition 25
1. Function tanh x =
sinh x
is called hyperbolic tangent.
cosh x
2. Function coth x =
cosh x
is called hyperbolic cotangent.
sinh x
Hyperbolic Identities
sinh(−x) = − sinh x
cosh(−x) = cosh x
cosh2 x − sinh2 x = 1
cosh2 x + sinh2 x = cosh 2x
sinh(x + y) = sinh x cosh y + cosh x sinh y
cosh(x + y) = cosh x cosh y + sinh x sinh y
IV
1
New Functions from Old Functions
Vertical and Horizontal Shifts
Proposition 4.1
Suppose c>0. To obtain the graph of
1. y = f (x) + c, shift the graph of y = f (x) a distance c units upward
2. y = f (x) − c, shift the graph of y = f (x) a distance c units downward
3. y = f (x + c), shift the graph of y = f (x) a distance c units to the left
4. y = f (x − c), shift the graph of y = f (x) a distance c units to the right
25
25
Chapter 1. Functions
2
Vertical and Horizontal Stretching and Reflecting
Proposition 4.2
Suppose c>1. To obtain the graph of
1. y = c f (x), stretch the graph of y = f (x) vertically by a factor of c
2. y =
1
f (x), shrink the graph of y = f (x) vertically by a factor of c
c
3. y = f (cx), shrink the graph of y = f (x) horizontally by a factor of c
µ ¶
1
4. y = f
x , stretch the graph of y = f (x) horizontally by a factor of c
c
5. y = − f (x), reflect the graph of y = f (x) about the x−axis
6. y = f (−x), reflect the graph of y = f (x) about the y−axis
26
26
IV. New Functions from Old Functions
Example 19
Given the graph of y = x 2 , use transformations to graphs
1. y = f (x) = (x − 2)2 − 3
2. y = | f (x)| = |(x − 2)2 − 3|
3. y = f (|x|) = (|x| − 2)2 − 3
SOLUTION. The graph of y = f (x) = (x − 2)2 − 3 is a translation of y = x 2 . The parent graph
has been translated right 2 units and down 3 units.
The graph of y = | f (x)| = |(x − 2)2 − 3| is a translation of y = f (x) = (x − 2)2 − 3. This transformation reflects any portion of the parent graph that is below the x−axis so that it is above
the x−axis.
27
27
Chapter 1. Functions
The graph of y = f (|x|) = (|x| − 2)2 − 3 is a translation of y = f (x) = (x − 2)2 − 3. This transformation results in the portion of the parent graph on the left on the y−axis being replaced
by a reflection of the portion on the right of the y−axis.
Example 20
Given the graph of y = cos x, use transformations to graphs
1. y = 2 cos x
2. y =
1
· cos x
2
3. y = cos(2x)
µ
¶
1
4. y = cos · x
2
28
28
V. Mathematical Models
V
1
Mathematical Models
Definition
Real-world problems (phenomenon) can be
1. The size of a population
2. The demand for a product
3. The speed of a falling object
4. The concentration of a product in a chemical reaction
5. The life expectancy of a person at birth
6. The cost of emission reductions
29
29
Chapter 1. Functions
2
The modeling process
Proposition 5.1
1. Formulate a mathematical model
2. Apply mathematics to the mathematical model in order to derive mathematical
conclusions
3. Interpret mathematical conclusions as information about the original realworld phenomenon by way of offering explanations or making predictions
4. Test our predictions by checking against new real data
3
Linear Models
Example 21
How fast has tuition at public colleges been increasing in recent years, and how well
we predict tuition in the future?
The table lists the average annual cost (in dollars) of tuition and fees at public
four-year colleges for selected years.
1. Plot the cost of public colleges by letting t = 0 correspond to 2000. Are the data
exactly linear? Could the data be approximated by a linear equation?
2. Use the points (0, 3508) and (9, 7020) to determine an equation that models
the data.
3. Discuss the accuracy of using this equation to estimate the cost of public
colleges in the year 2030.
30
30
V. Mathematical Models
SOLUTION
1. Although the scatter plot for Tuition and Fees is not exactly linear, it is approximately
linear and could be approximated by a linear equation.
2. The linear equation: y =
3512
· t + 3508.
9
3. t = 30 ⇒ y ≈ 15214. The year 2030 is many years in the future, many factors could
effect the tuition and the actual figure for 2030 could turn out to be different from
our prediction.
4
Polynomial Functions
Example 22
Many grandparents invest in the stock market for their grandchildren’s college fund.
18 years ago, Della Brooks purchased 1000$ worth of merchandising stocks at the
birth of her first grandchild Owen. 10 years ago, she purchased 500$ worth of
transportation stocks, and 5 years ago, she purchased 250$ worth of technology
stocks. The stocks will be used to help pay for Owen’s college education. If the stocks
appreciate at an average annual rate of 12.25%, determine the current value of the
college fund.
SOLUTION The formula is A = P (1 + r )t , where P is the original amount of money invested,
r is the interest rate, and t is the time invested (in years).
Total=merchandising+transportation+technology
T (r ) = 1000(1 + r )18 + 500(1 + r )10 + 250(1 + r )5
r = 0.1225 ⇒ T (0.1225) ≈ 10038.33
The present value of Owen’s college fund is about 10038.33$
31
31
Chapter 1. Functions
5
Trigonometric Functions
Example 23
On May 18, 1980, Mount Saint Helens, a volcano in Washington, erupted with such
force that the top of the mountain was blown off. To determine the new height at
the summit of Mount Saint Helens, a surveyor measured the angle of elevation to
the top of the volcano to be 37o 46′ . The surveyor then moved 1000 feet closer to the
volcano and measured the angle of elevation to be 40o 30′ . Determine the new height
of Mount Saint Helens.
SOLUTION
1. h represents the height of the volcano,
2. x represents the distance from the surveyor’s second position to the center of the
base of the volcano.
h
tan 37o 46′ =
⇒ h = (1000 + x) tan 37o 46′
(1.17)
1000 + x
tan 40o 30′ =
⇒x=
h
⇒ h = x tan 40o 30′
x
(1.18)
1000 tan 37o 46′
≈ 9765.826092
tan 40o 30′ − tan 37o 46′
⇒ h = x tan 40o 30′ ≈ 8340.803443
The new height of Mount Saint Helens is about 8341 feet.
6
Exponential Functions
Example 24
In recent years, beekeepers have experienced a serious decline in the honeybee
population in the United States. One of the causes for the decline is the arrival
of varroa mites. Experts estimate that as much as 90% of the wild bee colonies
have been wiped out. The graph shows typical honeybee and varroa populations
over several months. A graph of varroa population growth from April to September
resembles an exponential curve.
Suppose that a researcher estimates that the initial population of varroa in a colony
is 500. They are increasing at a rate of 14% per week. What is the expected population
in 22 weeks?
32
32
V. Mathematical Models
SOLUTION
1. Exponential growth or Decay: N = N0 (1 + r )t , where N is the final amount, N0 is the
initial amount, r is the rate of growth or decay per time period, and t is the number
of time periods.
2. N = 500(1 + 0.14)22 ≈ 8930.519719
There will be about 8931 varroa in the colony in 22 weeks.
7
Logarithmic Functions
Example 25
The intensity of an earthquake is described by a number on the Richter scale.
³ a ´ The
Richter scale number R of an earthquake is given by the formula R = log
+ B,
T
where a is the amplitude of the vertical ground motion in microns, T is the period of
the seismic wave in seconds, and B is a factor that accounts for the weakening of
seismic waves. Find the intensity of an earthquake to the nearest tenth if a recording
station measured the amplitude as 200 microns and the period as 1.6 seconds, and
B = 4.2.
SOLUTION
R = log
³a´
T
+ B,
where a = 200, T = 1.6, B = 4.2
¶
200
⇒ R = log
+ 4.2 ≈ 6.3.
1.6
µ
33
33
Chapter 1. Functions
VI
1
Sequences and Their Limits
Sequences
Definition 26
A function f : N −→ R whose domain is a set of non-negative integers and whose
range is a subset of the real numbers R is called a sequence.
f : N −→ R
x n = f (n),
n ∈ N.
(1.19)
Example 26
If the sequence {x n } is given by x n =
2
1
1
1
then x 1 = 1, x 2 = , . . . , x n = , . . .
n
2
n
Limits of Sequences
Definition 27
The number a ∈ R is called the limit of the sequence {x n } ⊂ R, if for every ε > 0, there
is an integer N = N (ε) such that |x n − a| < ε whenever n > N .
Definition 28
1. If a sequence {x n } ⊂ R has a finite limit a ∈ R then it converges to a and we
write x n → a.
2. A sequence {x n } ⊂ R is called divergent if every number a ∈ R is NOT a limit of
this sequence, i.e. a does not exist or is equal to ∞.
Theorem VI.1 (Uniqueness of limit)
If a sequence {x n } ⊂ R is convergent then it has the unique limit.
Proposition 6.1
If sequences {x n } ⊂ R and {y n } ⊂ R have finite limits a and b respectively, then
1. lim |x n | = |a|.
n→∞
2. lim (x n ± y n ) = a ± b
n→∞
3. lim (x n .y n ) = a.b.
n→∞
34
34
VI. Sequences and Their Limits
4. In addition that b ̸= 0 then we have lim
n→∞
xn a
= .
yn b
Proposition 6.2
Some basic limits
1. If |q| < 1 then lim q n = 0.
6. lim
1
2. lim α = 0,
n→∞ n
7. lim
n→∞
n→∞
1
= 0,
n→∞ lnα n
1
= 0.
n→∞ e n
p
n
5. lim n p = 1,
α > 0.
4. lim
n→∞
a = 1,
a > 0.
np
= 0, ∀p.
n→∞ e n
¶
µ
1 n
8. lim 1 +
= e.
n→∞
n
³
a ´n
9. lim 1 +
= ea,
n→∞
n
α > 0.
3. lim
p
n
lnp n
= 0,
n→∞ n α
10. lim
∀p.
∀a.
∀p, ∀α > 0.
NOTE. For p, α > 0, a > 1, as n → ∞ then
lnp n << n α << a n << n!
Lemma VI.1
If lim f (x) = a and f (n) = x n when n is an integer, then lim x n = a
x→+∞
n→+∞
Example 27
1
Evaluate I = lim n sin ·
n→+∞
n
1
1
sin x
= 1 or lim x sin = 1. If x n = n sin then I = 1.
x→+∞
x→0 x
x
n
SOLUTION. We know that lim
Theorem VI.2 (Squeeze theorem)
If
1. x n É y n É z n ,
∀n > n 0
2. lim x n = lim z n = a
n→∞
n→∞
then lim y n = a.
n→∞
Example 28
7n
.
n→∞ n n
Evaluate lim
35
35
Chapter 1. Functions
SOLUTION
We have
µ ¶n
7n
7
,
0< n <
n
8
∀n > 8.
µ ¶n
7
7n
and lim
= 0 then lim n = 0.
n→∞ 8
n→∞ n
Definition 29
lim x n = +∞(−∞; ∞) means that for every positive number M > 0 there is an
n→+∞
integer N = N (M ) > 0 such that if n > N then x n > M (x n < −M ; |x n | > M ).
Proposition 6.3
1. If lim |x n | = 0, then lim x n = 0.
n→∞
n→∞
2. If lim x n = a, and the function f is continuous at a, then lim f (x n ) = f (a).
n→∞
n→∞
Example 29
π
Find lim sin ·
n→∞
n
SOLUTION
Because the sine function is continuous at 0 then
³
π´
π
= sin 0 = 0.
lim sin = sin lim
n→∞ n
n→∞
n
3
Bounded, monotonic sequences
Definition 30
1. A sequence (x n ) is called increasing if x n < x n+1 for all n ∈ N.
2. It is called decreasing if x n > x n+1 for all n ∈ N.
3. A sequence is monotonic if it is either increasing or decreasing.
Example 30
n o
The sequence x n where x n =
Proof. Because x n =
3
, (n ∈ N) is decreasing.
n +5
3
3
>
= x n+1 so the sequence is decreasing.
n + 5 (n + 1) + 5
Definition 31
n o
1. A sequence x n is bounded above if there is a number M such that x n É
36
36
VI. Sequences and Their Limits
M , ∀n ∈ N.
n o
2. A sequence x n is bounded below if there is a number m such that x n Ê
m, ∀n ∈ N.
n o
3. If it is bounded above and below, then x n is a bounded sequence.
Proposition 6.3
n o
Every convergent sequence x n is bounded.
Theorem VI.3 (Weierstrass Theorem)
1. If a sequence is increasing and bounded above then it is convergent.
x1 É x2 É . . . É xn É . . . É M
2. If a sequence is decreasing and bounded below then it is convergent.
x1 Ê x2 Ê . . . Ê xn Ê . . . Ê m
37
37
Chapter 1. Functions
VII
1
Exercises
Essay Questions
Domain and range of a function
1
The graph of a function is given
1. State the value of f (1).
2. For what values of x is f (x) = 1.
3. State the domain and range of f .
L SOLUTION.
1. The value of f (1) is 3.
2. f (x) = 1 ⇔ x = 0.
3. The domain of f is [−2; 4]. The range of f is the interval [−1; 3].
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
2
Find the domain and range of the following functions
p
2x + 1
p
3
2. f (x) = 2x − 1
1. f (x) =
3. f (x) =
2x 3 − 5
x2 + x − 6
L SOLUTION.
1
1. The domain of f consists of all values of x such that 2x + 1 Ê 0 ⇔ x Ê − , so the
2
·
¶
1
domain is the interval − , +∞ . The range of f consists of all values of y such that
2
p
y = 2x + 1 Ê 0, so the range is the interval [0, +∞)
38
38
VII. Exercises
2. The domain of f consists of all real values of x, so the domain is the interval
(−∞, +∞). The range of f consists of all real values of y, so the range is the interval
(−∞, +∞)
½
x ̸= −3
So
x ̸= 2
the domain is the set R \ {−3, 2}. The range of f consists of all real values of y, so the
range is the interval (−∞, +∞)
2
3. The domain of f consists of all values of x, such that x + x − 6 ̸= 0 ⇒
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
3
Find the domain of the following functions
1. f (x) =
2. f (x) =
3. f (x) =
1
4. f (x) = ln(x − 1) − 1
x2 − x
1−e
x
s
2
5. f (x) =
2
1 − e 1−x
1+x
e cos x
µ
¶
1
ln 1 +
x
6. f (x) = arcsin
3x
x2 + 2
L SOLUTION.
½
2
1. The domain of f consists of all values of x, such that x − x ̸= 0 ⇒
domain is the set R \ {0, 1}.
x ̸= 0
So the
x ̸= 1
2
2. The domain of f consists of all values of x, such that 1 − e 1−x ̸= 0 ⇒ 1 − x 2 ̸= 0 ⇔ x ̸=
±1 so the domain is the set R \ {−1, 1}
3. The domain of f consists of all values of x, such that e cos x ̸= 0 ⇒ x ∈ R so the domain
is the set R
4. The domain of f consists of all values of x, such that x − 1 > 0 ⇒ x > 1 so the domain
is the interval (1, +∞)
5. The domain of f consists of all values of x, such that

1

 1 + Ê e0 = 1
1
1
1
x
ln 1 +
Ê0⇔
⇔ 1 + Ê 1 ⇔ Ê 0 ⇔ x > 0.
1

x
x
x

1+ > 0
x
µ
¶
So the domain is the interval (0, +∞)
6. The domain of f consists of all values of x, such that
3x
É 1 ⇔ −x 2 − 2 É 3x É x 2 + 2 ⇔
−1 É 2
x +2
½
So the domain is (−∞, −2] ∪ [−1, 1] ∪ [2, +∞)
39
39
x 2 + 3x + 2 Ê 0
⇔
x 2 − 3x + 2 Ê 0
½
x Ê −1 ∨ x É −2
x Ê 2∨x É 1
Chapter 1. Functions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
Composite functions
4
1
x +1
, find the composite function f ◦ g , g ◦ f , f ◦ f
If f (x) = x + , and g (x) =
x
x +2
and g ◦ g and their domains.
L SOLUTION.
We have
¶
x +1
1
x +1
=
+ x+1 =
( f ◦ g )(x) = f (g (x)) = f
x +2
x + 2 x+2
µ
=
x + 1 x + 2 (x + 1)2 + (x + 2)2
2x 2 + 6x + 5
+
=
=
x +2 x +1
(x + 1)(x + 2)
(x + 1)(x + 2)
Domain: R \ {−2, −1}
µ
¶
1
(g ◦ f )(x) = g ( f (x)) = g x +
=
x
=
x + x1 + 1
x + x1 + 2
=
x2 + 1 + x
x2 + x + 1
=
x 2 + 1 + 2x
(x + 1)2
Domain: R \ {−1, 0}
µ
¶
1
1
1
( f ◦ f )(x) = f ( f (x)) = f x +
=x+ +
=
x
x x + x1
=
Domain: R \ {0}
x2 + 1
x
(x 2 + 1)2 + x 2 x 4 + 3x 2 + 1
+ 2
=
=
x
x +1
x(x 2 + 1)
x(x 2 + 1)
¶
x +1
(g ◦ g )(x) = g (g (x)) = g
=
x +2
µ
=
½
¾
5
Domain: R \ − , −2
3
x+1
x+2
x+1
x+2
+1
+2
=
x +1+x +2
2x + 3
=
x + 1 + 2x + 4 3x + 5
...................................................................□
Inverse functions
5
If f (x) = x 5 + x 3 + x, find f −1 (3) and f ( f −1 (2))
L SOLUTION.
1. f (x) = 3 ⇔ x 5 + x 3 + x = 3 ⇔ x = 1 ⇒ f −1 (3) = 1
2. f ( f −1 (2)) = 2
40
40
VII. Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
6
Find a formula for the inverse of the following functions
1. f (x) =
p
2. f (x) = 1 + 2 + 3x.
4x − 1
.
2x + 3
3. f (x) = x 2 − x, x Ê
1
2
L SOLUTION.
1. f (x) =
4x − 1
4x − 1
. Let y =
⇔ 4x − 1 = (2x + 3)y ⇔ (2y − 4)x = −1 − 3y
2x + 3
2x + 3
⇔x=
−1 − 3y
·
2y − 4
−1 − 3x
2x − 4
p
p
p
2. f (x) = 1 + 2 + 3x. Let y = 1 + 2 + 3x ⇔ y − 1 = 2 + 3x ⇔ (y − 1)2 = 2 + 3x
Therefore, y = f −1 (x) =
⇔x=
Therefore, y = f −1 (x) =
(y − 1)2 − 2
·
3
(x − 1)2 − 2
3

p
1 + 4y
x=

1
p2
3. f (x) = x 2 − x, x Ê · Let y = x 2 − x ⇔ x 2 − x − y = 0 ⇔ 

1 + 1 + 4y
2
x=
2
p
1 + 1 + 4y
⇒x=
2
p
1
1 + 1 + 4x
−1
because x Ê · Therefore, y = f (x) =
2
2
1−
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
Symmetry
7
Determine whether f is even, odd, or neither
1. f (x) =
x
4. f (x) = x|x|
x2 + 1
x2
x4 + 1
x
3. f (x) =
x +1
5. f (x) = 1 + 3x 2 − x 4
2. f (x) =
6. f (x) = 1 + 3x 3 − x 5
L SOLUTION.
41
41
Chapter 1. Functions
1. f (x) =
x
x2 + 1
- odd f (−x) = − f (x)
2. f (x) =
x2
- even f (−x) = f (x)
x4 + 1
3. f (x) =
x
−x
- neither f (−x) =
x +1
−x + 1
4. f (x) = x|x| - odd f (−x) = − f (x)
5. f (x) = 1 + 3x 2 − x 4 - even f (−x) = f (x)
6. f (x) = 1 + 3x 3 − x 5 - neither f (−x) = 1 − 3x 3 + x 5
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
Elementary functions
8
Recent studies indicate that the average surface temperature of the earth has
been rising steadily. Some scientists have modeled the temperature by the linear
function T = 0.02t + 8.50, where T is temperature in o C and t represents years since
1900.
1. What do the slope and T −intercept represent?
2. Use the equation to predict the average global surface temperature in 2100
L SOLUTION.
1. The slope is m = 0.02 and T −intercept is 8.5
2. T (2100) = 0.02 × 2100 + 8.50 = 50.5o C
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
9
Find an expression for a cubic function f if f (1) = 6 and f (−1) = f (0) = f (2) = 0.
L SOLUTION.
Suppose f (x) = ax 3 + bx 2 + c x + d . Since f (1) = 6 and f (−1) = f (0) = f (2) = 0, we have


f (1) = a.13 + b.12 + c.1 + d = 6



f (−1) = a.(−1)3 + b.(−1)2 + c.(−1) + d = 0

f (0) = a.03 + b.02 + c.0 + d = 0



f (2) = a.23 + b.22 + c.2 + d = 0

a +b +c = 6
a = −3



−a + b − c = 0
b=3
⇔
⇔


8a + 4b + 2c = 0
c =6




d =0
d =0




Therefore f (x) = −3x 3 + 3x 2 + 6x.
.......................................................□
42
42
VII. Exercises
Limits of Sequences
10 Find the limit of the following sequences
p
n2 + 1 − n
4. lim p
p ·
n→∞ n + 1 − n
p
n2 + 1 − n
5. lim p
p ·
n→∞ n 3 + 1 − n n
p
p
4
n3 + n − n
6. lim
·
p
n→∞ n + 2 + n + 1
¶
n3
n2
− 2
·
1. lim
n→∞ n + 1
n +1
µ
(n + 1)4 − (n − 1)4
·
n→∞ (n 2 + 1)2 − (n 2 − 1)2
2. lim
3. lim
1
·
p
n 2 − 1 − n)
n→∞ n(
L SOLUTION.
¶
n2
n3
n2 − n3
n 2 (n 2 + 1) − n 3 (n + 1)
1. lim
− 2
=
lim
=
= lim
n→∞ n + 1
n→∞ (n + 1)(n 2 + 1)
n→∞
n +1
(n + 1)(n 2 + 1)
µ
= lim
n→∞ (1 +
1
−1
n
1
1
n )(1 + n 2 )
= −1.
(n + 1)4 − (n − 1)4
(n + 1 − n + 1)(n + 1 + n − 1)((n + 1)2 + (n − 1)2 )
=
=
lim
n→∞ (n 2 + 1)2 − (n 2 − 1)2
n→∞
(n 2 + 1 − n 2 + 1)(n 2 + 1 + n 2 − 1)
2. lim
2n(n 2 + 1)
= ∞.
n→∞
n2
= lim
p
1
n2 − 1 + n
= lim
= lim
3. lim
p
n→∞ n( n 2 − 1 − n)
n→∞ n(n 2 − 1 − n 2 )
n→∞
q
1 − n12 + 1
−1
= −2.
q
q
p
p
1
1
1
p
2
2
+
+
2
(n + 1 − n )( n + 1 + n)
n +1−n
n
n
n2
4. lim p
lim
= lim q
= 0.
p
p == n→∞
n→∞ n + 1 − n
n→∞
(n + 1 − n)( n 2 + 1 + n)
1 + n12 + 1
q
p
p
p
p
2
2
n + n13 + n
2
3
n +1−n
(n + 1 − n )( n + 1 + n n)
5. lim p
= ∞.
lim
= lim q
p
p = n→∞
n→∞ n 3 + 1 − n n
n→∞
1
(n 3 + 1 − n 3 )( n 2 + 1 + n)
1 + n2 + 1
q
q
p
1
1
p
4 1
4
n3 + n − n
n + n3 −
n
6. lim
= lim
= 0.
p
q
n→∞ n + 2 + n + 1
n→∞
1+ 2 + 1 + 1
n
n
n2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
43
43
Chapter 1. Functions
11 Find the limit of the following sequences
(−1)n .6n − 5n+1
n→∞ 5n − (−1)n .6n+1
1. lim
1 + 7n+2
n→∞ 3 − 7n
4. lim
2n+2 + 3n+3
2. lim
n→∞
2n + 3n
5. lim
2n + 3−n
n→∞ 2−n − 3n
5.2n − 3.5n+1
3. lim
n→∞ 100.2n + 2.5n
6. lim
(−1)n + n1
n→∞ 1
n2
− (−1)n
L SOLUTION.
1 + 7n+2
= lim
n→∞ 3 − 7n
n→∞
1. lim
1
+ 72
7n
3
−1
7n
2n+2 + 3n+3
2. lim
= lim
n→∞
n→∞
2n + 3n
4.2n
3n
2n
3n
5.2n − 3.5n+1
3. lim
= lim
n→∞ 100.2n + 2.5n
n→∞
1
= 0.
n→∞ 7n
= −49 since lim
+ 33
+1
2n
= 0.
n→∞ 3n
= 27 since lim
5.2n
− 3.5
5n
100.2n
5n + 2
=−
15
2n
since lim n = 0.
n→∞ 5
2
5.5n
1 − (−6)n
1
5n
(−1)n .6n − 5n+1
=
−
since
lim
4. lim n
lim
= 0.
n
n→∞ (−6)n
n→∞ 5 − (−1)n .6n+1 n→∞ 5
6
n −6
(−6)
2n + 3−n
= lim
5. lim −n
n→∞ 2
− 3n n→∞
6. lim
(−1)n + n1
n→∞ 1
n2
− (−1)n
1
2n
3n + 9 n
1
6n − 1
= lim
2n
1
1
=
lim
=
lim
= 0.
n→∞ 3n
n→∞ 9n
n→∞ 6n
= 0 vì lim
1 + (−1)
n
n
n→∞ (−1)
n2
n
−1
(−1)n
(−1)n
= lim
= 0.
n→∞
n→∞ n 2
n
= −1 since lim
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
2
Multiple-choice Questions
Domain and range of a function
s
Question 1 (L.O.1): Find the domain of function y =

3
x É−
A x >0
B 
2
x Ê0
3
3
D x É−
E − Éx <0
2
2
SOLUTION
44
44
µ
¶
6
ln 5 + .
 x
3
x É−
C 
2
x >0
VII. Exercises

µ
¶
3
6
6
4x + 6
x É−

The function is defined when ln 5 +
Ê 0 ⇒ 5+ Ê 1 ⇒
Ê0⇒
2
x
x
x
x >0
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Description of a function
Question 2 (L.O.2): A rectangular storage container with an open top has a volume of
19m 3 . The length of its base is triple its width. Material for the base costs 11 dollars per
square meter; material for the sides costs 8 dollars per square meter. Express the cost
of materials as a function of the width w of the base.
608
1216
1216
A 33w 2 +
B 24w 2 +
C 33w 2 +
3w
w
3w
1216
1672
2
2
D 33w +
E 24w +
3
3w
SOLUTION
Let w and 3w be the width and length of the base, respectively, and h be the height. The
total cost is
C = 11 × w × 3w + 8 × 2(w + 3w)h.
19
Using the fact that the volume is 19, we have w × 3w × h = 19 ⇒ h =
. Therefore,
3w 2
C (w) = 11 × w × 3w + 8 × 2(w + 3w) ×
1216
19
= 33w 2 +
2
3w
3w
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Inverse function
Question 3 (L.O.1): Let f (x) = x 5 +3x 3 +6x −5. What is the value of f −1 (63).
A 1
B 0
C 6
D 2
E 3
SOLUTION
We have f (x) = 5x + 9x + 6 > 0, ∀x ∈ R. Thus, f is one-to-one function and has inverse
function. Let f −1 (63) = x 0 ⇒ f (x 0 ) = 63 ⇒ x 05 + 3x 03 + 6x 0 − 5 = 63 ⇒ x 0 = 2.
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
′
4
2
45
45
Chapter 1. Functions
46
46
Chapter 2
The limit and continuity of a function
Learning Objectives
State some properties of the limit of a function.
Study some applications of limits.
Study the continuity of a function.
I
The limit of a function
1
Model real-world situations
According to the special theory of relativity developed by Albert Einstein, the length of a
moving object, as measured by an observer at rest, shrinks as its speed increases. If L 0 is
the length of the object when it is at rest, then its length L, as measured
by an observer at
s
rest, when traveling at speed v(m/s) is given by the formula L = L 0 .
1−
v2
, where c is the
c2
speed of light.
Question: If the space shuttle were able to approach the speed of light, what would happen
to its length L?
47
Chapter 2. The limit and continuity of a function
We need to find
s
lim L 0 .
v→c−0
2
1−
v
= L0.
c2
s
1−
c2
=0
c2
Remark 3
Conclusion: The closer the speed of the shuttle is to the speed of light, the closer
the length of the shuttle, as seen by an observer at rest, gets to 0.
2
Definitions
Let f (x) be defined on some open interval that contains the number a, except possibly at
a itself.
Definition 32
The number L ∈ R is called the limit of f (x) as x approaches a, and we write
lim f (x) = L
x→a
if for every number ε > 0 there is a number δ > 0 such that if 0 < |x − a| < δ then
| f (x) − L| < ε.
Remark 4
lim f (x) = L means that the values of f (x) can be made as close as we please to L by
x→a
taking x close enough to a (but not equal to a).
48
48
I. The limit of a function
3
Calculating Limits using the limit laws
Theorem I.1 (The limit laws)
Suppose that lim f (x) = A ∈ R and lim g (x) = B ∈ R. Then
x→a
x→a
1. lim [ f (x) ± g (x)] = A ± B
x→a
2. lim [c. f (x)] = c.A, where c is a constant
x→a
3. lim [ f (x).g (x)] = A.B
x→a
4. lim
x→a
f (x) A
= if B ̸= 0.
g (x) B
Example 31
x 3 − 5x + 4
x→3
x2 − 2
Evaluate lim
SOLUTION
x 3 − 5x + 4
=
lim
x→3
x2 − 2
lim (x 3 − 5x + 4)
x→3
lim (x 2 − 2)
=
x→3
33 − 5 × 3 + 4 16
=
32 − 2
7
Theorem I.2 (The squeeze theorem)
If
1. f (x) É g (x) É h(x) where x is near a (except possibly at a)
2. lim f (x) = L = lim h(x).
x→a
x→a
then lim g (x) = L.
x→a
Example 32
Evaluate I = lim x 2 . cos
x→0
1
x
SOLUTION
−1 É cos
49
49
1
É1
x
Chapter 2. The limit and continuity of a function
⇒ −x 2 É x 2 cos
1
É x2
x
and lim (−x 2 ) = lim x 2 = 0. Therefore I = 0.
x→0
4
x→0
One-sided limits
Definition 33
The number L ∈ R is called the limit of f (x) as x approaches a from the left if for
every number ε > 0 there is a number δ > 0 such that
if a − δ < x < a then | f (x) − L| < ε
Definition 34
The number L ∈ R is called the limit of f (x) as x approaches a from the right if for
every number ε > 0 there is a number δ > 0 such that
if a < x < a + δ then | f (x) − L| < ε
Theorem I.3
(
lim f (x) = L if and only if
x→a
lim f (x) = L
x→a+
lim f (x) = L
x→a−
50
50
II. Limits involving infinity. Asymptotes
Example 33
Evaluate lim f (x), where
x→0
f (x) = signx =


1,
0,

−1,
x >0
x =0
x <0
SOLUTION
lim f (x) = 1;
x→0+
lim f (x) = −1.
x→0−
It now follows that lim f (x) does not exist.
x→0
Example 34
Evaluate lim |x|
x→0
SOLUTION
lim |x| = lim x = 0;
x→0+
x→0+
lim |x| = lim −x = 0
x→0−
x→0−
Therefore, lim |x| = 0.
x→0
Example 35
½ p
Evaluate lim f (x) where f (x) =
x→4
x − 4, i f x Ê 4
8 − 2x, i f x < 4
SOLUTION
lim f (x) = lim
x→4+
x→4+
p
x − 4 = 0;
lim f (x) = lim (8 − 2x) = 0
x→4−
x→4−
Therefore, lim f (x) = 0.
x→4
II
1
Limits involving infinity. Asymptotes
Infinite limits.
Definition 35
Let f (x) be a function defined on some open interval that contains the number a,
except possibly at a itself. Then
lim f (x) = +∞
x→a
(2.1)
means that for every positive number M > 0 there is a positive number δ > 0 such
51
51
Chapter 2. The limit and continuity of a function
that
if 0 < |x − a| < δ then f (x) > M .
Remark 5
lim f (x) = +∞ means that the values of f (x) can be made arbitrarily large (larger
x→a
than any given number M ) by taking x close enough to a (but not equal to a).
Definition 36
Let f (x) be a function defined on some open interval that contains the number a,
except possibly at a itself. Then
lim f (x) = −∞
x→a
(2.2)
means that for every negative number N < 0 there is a positive number δ > 0 such
that
if 0 < |x − a| < δ then f (x) < N .
Remark 6
lim f (x) = −∞ means that the values of f (x) can be made arbitrarily small (smaller
x→a
than any negative number N ) by taking x close enough to a (but not equal to a).
52
52
II. Limits involving infinity. Asymptotes
Example 36
1
= +∞
x→0 x 2
1. lim
2. lim
x→3+
2x
= +∞
x −3
2x
= −∞
x→3− x − 3
3. lim
2
Vertical asymptote
Definition 37
The line x = a is called a vertical asymptote of the curve y = f (x) if at least one of
the following statements is true
lim f (x) = +∞; lim f (x) = +∞; lim f (x) = +∞
(2.3)
lim f (x) = −∞; lim f (x) = −∞; lim f (x) = −∞
(2.4)
x→a
x→a
x→a+
x→a−
x→a+
x→a−
53
53
Chapter 2. The limit and continuity of a function
Example 37
Find the vertical asymptotes of the curve of function f (x) =
2x
x −3
SOLUTION
2x
= +∞
x→3+ x − 3
lim f (x) = lim
x→3+
and
2x
= −∞
x→3− x − 3
lim f (x) = lim
x→3−
The line x = 3 is a vertical asymptote.
3
Limits at infinity
Definition 38
Let f be a function defined on some interval (a, +∞). Then
lim f (x) = L
x→+∞
means that for every ε > 0 there is a corresponding number N such that
if x > N then | f (x) − L| < ε
54
54
(2.5)
II. Limits involving infinity. Asymptotes
Remark 7
lim f (x) = L means that the values of f (x) can be made arbitrarily close to L
x→+∞
(within a distance ε, where ε is any positive number) by taking x sufficiently large
(larger than N , where N depends on ε).
Definition 39
Let f be a function defined on some interval (−∞, a). Then
lim f (x) = L
(2.6)
x→−∞
means that for every ε > 0 there is a corresponding number N such that
if x < N then | f (x) − L| < ε
Remark 8
lim f (x) = L means that the values of f (x) can be made arbitrarily close to L
x→−∞
(within a distance ε, where ε is any positive number) by taking x sufficiently small
(smaller than N , where N depends on ε).
Example 38
3x 2 − x − 2
x→∞ 5x 2 + 4x + 1
Evaluate lim
SOLUTION
Devide both the numerator and denominator by the highest power of x that occurs in the
denominator
3 − x1 − x22 3 − 0 − 0 3
3x 2 − x − 2
lim
= lim
=
=
x→∞ 5x 2 + 4x + 1
x→∞ 5 + 4 + 1
5
+
0
+
0
5
2
x
x
55
55
Chapter 2. The limit and continuity of a function
4
Horizontal asymptotes
Definition 40
The line y = L is called a horizontal asymptote of the curve y = f (x) if either
lim f (x) = L
(2.7)
lim f (x) = L.
(2.8)
x→+∞
or
x→−∞
Example 39
p
2x 2 + 1
Find the horizontal asymptotes of f (x) =
3x − 5
SOLUTION
q
p
p
p
2 + x12
2
2x + 1
2+0
2
= lim
=
1. lim
=
5
x→+∞ 3x − 5
x→+∞ 3 −
3−0
3
x
q
p
p
p
p
2 + t12
2x 2 + 1
2(−t )2 + 1
2+0
2
2. lim
=
lim
= lim
=
=
−
x→−∞ 3x − 5
t =−x,t →+∞ 3(−t ) − 5
t →+∞ −3 − 5
−3 − 0
3
t
p
p
2
2
Horizontal asymptotes: y =
, y =−
3
3
56
56
II. Limits involving infinity. Asymptotes
5
Slant asymptotes
Definition 41
If
h
i
lim f (x) − (mx + b) = 0
x→∞
(2.9)
then the line y = mx + b is called a slant asymptote.
Proposition 2.1
α(x) = f (x) − mx − b ⇒
α(x) f (x)
b x→∞
=
− m − −→ 0
x
x
x
Therefore
m = lim
x→∞
f (x)
;
x
h
i
b = lim f (x) − mx
x→∞
Example 40
Find the slant asymptotes of f (x) =
x3
x2 + 1
57
57
(2.10)
Chapter 2. The limit and continuity of a function
SOLUTION
1. m = lim
x→∞
f (x)
x3
1
= lim 2
= lim
=1
x→∞ (x + 1)x
x→∞ 1 + 1
x
2
x
µ
2. b = lim [ f (x) − mx] = lim
x→∞
= lim
− x1
x→∞ 1 + 1
x2
x→∞
¶
x3
x 3 − x(x 2 + 1)
−x
−
x
=
lim
=
lim
=
x→∞
x→∞ x 2 + 1
x2 + 1
x2 + 1
=0
So the line y = x is a slant asymptote.
6
Infinite limits at infinity
Definition 42
Let f be a function defined on some interval (a, +∞). Then
lim f (x) = +∞
x→+∞
(2.11)
means that for every M > 0 there is a corresponding number N > 0 such that
if x > N then f (x) > M
Remark 9
lim f (x) = +∞ means that the values of f (x) can be made arbitrarily large (larger
x→+∞
than M , where M is any positive number) by taking x sufficiently large (larger than
N , where N depends on M ).
Example 41
Find lim (x 2 − x)
x→+∞
SOLUTION It would be wrong to write
lim (x 2 − x) = lim x 2 − lim x = ∞ − ∞
x→+∞
x→+∞
x→+∞
The Limit Laws can not be applied to infinite limits because ∞ is not a number. However,
we can write
lim (x 2 − x) = lim x(x − 1) = ∞ × ∞ = ∞
x→+∞
x→+∞
58
58
III. Infinitely Small (infinitesimals) and Infinitely Large (infinities) Quantities
because both x and x − 1 become arbitrarily large and so their product does too.
Example 42
x2 + x
x→+∞ 3 − x
Find lim
SOLUTION We divide the numerator and denominator by the highest power of x in the
denominator
x2 + x
x + 1 +∞
= lim 3
= −∞.
=
x→+∞ 3 − x
x→+∞ − 1
−1
x
lim
III
Infinitely Small (infinitesimals) and Infinitely
Large (infinities) Quantities
1
Definition
α(x) 0
f (x) ∞
= , lim
=
x→a β(x)
0 x→a g (x) ∞
Consider the indeterminate form lim
Let y = f (x) be defined on X ⊂ R and a is a finite number or infinity.
Definition
The function f (x) is called an infinitesimal as x → a, if
lim f (x) = 0
The function f (x) is called an infinite
as x → a if
lim | f (x)| = +∞
(2.12)
x→a
x→a
(2.13)
Example 43
Which function among these functions is an infinitesimal?
1. f (x) = 2x 2 − 3 sin x is an infinitesimal as x → 0 since
lim f (x) = lim 2x 2 − 3 sin x = 0.
x→0
x→0
1
1
is NOT an infinitesimal as x → 0 since lim
= −1 ̸= 0. Howx→0 x − 1
x −1
1
ever, f (x) is an infinitesimal as x → ∞ since lim
= 0.
x→∞ x − 1
2. f (x) =
59
59
Chapter 2. The limit and continuity of a function
Example 44
Which function among these functions is an infinite?
1. f (x) = 2x 2 − 3 sin x is an infinite as x → ∞ since
lim |2x 2 − 3 sin x| = +∞.
x→∞
¯
¯
¯ 1 ¯
1
¯
¯ = +∞.
2. f (x) =
is an infinite as x → 1 since lim ¯
x→1 x − 1 ¯
x −1
Proposition 3.1
Using the infinitesimal to evaluate the limit of a function
(
lim g (x) = A ⇔
x→a
g (x) = A + f (x)
lim f (x) = 0
(2.14)
x→a
Example 45
2x + 3
·
x→∞ x + 1
Find I = lim
SOLUTION
1
2x + 3
= 2+
x +1
x +1
1
= 0 thus I = 2.
x→∞ x + 1
and lim
Theorem III.1
1. If f (x) is an infinite as x → a then
1
is an infinitesimal as x → a.
f (x)
2. If f (x) is an infinitesimal as x → a and f (x) ̸= 0, ∀x ∈ (a − ε, a + ε) then
1
is
f (x)
an infinite as x → a.
Example 46
The function f (x) = x is an infinite as x → ∞. Then,
x → ∞.
1
1
= is an infinitesimal as
f (x) x
Example 47
The function f (x) = x is an infinitesimal as x → 0. Then,
(−ε, ε) is an infinite as x → 0.
60
60
1
1
= where x ̸= 0, ∀x ∈
f (x) x
III. Infinitely Small (infinitesimals) and Infinitely Large (infinities) Quantities
Proposition 3.2
½
f (x) − i n f i ni t esi mal as x → a
⇒ f (x) ± g (x) − i n f i ni t esi mal as x → a
g (x) − i n f i ni t esi mal as x → a
½
f (x) −i n f i ni t esi mal as x → a
2o
⇒ f (x).g (x) − i n f i ni t esi mal as x → a
g (x) −i n f i ni t esi mal as x → a

| f (x)| É M = const ,

∀x ∈ (a − ε, a + ε) ⇒ f (x).g (x) − i n f i ni t esi mal as x → a
3o

g (x) − i n f i ni t esi mal as x → a
o
4 If f (x) is an infinitesimal as x → a then for all c ∈ R the product c. f (x) is also an
infinitesimal as x → a.
1
o
Definition 43
If lim f (x) = +∞ then f (x) is called an positive infinite, and if lim f (x) = −∞ then
x→a
x→a
f (x) is called an negative infinite.
Example 48
1
is an positive infinite as x → 0 + 0 and an negative infinite as
x
1
x → 0 − 0. The function g (x) = 2 is an positive infinite as x → 0 or as x → 0 + 0 or as
x
x → 0 − 0.
The function f (x) =
Proposition 3.3
1a o f (x), g (x) are positive infinites ( positive infinites) as x → a
⇒ f (x) + g (x) − i n f i ni t e as x → a


| f (x)| É M = const ,
∀x ∈ (a − ε, a + ε) ⇒ f (x) + g (x) − i n f i ni t e as x → a
1b

g (x) − i n f i ni t e as x → a
½
f (x) −i n f i ni t e as x → a
o
2
⇒ f (x).g (x) − i n f i ni t e as x → a
g (x) −i n f i ni t e as x → a

| f (x)| > M = const ,

∀x ∈ (a − ε, a + ε) ⇒ f (x).g (x) − i n f i ni t e as x → a
3o

g (x) − i n f i ni t e as x → a
4o If f (x) is an infinite as x → a then for all c ∈ R the product c. f (x) is also an infinite
as x → a.
o
Example 49
Functions x +sin x and x +cos x are infinites as x → ∞ since x is an infinite as x → ∞
and | sin x| É 1, | cos x| É 1.
NOTE. The sum of 2 infinites f (x), g (x) as x → a depends on the sign of infinites f (x), g (x).
61
61
Chapter 2. The limit and continuity of a function
Example 50
1. 2 functions f (x) = x, g (x) = 2x are infinites as x → ∞ and f (x) + g (x) = 3x is
also infinite as x → ∞.
2. 2 functions f (x) = x, g (x) = −x are infinites as x → ∞ however, f (x) + g (x) = 0
is an infinitesimal as x → ∞.
3. 2 functions f (x) = x + sin x, g (x) = −x are infinites as x → ∞ however, f (x) +
g (x) = sin x does not have limit as x → ∞.
Definition 44
Let f (x) and g (x) be infinitesimals as x → a, if
f (x)
= 0 then f (x) is an infinitesimal of higher order than g (x). Denote it
x→a g (x)
by f (x) = o(g (x)).
1. lim
2. lim
x→a
f (x)
= c ̸= 0(c ∈ R) then f (x), g (x) are infinitesimals of the same order.
g (x)
f (x)
= ∞ then f (x) is an infinitesimal of lower order than g (x).
x→a g (x)
3. lim
f (x)
does not exist (finite or infinity) then f (x), g (x) are called incompag (x)
rable infinitesimals.
4. lim
x→a
Definition 45
Let f (x) and g (x) be infinites as x → a, if
f (x)
= 0 thì f (x) is an infinite of lower order than g (x). Denote it by
x→a g (x)
f (x) << g (x).
1. lim
2. lim
x→a
f (x)
= c ̸= 0(c ∈ R) then f (x), g (x) are infinites of the same order.
g (x)
f (x)
= ∞ then f (x) is an infinite of higher order than g (x).
x→a g (x)
3. lim
f (x)
does not exist (finite or infinity) then f (x), g (x) are called incompag (x)
rable infinites.
4. lim
x→a
2
Some basic limits of the form
Proposition 3.4
62
62
0
0
III. Infinitely Small (infinitesimals) and Infinitely Large (infinities) Quantities
sin x
=1
x→0 x
1. lim
2. lim
loga (1 + x)
x→0
x
= loga e =
1
, (a > 0, a ̸= 1)
ln a
ln(1 + x)
=1
x→0
x
3. lim
ax − 1
= ln a, (a > 0, a ̸= 1)
x→0
x
4. lim
ex − 1
=1
x→0
x
5. lim
(1 + x)µ − 1
= µ, (µ ∈ R)
x→0
x
p
n
1+x −1 1
7. lim
= , (n ∈ N)
x→0
x
n
p
1+x −1 1
8. lim
=
x→0
x
2
6. lim
arcsin x
arctan x
= 1, lim
=1
x→0
x→0
x
x
9. lim
sinh x
cosh x − 1 1
= 1, lim
=
x→0
x→0
x
x2
2
10. lim
3
Some basic limits of the form
∞
∞
Proposition 3.5
1.
2.
xα
= 0 (a > 1)
x→+∞ a x
lim
lim
x→+∞
lnα x
xβ
= 0 (∀α > 0, β > 0)
Therefore, the order of infinites as x → +∞ is as follows
lnα x << x β << a x , (α, β > 0, a > 1)
Example 51
2 functions f (x) = x and g (x) = sin 2x are infinitesimals of the same order as x → 0
since
f (x)
x
1
2x
1
lim
= lim
= · lim
= ·
x→0 g (x)
x→0 sin 2x
2 x→0 sin 2x 2
63
63
Chapter 2. The limit and continuity of a function
Example 52
The function f (x) = 1 − cos x is an infinitesimal of higher order than g (x) = x as
x → 0 since
1 − cos x
sin2 (x/2)
f (x)
= lim
= lim
=
lim
x→0
x→0
x→0 g (x)
x
x/2
= lim
x→0
sin(x/2)
· lim sin(x/2) = 1 × 0 = 0.
x→0
x/2
Example 53
The function f (x) =
since
p
3
x is an infinitesimal of lower order than g (x) = x as x → 0
p
3
f (x)
1
x
lim
= lim
= lim p
= ∞.
3
x→0 g (x)
x→0 x
x→0 x 2
Example 54
2 functions f (x) = x sin(1/x) and g (x) = x are incomparable infinitesimals as x → 0
since
f (x)
x sin(1/x)
lim
= lim
= lim sin(1/x).
x→0 g (x)
x→0
x→0
x
and lim sin(1/x) does not exist.
x→0
Example 55
2 functions f (x) = 2x 2 and g (x) = x 2 + x are infinites of the same order as x → ∞
since
2x 2
2
f (x)
= lim 2
= lim
lim
= 2 ̸= 0.
x→∞ x + x
x→∞ 1 + 1/x 2
x→∞ g (x)
Example 56
The function f (x) = 2x is an infinite of higher order than g (x) = x as x → +∞ since
f (x)
2x
= lim
= +∞.
x→+∞ g (x)
x→+∞ x
lim
Example 57
2 functions f (x) = x(2 + sin x) and g (x) = x are incomparable infinites as x → ∞
since
f (x)
x(2 + sin x)
lim
= lim
= lim (2 + sin x).
x→∞ g (x)
x→∞
x→∞
x
and lim (2 + sin x) does not exist.
x→∞
64
64
III. Infinitely Small (infinitesimals) and Infinitely Large (infinities) Quantities
4
Equivalent functions
Equivalent functions
The infinitesimals f (x) and g (x) as x →
a are called equivalent if
lim
x→a
f (x)
=1
g (x)
The infinites f (x) and g (x) as x → a are
called equivalent if
(2.15)
lim
x→a
x→a
f (x)
=1
g (x)
(2.16)
x→a
Denote it by f (x) ∼ g (x).
Denote it by f (x) ∼ g (x).
Using equivalent functions to find limits
As x → a infinitesimals f (x) ∼ f (x) and
infinitesimals g (x) ∼ g (x). Then
lim
x→a
f (x)
f (x)
= lim
x→a
g (x)
g (x)
As x → a infinites f (x) ∼ f (x) and infinites g (x) ∼ g (x). Then
(2.17)
lim
x→a
if at least one of the 2 limits above exists.
f (x)
f (x)
= lim
x→a
g (x)
g (x)
if at least one of the 2 limits above exists.
Replacing infinitesimals of higher order
the sum of infinitesimals of different orders
=
x→a the sum of infinitesimals of different orders
the term whose order is lower in numerator
= lim
x→a the term whose order is lower in denominator
lim
Replacing infinites of lower order
the sum of infinites of different orders
=
x→a the sum of infinites of different orders
the term whose order is higher in numerator
= lim
x→a the term whose order is higher in denominator
lim
5
(2.18)
Some equivalent infinitesimals
Proposition 3.6
As x → 0 the following infinitesimals are equivalent
1
1. sin x ∼ x, tan x ∼ x, 1 − cos x ∼ x 2
2
65
65
Chapter 2. The limit and continuity of a function
2. arctan x ∼ x, arcsin x ∼ x
3. a x − 1 ∼ x. ln a, (a > 0, a ̸= 1), e x − 1 ∼ x
x
, (a > 0, a ̸= 1), ln(1 + x) ∼ x
ln a
p
x p
x
n
5. (1 + x)µ − 1 ∼ µ.x, (µ ∈ R), 1 + x − 1 ∼ , 1 + x − 1 ∼ , (n ∈ N)
2
n
4. loga (1 + x) ∼ x loga e =
6. sinh x ∼ x, cosh x − 1 ∼
6
x2
2
Evaluating limits using equivalent infinitesimals
Theorem III.2
If
½
u(x) → 0 as x → a
f (x) ∼ g (x) as x → 0
then f (u(x)) ∼ g (u(x)) as x → a.
Theorem III.3
If g (x) → A ̸= 0 and f (x) ∼ f (x) as x → a then f (x).g (x) ∼ A. f (x) as x → a
Example 58
p
sin 1 + x 3 − sin 1
Evaluate I = lim p
·
x→0 5 1 − 2x ln cos x − 1
SOLUTION
1
2x
x→0 1
x→0 2x
x→0
(1−2x ln cos x) 5 −1 ∼ (−2x ln cos x) = − ln(1+(cos x − 1)) ∼ − (cos x − 1) ∼
5
5
5
µ 2¶
2x
x
1
−
−
= x3
5
2
5
p
p
p
p
3 +1
3 −1
1
+
x
1
+
x
1 + x 3 − 1 x→0
x→0
sin
sin 1 + x 3 −sin 1 = 2 cos
∼ 2cos 1·
∼ 2 cos 1·
2
2
2
3
x /2 1
= cos 1.x 3
2
2
1
3
5
2 cos 1.x
Therefore, I = lim
= cos 1.
1
3
x→0
2
x
5
Example 59
2
e x − cos x
Evaluate I = lim
·
x→0
x2
SOLUTION
2
(e x − 1) + (1 − cos x)
x 2 + x 2 /2 3
I = lim
=
lim
= ·
x→0
x→0
x2
x2
2
66
66
IV. Continuity
7
Evaluating limits using equivalent infinites
Example 60
p
Evaluate I = lim
x→+∞
p
x 2 + 4 + 2x + 3 x
·
p
x2 − 4 + x
SOLUTION
p
p
x→+∞
x→+∞
As x → +∞ we have x 2 + 4 ∼ x, x 2 − 4 ∼ x so
3x 3
= ·
x→+∞ 2x
2
I = lim
Example 61
p
Evaluate I = lim
x→+∞
p
x2 + 4 − x2 + x
·
p
x2 + 1 − x
SOLUTION
p
−x.2x
(4 − x)( x 2 + 1 + x)
= lim
= lim −x = −∞
I = lim
p
p
x→+∞ 2x
x→+∞
x→+∞ (x 2 + 1 − x 2 )( x 2 + 4 + x 2 + x)
IV
1
Continuity
Continuity at a number
Definition 46
A function f (x) is continuous at a number a if
lim f (x) = f (a)
x→a
Proposition 4.1
If f (x) is continuous at a then:
1. f (a) is defined
2. lim f (x) exists
x→a
3. lim f (x) = f (a).
x→a
Proposition 4.2
67
67
(2.19)
Chapter 2. The limit and continuity of a function
The elementary functions such as polynomials, x α , sin x, cos x, a x , l og a x(x > 0) are
continuous at every number in their domains.
Example 62
Evaluate lim (x 3 − 5x 2 + 7x − 10)
x→3
SOLUTION
Since f (x) = x 3 − 5x 2 + 7x − 10 is a polynomial function, it is continuous at every number.
So the limit as x approaches 3 is the same as the value of f (x) at x = 3
lim (x 3 − 5x 2 + 7x − 10) = 33 − 5 × 32 + 7 × 3 − 10 = −7
x→3
Example 63
x 2 − 2x − 8
x→4 x 2 − 4x
Evaluate lim
SOLUTION
(x + 2)(x − 4)
x +2 4+2 3
x 2 − 2x − 8
= lim
= lim
=
=
2
x→4
x→4 x
x→4 x − 4x
x(x − 4)
4
2
lim
2
Continuity from one-side
Definition 47
A function f (x) is continuous from the right at a number a if
lim f (x) = f (a)
x→a+
(2.20)
Definition 48
A function f (x) is continuous from the left at a number a if
lim f (x) = f (a)
x→a−
(2.21)
Theorem IV.1
A function f (x) is continuous at a number a if and only if
lim f (x) = lim f (x) = f (a)
x→a+
x→a−
68
68
(2.22)
IV. Continuity
Theorem IV.2
1. If f (x) is continuous at every number on an open interval (a, b), then f (x) is
continuous on (a, b)
2. f (x) is continuous on the closed interval [a, b], if f (x) is continuous on the
open interval (a, b) and
lim f (x) = f (a); lim f (x) = f (b)
x→a+
x→b−
(2.23)
Example 64
Show that the function f (x) = 1 −
p
1 − x 2 is continuous on the interval [−1, 1].
SOLUTION
1. If −1 < a < 1, then we have
lim f (x) = lim (1 −
x→a
2.
x→a
p
p
1 − x 2 ) = 1 − 1 − a 2 = f (a)
lim f (x) = 1 = f (−1) and lim f (x) = 1 = f (1)
x→−1+
x→1−
Therefore, f is continuous on [−1, 1].
3
Limit of composite function
Theorem IV.3
If f is continuous at b and lim g (x) = b, then lim f (g (x)) = f (b). In other words,
x→a
x→a
³
´
lim f (g (x)) = f lim g (x)
x→a
(2.24)
x→a
Corollary 4.2
Suppose that g is continuous at a and f is continuous at g (a). Then, the composition
f ◦ g is continuous at a.
³
´
lim ( f ◦ g )(x) = lim f (g (x)) = f lim g (x) = f (g (a)) = ( f ◦ g )(a)
x→a
x→a
x→a
Example 65
lim sin(x 2 + 2x + 3) = sin 3 because lim (x 2 + 2x + 3) = 3 and lim sin y = sin 3.
x→0
x→0
y→3
69
69
(2.25)
Chapter 2. The limit and continuity of a function
Example 66
Determine where h(x) = cos(x 2 − 5x + 2) is continuous.
SOLUTION
We have h(x) = f (g (x)), where g (x) = x 2 − 5x + 2 and f (x) = cos x. Since both f and g are
continuous for all x, h is continuous for all x.
V
1
Discontinuity
Discontinuity at a number
Definition 49
Function f (x) is discontinuous at a if f (x) is not continuous at a.
Remark 10
If f (x) is discontinuous at a then at least one of the following equality is not true
lim f (x) = lim f (x) = f (a).
x→a+
x→a−
This means that:
1. at least one of the following limits lim f (x) and lim f (x) does not exist or is
x→a+
x→a−
equal ∞.
2. both limits lim f (x) and lim f (x) exist but at least one of the above equality
is not true.
2
x→a+
x→a−
Removable discontinuity
Definition 50
A function f has a removable discontinuity at a if lim f (x) and lim f (x) exist and
x→a+
x→a−
either f (a) is undefined or
lim f (x) = lim f (x) ̸= f (a).
x→a+
x→a−
70
70
(2.26)
V. Discontinuity
Example 67
Find all discontinuities of function
½
f (x) =
|x|,
1,
x ̸= 0
x =0
SOLUTION
lim f (x) = lim |x| = lim x = 0
x→0+
x→0+
x→0+
lim f (x) = lim |x| = lim (−x) = 0.
x→0−
x→0−
x→0−
So lim f (x) = lim f (x) = 0 ̸= 1 = f (0). Therefore, f has a removable discontinuity at a = 0.
x→0+
3
x→0−
Jump discontinuity
Definition 51
A function f has a jump discontinuity at a if lim f (x) and lim f (x) exist and
x→a+
x→a−
lim f (x) ̸= lim f (x).
x→a+
x→a−
Example 68
Find all discontinuities of function


1,
0,
f (x) = sign(x) =

−1,
x >0
x =0
x <0
SOLUTION
lim f (x) = lim 1 = 1
x→0+
x→0+
lim f (x) = lim (−1) = −1.
x→0−
x→0−
So lim f (x) ̸= lim f (x). Therefore, f has a jump discontinuity at a = 0.
x→0+
x→0−
71
71
(2.27)
Chapter 2. The limit and continuity of a function
4
Infinite discontinuity
Definition 52
A function f has a infinite discontinuity at a if at least one of the limits lim f (x) or
x→a+
lim f (x) is equal ∞.
x→a−
Example 69
Find all discontinuities of function

 1
,
f (x) =
 x0,
x ̸= 0
x =0
SOLUTION
1
= +∞
x→0+ x
1. lim f (x) = lim
x→0+
1
= −∞.
x→0− x
2. lim f (x) = lim
x→0−
f has a infinite discontinuity at a = 0.
VI
1
Exercises
Essay Questions
Limit of a function
72
72
VI. Exercises
12 Evaluate the following limits
1. lim (x 4 − 3x)(x 2 + 5x + 3)·
5. lim
x→−1
s
2. lim
x→2
p
x→0
2x 2 + 1
3x − 2
6. lim+
x→0
x 3 + x 2 sin
p
π
x
¸
·
2 2π
x 1 + sin
x
2−x
x→1 (x − 1)2
x 2 + 2x − 8
x→2 x 2 − 5x + 6
p
4x + 1 − 3
4. lim
x→2
x −2
7. lim
3. lim+
8. lim−
x→2
x 2 − 2x
x 2 − 4x + 4
SOLUTION
1. We have
lim [(−1)4 − 3.(−1)][(−1)2 + 5.(−1) + 3] = 4 × (−1) = −4
x→−1
2. We have
s
lim
x→2
3. We have
lim+
x→2
2x 2 + 1
=
3x − 2
2 × 22 + 1 3
=
3×2−2
2
x 2 + 2x − 8
(x − 2)(x + 4)
= lim+
2
x − 5x + 6 x→2 (x − 2)(x − 3)
= lim+
x→2
4. We have
s
x +4 2+4
=
= −6
x −3 2−3
p
4x + 1 − 3
4x + 1 − 9
lim
= lim
p
x→2
x→2 (x − 2)( 4x + 1 + 3)
x −2
4 2
4(x − 2)
4
= =
=p
p
x→2 (x − 2)( 4x + 1 + 3)
4×2+1+3 6 3
= lim
5. We have
Since lim
x→0
¯p
π ¯¯ p
¯
0 É ¯ x 3 + x 2 sin ¯ É x 3 + x 2
x
p
x 3 + x 2 = 0 then
¯p
π ¯¯
¯
3
2
lim ¯ x + x sin ¯ = 0
x→0
x
In addition, we have
¯p
π ¯¯ p 3
π ¯¯p 3
π ¯¯
¯
3
2
2
2
− ¯ x + x sin ¯ É x + x sin É ¯ x + x sin ¯
x
x
x
p
π
So lim x 3 + x 2 sin = 0.
x→0
x
73
73
Chapter 2. The limit and continuity of a function
6. We have
2π
1 É 1 + sin2
É2
x
¸
·
p
p
p
2 2π
⇒ x É x 1 + sin
É2 x
x
p
p
Since lim+ x = 0 and lim+ 2 x = 0 then
x→0
x→0
lim
x→0+
7. We have
p
·
¸
2 2π
x 1 + sin
=0
x
2−x
1
= + = +∞
2
x→1 (x − 1)
0
lim
8. We have
lim−
x→2
x 2 − 2x
x(x − 2)
=
lim
=
x 2 − 4x + 4 x→2− (x − 2)2
2
x
= − = −∞
= lim−
x→2 x − 2
0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
One-sided limit
13 Evaluate the following limits
1. lim−
x→1
x −1
|x 3 − x 2 |
2 − |x|
x→−2 2 + x
2. lim
SOLUTION
1. We have
x −1
x −1
= lim− 2
=
3
2
x→1
|x − x |
x |x − 1|
x −1
1
1
= lim− 2
= lim−
=
= −1
2
x→1 x (1 − x)
x→1 −x
−1
lim
x→1−
2. We have
2 − |x|
2+x
= lim
=1
x→−2 2 + x
x→−2 2 + x
lim
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
Asymptotes
14
x2 + 1
Find the vertical asymptotes of the function y =
3x − 2x 2
74
74
VI. Exercises
SOLUTION
½
¾
3
Domain: D = R \ 0,
2
x2 + 1
1
= =∞
2
x→0 3x − 2x
0
2
x +1
9/4 + 1
lim
=
=∞
2
x→3/2 3x − 2x
0
3
Vertical asymptotes: x = 0, x = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
2
lim
15
Find the vertical and horizontal asymptotes of the function y =
2x 2 + x − 1
x2 + x − 2
SOLUTION
Domain: D = R \ {1, −2}
2x 2 + x − 1 2
2x 2 + x − 1 5
=
=
∞,
lim
= =∞
x→1 x 2 + x − 2
x→−2 x 2 + x − 2
0
0
lim
Vertical asymptotes: x = 1, x = −2.
2 + x1 − x12 2
2x 2 + x − 1
=
lim
= =2
x→∞ 1 + 1 − 2
x→∞ x 2 + x − 2
1
2
x
lim
x
Horizontal asymptote: y = 2
............................................................ □
16
Find the asymptotes of the function y =
x2 + 1
x +1
SOLUTION
1. Domain: D = R \ {−1}
x2 + 1 2
= =∞
x→−1 x + 1
0
lim
Vertical asymptote: x = −1.
2. We have
x + x1 ∞
x2 + 1
lim
= lim
=
=∞
x→∞ x + 1
x→∞ 1 + 1
1
x
Horizontal asymptote does not exist.
3. Slant asymptote y = mx + b
m = lim
x→∞
1 + x12
f (x)
x2 + 1
= lim
= lim
=1
x→∞ (x + 1)x
x→∞ 1 + 1
x
2
x
2
¶
x +1
b = lim [ f (x) − mx] = lim
−x =
x→∞
x→∞ x + 1
µ
1
−1
x2 + 1 − x2 − x
1−x
= lim
= lim x 1 = −1
x→∞
x→∞ x + 1
x→∞ 1 +
x +1
x
= lim
Slant asymptote y = x − 1.
75
75
Chapter 2. The limit and continuity of a function
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
Continuity of a function
17 Find the value a that makes function f continuous at x0 = 0


1
x sin , x ̸= 0
f (x) =
x

a, x = 0.
SOLUTION
¯
¯
¯
1 ¯¯
¯
Since 0 É ¯x sin ¯ É |x| so
x
¯
¯
¶
µ
¯
1
1 ¯¯
¯
= 0.
lim x sin ¯ = 0 ⇒ lim x sin
x→0 ¯
x→0
x
x
In order to function f is continous at x 0 = 0 we have a = 0.
............................ □
18 Find the value a that makes function f continuous at x0 = 0
ax 2 + 1, x > 0
−x, x É 0.
½
f (x) =
SOLUTION
We have
lim f (x) = lim+ (ax 2 + 1) = 1 ̸= lim− f (x) =
x→0+
x→0
x→0
= lim− (−x) = 0 = f (0).
x→0
Answer: a does not exist
................................................................ □
19 Find the value a that makes function f continuous at x0 = 0
½
f (x) =
cos x, x É 0
a(x − 1), x > 0.
SOLUTION
We have
lim f (x) = lim− cos x = 1 = f (0)
x→0−
x→0
and
lim f (x) = lim+ (a(x − 1)) = −a = f (0).
x→0+
x→0
In order to function f is continuous at x 0 = 0 we have −a = 1 or a = −1.
...............□
20 Find the values a, b that make function f continuous on its domain

3
x É0
 (x − 1) ,
ax + b, 0 < x < 1
f (x) =
p

x,
x Ê 1.
76
76
VI. Exercises
SOLUTION
We have
lim f (x) = lim+ ax + b = b
x→0+
x→0
and
lim f (x) = lim− (x − 1)3 = −1 = f (0).
x→0−
x→0
In order to function is continuous at x 0 = 0 we have b = −1
lim+ f (x) = lim+
x→1
p
x→1
x = 1 = f (1)
and
lim f (x) = lim− ax + b = a + b.
x→1−
x→1
In order to function is continuous at x 0 = 1 we have a + b = 1 ⇒ a = 2 (since b = −1)
Answer: a = 2, b = −1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
21 Find the values a, b that make function f continuous on its domain
½
f (x) =
x, |x| É 1
x + ax + b, |x| > 1.
2
SOLUTION
1. We have
lim f (x) = lim+ x 2 + ax + b = 1 + a + b
x→1+
x→1
and
lim f (x) = lim− x = 1 = f (1).
x→1−
x→1
In order to function is continuous at x 0 = 1 we have a + b + 1 = 1 or a + b = 0.
2. Moreover,
lim
x→(−1)+
f (x) =
lim x = −1 = f (−1)
x→(−1)+
and
lim
x→(−1)−
f (x) =
lim x 2 + ax + b = 1 − a + b.
x→(−1)−
In order to function is continuous at x 0 = −1 we have 1 − a + b = −1 or −a + b = −2
3. Solve the simultaneous of equations
½
a +b = 0
⇔ a = 1, b = −1.
−a + b = −2
Answer: a = 1, b = −1.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
77
77
Chapter 2. The limit and continuity of a function
2
Multiple-choice Questions
Limits
µ
Question 1 (L.O.1): Find m
µ
¶
ln(1 + 6.04x)
lim
.m = 18.62.
x→0
x
A 1.6481
B 1.2026
>
0
such
that
C 1.2978
¶
e 3.19x − cos(x)
.m 2 +
lim
x→0
x
D 1.3338
E 2.3985
SOLUTION
Using the L’Hospital rule, we have
e 3.19x − cos(x)
3.19e 3.19x + sin(x)
= lim
= 3.19,
x→0
x→0
x
1
lim
and
ln(1 + 6.04x)
lim
= lim
x→0
x→0
x
6.04
1+6.04x
1
= 6.04.
Therefore, we receive 3.19m 2 +6.04m = 18.62. Choosing m > 0, we get m = 1.64814172658632 ≈
1.6481.
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
¶6x
6x + m 2
Question 2 (L.O.1): Find m > 1 such that lim
= e 26.55m .
x→∞ 6x − 1
A 26.5123
B 26.5475
C 27.3814
D 26.8591
E 26.4543
µ
SOLUTION
We have
6x + m 2
lim
x→∞ 6x − 1
µ
¶6x
m2 + 1
= lim 1 +
x→∞
6x − 1
µ
2 +1)
¶ 6x−1
· 6x(m
2
6x−1
m +1
= em
2
+1
= e 26.55m ⇒ m 2 + 1 = 26.55m.
Choosing m > 1, we receive m = 26.5122816318155 ≈ 26.5123.
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
³
´ 1
2
Question 3 (L.O.1): Find lim 1 + 2x 2 e 7x 9x .
x→0
A e −7/9
B 1
C −e 2/9
D 0
E e 2/9
SOLUTION
³
´ 1
2 7x
ln(1
+
2x
e )
2
Let y = 1 + 2x 2 e 7x 9x ⇒ ln y =
· Then
2
9x
ln(1 + 2x 2 e 7x )
ln(1 + 2x 2 e 7x )
2x 2 e 7x
2 2
lim ln y = lim
=
lim
×
lim
=
1
×
=
x→0
x→0
x→0
x→0 9x 2
9x 2
2x 2 e 7x
9 9
Therefore, lim y = lim e ln y = e 2/9
x→0
x→0
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
78
78
VI. Exercises
Question 4 (L.O.1): Find S = 4a + 8b where a and b are the two real constants such
that
³
´
2
lim x e a/x + + b = 15.
x→+∞
x
A 43
B 40
C 48
D 44
E 42
SOLUTION
1
e at + 2 × t + b
Let t = then I = lim+
= 15. The constant b must satisfy the equation
t →0
x
t
e a×0 + 2 × 0 + b = 0 ⇒ b = −1.
Otherwise, I = ∞. When b = −1, using L’ Hospital rule, we obtain
ae at + 2
I = lim+
= a + 2 = 15 ⇒ a = 15 − 2 = 13.
t →0
1
Therefore, S = 4 × 13 − 1 × 8 = 44
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
e 4ax − 1 − 9x
·
x→0
x + 13x 2
9
B 4a − 9, ∀a ̸=
C 4a, ∀a ∈ R
4
E 4a − 9, ∀a ∈ R
Question 5 (L.O.1): Evalulate I = lim
9
4
D 4a − 9, ∀a ̸= 0
A 4a, ∀a ̸=
SOLUTION
Using the L’Hospital rule, we have
(4a − 9)x
= 4a − 9.
x→0
x
I = lim
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Asymptotes
Question 6 (L.O.1): Find all real numbers a so that the line of equation x = 0 is a
ln(7x + a − 8)
vertical asymptote of the curve y =
·
x +9
A 11
B 7
C 6
D 8
E 12
SOLUTION
The line of equation x = 0 is a vertical asymptote of the curve y =
ln(7x + a − 8)
when
x +9
ln(7x + a − 8) ln(7 × 0 + a − 8) ln(a − 8)
=
=
= ∞.
x→0
x +9
0+9
9
lim
Therefore, a = 8.
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
79
79
Chapter 2. The limit and continuity of a function
Question 7 (L.O.1): Find all asymptotes of the graph of the function y = 4x +
9x
− 5.
p
1 − e −x
A x = 0, y = 13x − 5
B y = 13x − 5
C y = 9x − 5
D x = 0, y = 4x − 5
E y = 4x − 5
SOLUTION
Domain: x > 0. We have
9x
9x
− 5 = −5 + lim+ p
= −5.
lim+ f (x) = lim+ 4x + p
−x
x→0
x→0
x→0
1−e
1 − e −x
The line x = 0 is not the vertical asymptote of the curve y = f (x).
On the other hand,
9x
− 5 = +∞.
lim 4x + p
x→+∞
1 − e −x
Thus, the function does not have horizontal asymptote.
Consider the asymptote of the form y = mx + b where
m = lim
x→+∞
9x
1−e −x
4x + p
x
−5
= lim 4 −
x→+∞
5
9
+p
= 13
x
1 − e −x
and
9x
9x
9x
b = lim 4x+ p
−5−mx = lim 4x+ p
−5−13x = lim −5+ p
−9x = −5.
x→+∞
x→+∞
x→+∞
1 − e −x
1 − e −x
1 − e −x
Therefore, the function has only 01 slant asymptote y = 13x − 5.
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Continuity
Question 8 (L.O.1): Find
real
values
of
a
such
½
ax − 2, if x É 13
p
is continuous at x = 13.
arctan( x − 13), if x > 13
2
24
37
50
A
B −
D −
C −
13
13
13
13
f (x)
that
E −
=
11
13
SOLUTION
The function f is continuous at x = 13 if
lim+ f (x) = 0 = lim− f (x) = f (13) = 13 × a − 2 ⇒ a =
x→13
x→13
2
13
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question
9 (L.O.1): Find
real
values
of
a
such

x 3 − 31

5 arctan
if x ̸= 3 is continuous at x = 3.
x −3

a, if x = 3
5π
3π
5π
A −
B
C
2
2
2
7π
D a does not exist
E
2
80
80
that
f (x)
=
VI. Exercises
SOLUTION
The function f is continuous at x = 3 if
lim f (x) = lim− f (x) = f (3)
x→3+
We have
lim+ f (x) = −
x→3
x→3
5π
5π
̸= lim− f (x) =
x→3
2
2
Therefore, there is no exist a such that f is continuous at x = 3.
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
81
81
Chapter 2. The limit and continuity of a function
82
82
Chapter 3
Derivatives and Differentials
Learning Objectives
State basic rules for calculating derivatives of a function.
Study some applications of differentiation.
Study parametric curve.
I
Derivatives
1
Tangents
Definition 53
The tangent line to the curve y = f (x) at the point P (a, f (a)) is the line through P
with slope
m = lim
x→a
f (x) − f (a)
x −a
provided that this limit exists.
Example 70
Find an equation of the tangent line to the parabola y = x 2 at the point P (1, 1).
83
(3.1)
Chapter 3. Derivatives and Differentials
SOLUTION The slope of tangent line to the parabola y = x 2 is
m = lim
x→1
f (x) − f (1)
x2 − 1
= lim
=
x→1 x − 1
x −1
(x − 1)(x + 1)
= lim (x + 1) = 1 + 1 = 2
x→1
x→1
x −1
The equation of the tangent line at (1, 1) is
= lim
y − 1 = 2(x − 1) ⇔ y = 2x − 1
2
Velocities
Suppose an object moves along a straight line according to an equation of motion s = f (t ),
where s is the directed distance of the object from the origin at the time t . In the time
interval from t = a to t = a+h the change in position is f (a+h)− f (a). The average velocity
over this time interval is
average velocity =
f (a + h) − f (a)
h
Now suppose we compute the average velocities over shorter and shorter time intervals
[a, a + h]. We let h approach 0. The instantaneous velocity v(a) at time t = a is defined by
v(a) = lim
h→0
f (a + h) − f (a)
h
(3.2)
Example 71
Suppose that a ball is dropped from the upper observation deck of CN Tower, 450 m
above the ground
1. What is the velocity of the ball after 5 seconds?
2. How fast is the ball travelling when it hits the ground?
SOLUTION
1
1. The equation of motion s = f (t ) = .g .t 2 = 4.9t 2 , where g − acceleration of gravity
2
f (a + h) − f (a)
4.9(a + h)2 − 4.9a 2
4.9(a 2 + 2ah + h 2 − a 2 )
= lim
= lim
=
h→0
h→0
h→0
h
h
h
4.9(2ah + h 2 )
lim
= = lim 4.9(2a + h) = 9.8a
h→0
h→0
h
2. v(a) = lim
84
84
I. Derivatives
3. The velocity after 5s is v(5) = 9.8 × 5 = 49m/s
4. Since the observation deck is 450 m above the ground,
r the ball will hit the ground
450
at the time t 1 when s(t 1 ) = 450 ⇒ 4.9.t 12 = 450 ⇒ t 1 =
≈ 9.6s ⇒ v(t 1 ) = 9.8t 1 ≈
4.9
94m/s. (the velocity of the ball as it hits the ground)
3
Definitions
Definition 54
The derivative of a function f at a number a, denoted by f ′ (a), (read: f prime of a)
is
f ′ (a) = lim
h→0
f (a + h) − f (a)
h
(3.3)
if this limit exists.
Remark 11
¯
d y ¯¯
d
Other Notations: f (a) = y (a) =
=
f (a).
¯
d x x=a d x
′
′
Example 72
Find the derivative of the function f (x) = x 2 − 8x + 9 at the number a
SOLUTION
f ′ (a) = lim
h→0
f (a + h) − f (a)
=
h
[(a + h)2 − 8(a + h) + 9] − (a 2 − 8a + 9)
=
h→0
h
= lim
a 2 + 2ah + h 2 − 8a − 8h + 9 − a 2 + 8a − 9
=
h→0
h
= lim
2ah + h 2 − 8h
= lim (2a + h − 8) = 2a − 8
h→0
h→0
h
= lim
Definition 55
The left-hand derivative of y = f (x) at a number a is the limit (if this limit exists)
f −′ (a) = lim
x→a−
f (x) − f (a)
x −a
(3.4)
The right-hand derivative of y = f (x) at a number a is the limit (if this limit exists)
f +′ (a) = lim
x→a+
f (x) − f (a)
x −a
85
85
(3.5)
Chapter 3. Derivatives and Differentials
Theorem I.1
A function y = f (x) is differentiable at a if and only if the left-hand and the righthand derivatives of f at a exist and are equal.
f ′ (a) = f −′ (a) = f +′ (a)
(3.6)
Definition 56
A function y = f (x) is differentiable on an open interval (a, b) [or (a, ∞) or (−∞, a)
or (−∞, ∞)] if it is differentiable at every number in the interval.
Example 73
½
Where is the function y = f (x) = |x| =
x, x Ê 0
not differentiable?
−x, x < 0
SOLUTION
x
|x| − |0|
= lim+ = 1
x→0 x
x→0
x −0
|x| − |0|
−x
f −′ (0) = lim−
= lim−
= −1
x→0
x→0 x
x −0
Since f +′ (0) = 1 ̸= −1 = f −′ (0), f ′ (0) does not exist. Thus f is not differentiable at a = 0.
f +′ (0) = lim+
1. When a > 0 we have
|x| − |a|
x −a
= lim
=1
x→a x − a
x→a x − a
f ′ (a) = lim
2. When a < 0 we have
|x| − |a|
−x − (−a)
= lim
= −1
x→a x − a
x→a
x −a
f ′ (a) = lim
Conclusion: f is not differentiable at a = 0.
4
The derivative of elementary functions
Proposition 1.1
1. Derivative of a constant function y = C = const ⇒ y ′ = 0.
2. Derivatives of power functions y = x α (x ̸= 0) ⇒ y ′ = αx α−1 .
Special cases:
a) y = x ⇒ y ′ = 1.
1
1
b) y = ⇒ y ′ = − 2 ·
x
x
p
1
c) y = x ⇒ y ′ = p ·
2 x
86
86
I. Derivatives
d) y =
p
n
x ⇒ y′ =
1
·
p
n
n x n−1
3. Derivatives of exponential functions y = a x (a > 0, a ̸= 1) ⇒ y ′ = a x ln a.
Special case: y = e x ⇒ y ′ = e x , since ln e = 1
4. Derivatives of logarithmic functions y = loga |x| (a > 0, a ̸= 1) ⇒ y ′ =
1
since ln e = 1
x
Derivatives of trigonometric functions
Special case: y = ln |x| ⇒ y ′ =
5. y = sin x ⇒ y ′ = cos x.
6. y = cos x ⇒ y ′ = − sin x.
7. y = tan x ⇒ y ′ =
1
cos2 x
1
8. y = cot x ⇒ y ′ = −
sin2 x
Derivatives of inverse trigonometric functions
1
9. y = arcsin x, (x ∈ (−1, 1)) ⇒ y ′ = p
1 − x2
1
10. y = arccos x, (x ∈ (−1, 1)) ⇒ y ′ = − p
1 − x2
11. y = arctan x ⇒ y ′ =
1
1 + x2
1
1 + x2
Derivatives of hyperbolic functions
12. y = ar ccot x ⇒ y ′ = −
13. y = sinh x ⇒ y ′ = cosh x
14. y = cosh x ⇒ y ′ = sinh x
15. y = tanh x ⇒ y ′ =
1
cosh2 x
16. y = coth x ⇒ y ′ = −
5
1
sinh2 x
Differentiation rules
Proposition 1.2
87
87
1
·
x ln a
Chapter 3. Derivatives and Differentials
1. The constant multiple rule
y = cu = cu(x) ⇒ y ′ = cu ′ (x), c − const .
(3.7)
2. The sum (difference) rule
y = u(x) ± v(x) ⇒ y ′ = u ′ (x) ± v ′ (x)
(3.8)
y = u(x).v(x) ⇒ y ′ = u ′ (x).v(x) + u(x).v ′ (x)
(3.9)
3. The product rule
4. The quotient rule
y=
6
u(x)
u ′ (x).v(x) − u(x).v ′ (x)
⇒ y′ =
v(x)
v 2 (x)
(3.10)
The chain rule
Theorem I.2 (The chain rule)
If function u = u(x) is differentiable at x and function y = f (u) is differentiable at
u(x) then the composite function y = f ◦ u = f (u) = f (u(x)) is differentiable at x and
y ′ is given by the product
y ′ (x) = f ′ (u(x)).u ′ (x).
Example 74
Differentiate y = sin5 (4x + 3)
SOLUTION
y ′ = 5 sin4 (4x + 3).[sin(4x + 3)]′ =
= 5 sin4 (4x + 3). cos(4x + 3).(4x + 3)′ =
= 20 sin4 (4x + 3) cos(4x + 3).
88
88
(3.11)
II. Higher derivatives
II
1
Higher derivatives
The second derivative
Definition 57
If f is a differentiable function, then its derivative f ′ is also a function. If f ′ (x)
has derivative on the interval (a, b) then the derivative of f ′ (x) is called the second
derivative of f (x). It is denoted by
f ′′ (x) = ( f ′ (x))′
(3.12)
Example 75
If f (x) =
2x + 3
, then find f ′′ (0).
x −2
SOLUTION
f ′ (x) =
−7
14
7
⇒ f ′′ (x) =
⇒ f ′′ (0) = −
2
3
(x − 2)
(x − 2)
4
Example 76
If s = s(t ) is the position function of an object that moves in a straight line, we know
that its first derivative represents the velocity v(t ) of the object as a function of time:
v(t ) = s ′ (t )
The instantaneous rate of change of velocity with respect to time is called the acceleration a(t ) of the object. Thus the acceleration function is the derivative of the
velocity function and is therefore the second derivative of the position function:
a(t ) = v ′ (t ) = s ′′ (t )
2
The n−th derivative
Definition 58
The n−th derivative of f (x) is obtained from f by differentiating n times.
f (n) (x) = ( f (n−1) (x))′ , n ∈ N.
(3.13)
Corollary 2.0
1. If f (x) and g (x) have n−th derivatives then c 1 f (x) + c 2 g (x), c 1 , c 2 ∈ R also has
89
89
Chapter 3. Derivatives and Differentials
n−th derivative and
(c 1 f (x) + c 2 g (x))(n) = c 1 f (n) (x) + c 2 g (n) (x)
(3.14)
2. (Leibniz’s formula). If f (x) and g (x) have n−th derivatives then f (x).g (x) also
has n−th derivative and
(n)
( f (x).g (x))
=
n
X
C nk f (n−k) (x)g (k) (x).
k=0
3
Some basic formulas
Corollary 2.0
1. (a x )(n) = a x . lnn a.
2. (e x )(n) = e x
³
nπ ´
3. (sin ax)(n) = a n sin ax +
2
³
nπ ´
4. (cos ax)(n) = a n cos ax +
2
5. ((ax + b)α )(n) = a n α(α − 1) . . . (α − n + 1)(ax + b)α−n
6. (loga |x|)(n) =
7. (ln |x|)(n) =
(−1)n−1 (n − 1)!
x n ln a
(−1)n−1 (n − 1)!
xn
Example 77
Find the n−th derivative of f (x) =
1
x2 − 4
SOLUTION
µ
¶
1
1
1
1
=
−
x2 − 4 4 x − 2 x + 2
It implies
µ
1
2
x −4
¶(n)
µ
¶
µ
¶
1
1 (n) 1
1 (n)
=
−
4 x −2
4 x +2
By substituting α = −1, a = 1, b = ±2, we have
µ
1
x ±2
¶(n)
= (−1)(−2) . . . (−1 − n + 1)(x ± 2)−1−n =
90
90
(−1)n n!
(x ± 2)n+1
(3.15)
III. Linear approximations and Differentials
Therefore
f
(n)
µ
¶
(−1)n n!
1
1
(x) =
−
4
(x − 2)n+1 (x + 2)n+1
Example 78
Find the n−th derivative of f (x) = x 2 cos 2x.
SOLUTION Using Leibniz’s formula, we have
(x 2 . cos 2x)(n) = C n0 x 2 (cos 2x)(n) +C n1 (x 2 )′ (cos 2x)(n−1) +C n2 (x 2 )′′ (cos 2x)(n−2)
On another hand, we have
³
nπ ´
(cos 2x)(n) = 2n cos 2x +
,
2
¶
µ
³
nπ ´
(n − 1)π
= 2n−1 sin 2x +
,
(cos 2x)(n−1) = 2n−1 cos 2x +
2
2
µ
¶
³
(n − 2)π
nπ ´
(n−2)
n−2
(cos 2x)
=2
cos 2x +
= −2n−2 cos 2x +
.
2
2
So
2
(x . cos 2x)
III
1
(n)
=2
n
µ
¶
³
³
nπ ´
nπ ´
n(n − 1)
+ 2n nx sin 2x +
cos 2x +
x −
4
2
2
2
Linear approximations and Differentials
Linear approximations
We use the tangent line at (a, f (a)) as an approximation to the curve y = f (x) when x is
near a.
Definition 59
1. f (x) ≈ f (a) + f ′ (a)(x − a) is called the linear approximation or tangent line
approximation of f at a.
91
91
Chapter 3. Derivatives and Differentials
2. The linear function whose graph is this tangent line, that is,
L(x) = f (a) + f ′ (a)(x − a)
is called the linearization of f at a.
Example 79
p
Find the linearization
of
the
function
f
(x)
=
x + 3 at a = 1 and use it to approxp
p
imate the numbers 3.98 and 4.05. Are these approximations overestimates or
underestimates?
SOLUTION
1
1
1
·
f ′ (x) = (x + 3)− 2 = p
2
2 x +3
1
7 x
⇒ L(x) = f (1) + f ′ (1)(x − 1) = 2 + (x − 1) = +
4
4 4
p
x +3 ≈
7 x
+
4 4
(when x is near 1)
p
p
7 0.98
3.98 = 0.98 + 3 ≈ +
= 1.995
4
4
p
⇒ 3.98 < 1.995
and
p
p
7 1.05
4.05 = 1.05 + 3 ≈ +
= 2.0125
4
4
p
⇒ 4.05 < 2.0125
Our approximates are overestimates.
2
The 1-st order differentials
Definition 60
The 1-st order differential d y of y = f (x) at a is defined in terms of d x by equation
d f (a) = f ′ (a)d x.
(3.16)
Example 80
If f (x) =
ex
, then find d f (1)
x2
SOLUTION f ′ (x) =
e x .x 2 − e x .2x e x (x − 2)
=
⇒ f ′ (1) = −e. So d f (1) = f ′ (1)d x = −ed x.
4
3
x
x
92
92
III. Linear approximations and Differentials
Example 81
Compare the values of ∆y and d y if y = f (x) = x 3 + x 2 − 2x + 1 and x changes
1. from 2 to 2.05
2. from 2 to 2.01
SOLUTION d y = f ′ (x)d x = (3x 2 + 2x − 2)d x.
1. f (2) = 9, f (2.05) = 9.717625 ⇒ ∆y = 0.717625. When x = 2 and d x = ∆x = 2.05 − 2 =
0.05, this becomes d y = [3(2)2 + 2(2) − 2]0.05 = 0.7
2. f (2) = 9, f (2.01) = 9.140701 ⇒ ∆y = 0.140701. When x = 2 and d x = ∆x = 2.01 − 2 =
0.01, this becomes d y = [3(2)2 + 2(2) − 2]0.01 = 0.14
∆y ≈ d y becomes better as ∆x becomes smaller.
3
The 2-nd order differentials
Definition 61
The 2-nd order differential of y = f (x) at a is defined in terms of d x 2 by equation
d 2 f (a) = f ′′ (a)d x 2 .
(3.17)
Example 82
If f (x) =
p
x 2 − 4x + 3, then find d 2 f (0)
SOLUTION f ′′ (x) =
4
−1
1
1
⇒ f ′′ (0) = − p . So d 2 f (0) = − p d x 2 .
p
3 3
3 3
(x 2 − 4x + 3) x 2 − 4x + 3
The n−th order differentials
Definition 62
The n−th order differential of y = f (x) at a is defined in terms of d x n by equation
d n f (a) = f (n) (a)d x n .
93
93
(3.18)
Chapter 3. Derivatives and Differentials
IV
1
Rates of change and Related rates
Rates of change in the natural and social sciences
Proposition 4.1
If x changes from x 1 to x 2 , then the change in x is ∆x = x 2 − x 1 and the corresponding
change in y is ∆y = f (x 2 ) − f (x 1 ). The difference quotient
∆y
f (x 2 ) − f (x 1 )
=
∆x
x2 − x1
is the average rate of change of y with respect to x over the interval [x 1 , x 2 ].
The instantaneous rate of change of y with respect to x or the slope of the tangent
line at P (x 1 , f (x 1 )) is
dy
∆y
= lim
d x ∆x→0 ∆x
Proposition 4.2
If s = f (t ) is the position function of a particle that is moving in a straight line, then
1.
∆s
represents the average velocity over a time period ∆t
∆t
2.
ds
∆s
= lim
represents the instantaneous velocity
d t ∆t →0 ∆t
3. the instantaneous rate of change of velocity with respect to time is acceleration:
a(t ) = v ′ (t ) = s ′′ (t ).
94
94
IV. Rates of change and Related rates
Example 83
The position of a particle is given by the equation
s = f (t ) = t 3 − 6t 2 + 9t ,
where t is measured in seconds and s in meters.
1. Find the velocity at time t . What is the velocity after 2s? When is the particle at
rest? When is the particle moving forward (that is, in the positive direction)
and backward?
2. Find the total distance travelled by the particle during the first five seconds.
3. Find the acceleration at time t and after 4s. When is the particle speeding up,
slowing down?
SOLUTION
1. The velocity function
v(t ) = s ′ (t ) = 3t 2 − 12t + 9 ⇒ v(2) = −3m/s.
·
2
The particle is at rest when v(t ) = 0 ⇔ 3t − 12t + 9 = 0 ⇔
t = 1s
The particle
t = 3s.
moves forward when
·
v(t ) > 0 ⇔
t >3
t <1
It moves backward then 1 < t < 3.
2. We need to calculate the distances travelled by the particle during the time intervals
[0, 1], [1, 3] and [3, 5] separately.
| f (1) − f (0)| + | f (3) − f (1)| + | f (5) − f (3)| =
= |4 − 0| + |0 − 4| + |20 − 0| = 4 + 4 + 20 = 28m.
3. The acceleration is the derivative of the velocity function:
a(t ) = v ′ (t ) = s ′′ (t ) = 6t − 12 ⇒ a(4) = 12m/s 2
95
95
Chapter 3. Derivatives and Differentials
4. The particle speeds up when the velocity is positive and increasing (it means v(t )
and a(t ) are both positive) and also when the velocity is negative and decreasing (it
means v(t ) and a(t ) are both negative). In other words, the particle speeds up when
the velocity and acceleration have the same sign.
v(t ).a(t ) > 0 ⇔ (3t 2 − 12t + 9)(6t − 12) > 0
·
t >3
⇔ 18(t − 1)(t − 3)(t − 2) > 0 ⇔
1<t <2
5. The particle slows down when v(t ) and a(t ) have opposite signs
·
v(t ).a(t ) < 0 ⇔
2
2<t <3
0<t <1
Related rates
Example 84
Air is being pumped into a spherical balloon so that its volume increases at a rate
of 100cm 3 /s. How fast is the radius of the balloon increasing when the diameter is
50cm
SOLUTION Let V (t ) be the volume of the balloon and let r (t ) be its radius. We start by
identifying two things
1. the given information: the rate of increase of the volume of air is 100cm 3 /s ⇒
100cm 3 /s
2. the unknown: the rate of increase of the radius when the diameter is 50cm ⇒
when r = 25cm.
4
3. Equation that relates V (t ) and r (t ) is V = πr 3
3
4. Use the Chain Rule to differentiate both sides with respect to time
dV
dV d r
dr
=
·
= 4πr 2 ·
dt
dr dt
dt
⇒
If we put r = 25 and
dr
1
dV
=
·
d t 4πr 2 d t
dV
= 100 in this equation, we obtain
dt
dr
1
1
=
· 100 =
≈ 0.0127cm/s.
2
d t 4π(25)
25π
V
Indeterminate forms and L’ Hospital’s rule
96
96
dV
=
dt
dr
=?
dt
V. Indeterminate forms and L’ Hospital’s rule
Proposition 5.1
L’ Hospital’s rule is applied to the following indeterminate forms
7 indeterminate forms
∞ 0
; ; ∞ − ∞ ; 0.∞ ;1∞ ; ∞0 ; 00
∞ 0
1
Indeterminate form of type
0
0
Definition 63
f (x)
, where lim f (x) = 0 and lim g (x) = 0. This kind of limit is called
x→a
x→a
g (x)
0
an indeterminate form of
0
Evaluate lim
x→a
Theorem V.1 (L’ Hospital’s rule)
Suppose that
1. f (x) and g (x) are defined on the interval (a, b](a < b)
2. lim f (x) = 0, lim g (x) = 0
x→a
x→a
3. f ′ (x) and g ′ (x) exist on the interval (a, b] and g ′ (x) ̸= 0
f ′ (x)
= K (K may be a real number or infinity)
x→a g ′ (x)
4. lim
f (x)
=K
x→a g (x)
Then lim
Example 85
tan x − x
x→0 x − sin x
Evaluate I = lim
SOLUTION
1
−1
(tan x − x)′
cos2 x
I = lim
=
lim
=
x→0 1 − cos x
x→0 (x − sin x)′
(1 − cos x)(1 + cos x)
1 + cos x 2
=
lim
= =2
x→0 cos2 x(1 − cos x)
x→0 cos2 x
1
= lim
97
97
Chapter 3. Derivatives and Differentials
Theorem V.2 (L’ Hospital’s rule)
Suppose that
1. f (x) and g (x) are defined on the interval [c, +∞)
2.
lim f (x) = lim g (x) = 0
x→+∞
x→+∞
3. f ′ (x) and g ′ (x) exist on the interval [c, +∞) and g ′ (x) ̸= 0
4.
f ′ (x)
= K (K may be a real number or infinity)
x→+∞ g ′ (x)
lim
Then lim
x→+∞
f (x)
=K.
g (x)
Example 86
1
Evaluate I = lim
e x −1
x→+∞
SOLUTION
1
x
´′
³ 1
1
e x −1
e x .(− x12 )
1
= lim e x = 1
I = lim
= lim
¡
¢
′
1
1
x→+∞
x→+∞
x→+∞
− x2
x
Proposition 5.2
Basic indeterminate form of type
0
0
sin x
=1
x→0 x
1. lim
2. lim
x→0
loga (1 + x)
x
= loga e =
1
, (a > 0, a ̸= 1)
ln a
ln(1 + x)
=1
x→0
x
3. lim
ax − 1
= ln a, (a > 0, a ̸= 1)
x→0
x
4. lim
ex − 1
=1
x→0
x
5. lim
(1 + x)α − 1
= α, (α ∈ R)
x→0
x
p
n
1+x −1 1
7. lim
= , (n ∈ N)
x→0
x
n
6. lim
98
98
V. Indeterminate forms and L’ Hospital’s rule
p
1+x −1 1
=
8. lim
x→0
x
2
arctan x
arcsin x
= 1, lim
=1
x→0
x→0
x
x
9. lim
sinh x
cosh x − 1 1
= 1, lim
=
x→0
x→0
x
x2
2
10. lim
2
Indeterminate form of type
∞
∞
Definition 64
f (x)
, where lim f (x) = +∞ and lim g (x) = +∞. This kind of limit is
x→a g (x)
x→a
x→a
∞
called an indeterminate form of
∞
Evaluate lim
Theorem V.3 (L’ Hospital’s rule)
Suppose that
1. f (x) and g (x) are defined on the interval (a, b](a < b)
2. lim f (x) = lim g (x) = +∞
x→a
x→a
3. f ′ (x) and g ′ (x) exist on the interval (a, b] and g ′ (x) ̸= 0
f ′ (x)
= K (K may be a real number or infinity)
x→a g ′ (x)
4. lim
f (x)
=K
x→a g (x)
Then lim
Example 87
Evaluate I = lim+
x→0
ln x
1
x
SOLUTION
1
x
x→0 − 12
x
(ln x)′
I = lim+ ¡ ¢′ = lim+
1
x→0
x
99
99
= lim+ (−x) = 0.
x→0
Chapter 3. Derivatives and Differentials
Theorem V.4 (L’ Hospital’s rule)
Suppose that
1. f (x) and g (x) are defined on the interval [c, +∞)
2.
lim f (x) = lim g (x) = +∞
x→+∞
x→+∞
3. f ′ (x) and g ′ (x) exist on the interval [c, +∞) and g ′ (x) ̸= 0
4.
f ′ (x)
= K (K may be a real number or infinity)
x→+∞ g ′ (x)
lim
Then lim
x→+∞
f (x)
=K.
g (x)
Example 88
x
x→+∞ 2x
Evaluate I = lim
SOLUTION
x
(x)′
1
=
lim
= lim x
=0
x
x
′
x→+∞ 2
x→+∞ (2 )
x→+∞ 2 ln 2
I = lim
Corollary 5.2
Basic indeterminate form of type
1.
2.
3
∞
∞
xα
= 0 (a > 1)
x→+∞ a x
lim
lim
x→+∞
lnα x
xβ
= 0 (∀α > 0, β > 0)
Indeterminate products 0.∞
Definition 65
h
i
Evaluate lim f (x).g (x) , where lim f (x) = 0 and lim g (x) = ∞. This kind of limit is
x→a
x→a
x→a
called an indeterminate form of 0.∞.
We can deal with it by writing the product f .g as a quotient
f (x).g (x) =
f (x)
1
g (x)
=
g (x)
1
f (x)
This method converts the given limit into an indeterminate form of type
use L’Hospital’s rule.
100
100
0
∞
or
so we can
0
∞
V. Indeterminate forms and L’ Hospital’s rule
Example 89
³
´
Evaluate I = lim+ x. ln x
x→0
SOLUTION
1
I = lim+
x→0
4
ln x
(ln x)′
=
lim
= lim x = lim (−x) = 0
x −1 x→0+ (x −1 )′ x→0+ −x −2 x→0+
Indeterminate form of type ∞ − ∞
Definition 66
h
i
If lim f (x) = lim g (x) = ∞, then the limit lim f (x) − g (x) is called an indetermix→a
x→a
x→a
nate form of type ∞ − ∞
We try to convert the difference into quotient (for instance, by using a common denominator, or rationalization, or factoring out a common factor) so that we have an indeterminate
∞
0
form of type or .
0
∞
1
1
− f (x)
1
1
g (x)
f (x) − g (x) = 1 − 1 = 1
1
f (x)
g (x)
f (x) . g (x)
Example 90
µ
Evaluate I = lim
x→π/2
1
− tan x
cos x
SOLUTION
µ
I = lim
x→π/2
¶
¶
1 − sin x
1
− tan x = lim
=
x→π/2 cos x
cos x
(1 − sin x)′
− cos x
= 0.
=
lim
x→π/2 (cos x)′
x→π/2 − sin x
= lim
5
Indeterminate powers 1∞ , 00 , ∞0
Several indeterminate forms arise from the lim [ f (x)]g (x)
x→a
Corollary 5.2
1. lim f (x) = 1 and lim g (x) = ∞ type 1∞
x→a
x→a
2. lim f (x) = 0 and lim g (x) = 0 type 00
x→a
x→a
3. lim f (x) = ∞ and lim g (x) = 0 type ∞0
x→a
Let y = f (x)
g (x)
x→a
, then ln y = g (x) ln f (x).
1. If lim ln y = K ∈ R then lim f (x)g (x) = e K
x→a
x→a
101
101
Chapter 3. Derivatives and Differentials
2. If lim ln y = +∞ then lim f (x)g (x) = +∞.
x→a
x→a
3. If lim ln y = −∞ then lim f (x)g (x) = 0.
x→a
x→a
Example 91
Evaluate I = lim (1 + sin 4x)cot x
x→0
SOLUTION
As x → 0, we have 1+sin 4x → 1 and cot x → ∞, so the given limit is indeterminate form type
1∞ . Let y = (1 + sin 4x)cot x then ln y = ln(1 + sin 4x)cot x = cot x ln(1 + sin 4x), so L’Hospital’s
rule gives
ln(1 + sin 4x)
= lim
x→0
x→0
tan x
lim ln y = lim
x→0
4 cos 4x
1+sin 4x
1
cos2 x
=
4 cos 4x. cos2 x 4
= = 4 ⇒ lim y = lim e ln y = e 4 .
x→0
x→0
x→0
1 + sin 4x
1
= lim
Example 92
Evaluate I = lim+ x x
x→0
SOLUTION
I = lim+ x x (type 00 ) = lim+ e x ln x = e 0 = 1
x→0
x→0
since
lim+ x ln x (type 0.∞) = lim+
x→0
x→0
1
x
x→0 − 12
x
(ln x)′
= lim+ ¡ ¢′ = lim+
1
x→0
VI
1
x
ln x
1
x
(type
∞
)=
∞
= lim+ (−x) = 0.
x→0
Applications of Differentiation
Global maximum, global minimum
Definition 67
Let c be a number in the domain D of a function f . Then f (c) is the
1. global (absolute) maximum value of f on D if f (c) Ê f (x) for all x ∈ D.
2. global (absolute) minimum value of f on D if f (c) É f (x) for all x ∈ D.
3. The global maximum and global minumum values of f are called extreme
values of f
102
102
VI. Applications of Differentiation
Theorem VI.1 (The extreme value theorem)
If f is continuous on a closed interval [a, b], then f attains a global maximum value
f (c) and a global minimum value f (d ) at some numbers c and d in [a, b].
2
Local extrema
Definition 68
The number f (c) is a
1. local maximum value of f on D if f (c) Ê f (x) when x is near c
2. local minimum value of f on D if f (c) É f (x) when x is near c
3. The local maximum and local minumum values of f are called local extreme
values of f
Example 93
If f (x) = x 2 , then f (x) Ê f (0) because x 2 Ê 0 for all x. Therefore f (0) = 0 is the global
minimum and local minimum value of f . This function has no maximum value.
Example 94
Function f (x) = x 3 has neither a global maximum value nor a global mininum value.
In fact, it has no local extreme values either.
103
103
Chapter 3. Derivatives and Differentials
Theorem VI.2 (Fermat’s theorem)
If
1. f (x) has a local maximum or local minimum at c,
2. and if f ′ (c) exists,
then
f ′ (c) = 0
(3.19)
Geometric meaning: at the local maximum and local minimum points the tangent lines of
function y = f (x) are horizontal and therefore each has slope 0.
We can’t expect to locate extreme values simply by setting f ′ (x) = 0 and solving for x.
Example 95
If f (x) = x 3 , then f ′ (x) = 3x 2 , so f ′ (0) = 0. But f has no local maximum or minimum
at 0. The fact that f ′ (0) = 0 simply means that the curve y = x 3 has a horizontal
tangent at (0, 0).
104
104
VI. Applications of Differentiation
Example 96
The function f (x) = |x| has its local minimum value at 0, but that value can’t be
found by setting f ′ (x) = 0, because f ′ (0) does not exist
3
Critical number
We should at least start looking for extreme values of f at the numbers c where f ′ (c) = 0 or
where f ′ (c) does not exist.
Definition 69
A critical number of a function f (x) is a number c in the domain of f such that
1. either f ′ (c) = 0
2. or f ′ (c) does not exist.
Example 97
Find the critical numbers of f (x) = x 3/5 (4 − x)
SOLUTION
3
3(4 − x)
f ′ (x) = x −2/5 (4 − x) + x 3/5 (−1) =
− x 3/5 =
2/5
5
5x
=
−5x + 3(4 − x)
1. f ′ (x) = 0 ⇔ 12 − 8x = 0 ⇔ x =
5x 2/5
3
2
2. f ′ (x) does not exist when x = 0.
Thus the critical numbers are
3
and 0.
2
105
105
=
12 − 8x
5x 2/5
Chapter 3. Derivatives and Differentials
4
The mean value theorem
Theorem VI.3 (Rolle’s theorem)
Let f be a function that satisfies the following three hypotheses:
1. f is continuous on the closed interval [a, b]
2. f is differentiable on the open interval (a, b)
3. f (a) = f (b)
Then there is a number c in (a, b) such that
f ′ (c) = 0.
106
106
(3.20)
VI. Applications of Differentiation
Theorem VI.4 (The mean value theorem)
Let f be a function that satisfies the following hypotheses:
1. f is continuous on the closed interval [a, b]
2. f is differentiable on the open interval (a, b)
Then there is a number c in (a, b) such that
f ′ (c) =
f (b) − f (a)
b−a
(3.21)
or, equivalently,
f (b) − f (a) = f ′ (c)(b − a)
5
(3.22)
Monotonicity
Theorem VI.5 (Increasing-decreasing test)
1. If f ′ (x) > 0 on an interval, then f is increasing on that interval.
2. If f ′ (x) < 0 on an interval, then f is decreasing on that interval.
Example 98
9
Find where the function f (x) = x 3 − x 2 + 6x is increasing and where it is decreasing
2
SOLUTION
Function f is defined at all value x ∈ R
f ′ (x) = 3x 2 − 9x + 6 = 3(x − 1)(x − 2)
½
⇒
f ′ (x) > 0, ∀x ∈ (−∞, 1) ∪ (2, +∞)
f ′ (x) < 0,
∀x ∈ (1, 2)
Therefore, f is increasing on the intervals (−∞, 1) and (2, +∞); is decreasing on the interval
(1, 2).
107
107
Chapter 3. Derivatives and Differentials
6
How to find the local maximum and minimum
How to find the local maximum and minimum
1. Find domain of function f (x)
2. Find f ′ (x)
3. Find x i (i = 1, 2, . . .) where f ′ (x i ) = 0 or does not exist
4. Establish the table of trend
If f ′ changes from positive to negative at x i and f (x i ) is defined, then f has a
local maximum at x i
If f ′ changes from negative to positive at x i and f (x i ) is defined, then f has a
local minimum at x i
If f ′ does not change sign at x i (for example, if f ′ is positive on both sides of x i
or negative on both sides), then f has no local maximum or minimum at x i .
Example 99
Find the local extrema of function y = |x 2 − 2x| + 3
SOLUTION
Domain D = R

½

2
2
x − 2x + 3, if x − 2x Ê 0
′
y=
⇒y =

−x 2 + 2x + 3, if x 2 − 2x < 0
2x − 2, if x 2 − 2x > 0
−2x + 2, if x 2 − 2x < 0 ⇒ y ′ = 0 ⇔
does not exist, if x 2 − 2x = 0
x = 1.
So the critical numbers are x = 0, x = 1, x = 2
108
108
VI. Applications of Differentiation
Example 100
Find the local extrema of function y =
p
3
(1 − x)(x − 2)2
SOLUTION
Domain D = R
4 − 3x
4
y′ = p
⇒ y′ = 0 ⇔ x = ·
3
3
3 (1 − x)2 (x − 2)
4
So the critical numbers are x = , x = 1, x = 2
3
109
109
Chapter 3. Derivatives and Differentials
7
How to find the global maximum and minimum
How to find the global maximum-global minimum
1. y = f (x) is continuous on the open interval (a, b) (a-may be −∞, b may be
+∞) ⇒ Establish table of trend ⇒ Conclusion
2. y = f (x) is continuous on the closed interval [a, b]
Find f ′ (x) ⇒ find the critical numbers x i (it means f ′ (x i ) = 0 or f ′ (x i ) does
not exist)
Exclude x i ∉ [a, b]. Find the values of f at the critical numbers x i ∈ [a, b]
Find the values of f at the endpoints of the interval: f (a), f (b)
Compare f (a), f (b) and f (x i ) where x i ∈ [a, b] ⇒ Global maximum value,
Global minimum value
Example 101
Find the global maximum and minimum values of the function y = (x − 3)e |x+1| on
the interval [−2, 6]
SOLUTION
½
y=
(x − 3)e x+1 , x Ê −1
(x − 3)e −x−1 , x < −1

x+1
, x > −1
 (x − 2)e
′
−x−1
⇒ y = (4 − x)e
, x < −1

Ø, x = −1
y ′ = 0 ⇔ x = 2 ⇒ f (−2) = −5e, f (6) = 3e 7 , f (2) = −e 3 , f (−1) = −4. So global maximum is
f (6) = 3e 7 , and global minimum is f (2) = −e 3 .
8
Concavity
Definition 70
1. If the graph of f lies above all of its tangents on an interval (a, b), then it is
called concave upward on (a, b).
2. If the graph of f lies below all of its tangents on an interval (a, b), then it is
called concave downward on (a, b).
110
110
VI. Applications of Differentiation
Definition 71
A point P on a curve y = f (x) is called an inflection point if f is continuous there
and the curve changes from concave upward to concave downward or from concave
downward to concave upward at P.
Discuss the curve with respect to concavity-points of inflection
1. Find domain
2. Find f ′′ (x)
3. Find x i where f ′′ (x i ) = 0 or does not exist
4. Consider the sign of f ′′ (x)
If f ′′ (x) > 0 for all x in interval (a, b), then the graph of f is concave upward
on (a, b).
If f ′′ (x) < 0 for all x in interval (a, b), then the graph of f is concave downward
on (a, b).
If f ′′ (x) changes sign at x i and f (x i ) is defined, then the graph has inflection
point at x i
9
The second derivative test
Theorem VI.6 (The second derivative test)
Suppose f ′′ is continuous near c
1. If f ′ (c) = 0 and f ′′ (c) > 0, then f has a local minimum at c.
2. If f ′ (c) = 0 and f ′′ (c) < 0, then f has a local maximum at c.
111
111
Chapter 3. Derivatives and Differentials
Example 102
Use the second derivative test to find the local extrema of f (x) = x 4 − 8x 2 + 10.
SOLUTION
f ′ (x) = 4x 3 − 16x = 4x(x 2 − 4) = 4x(x − 2)(x + 2)

x =0
′

x =2
⇒ f (x) = 0 ⇔
x = −2
Thus, the critical numbers are x = 0, x = 2 and x = −2.
f ′′ (x) = 12x 2 − 16 ⇒ f ′′ (0) = −16 < 0, f ′′ (−2) = 32 > 0,
f ′′ (2) = 32 > 0. So f (0) is a local maximum and f (−2), f (2) are local minima.
Example 103
Discuss the curve y = x 3 .e −x with respect to concavity, points of inflection, local
maxima and minima and asymptotes.
Domain D = R
y ′ = 3x 2 .e·−x + x 3 .(−e −x ) = (3x 2 − x 3 )e −x .
x =0
y′ = 0 ⇔
x =3
2
3
−x
3
2
−x
y ′′ = (6x −3x 2 )e −x + (3x
p − x )(−e ) = (x − 6x + 6x)e .
x = 3+ 3
p

′′
y = 0 ⇔  x = 3− 3
x = 0
112
112
VI. Applications of Differentiation
1. The function does not have vertical asymptote because its domain is R.
2.
x3
= 0 ⇒ y = 0 is horizontal asymptote from the right.
x→+∞ e x
3.
x 3 −∞
= + = −∞ ⇒ The function does not have horizontal asymptote from the
x→−∞ e x
0
left.
lim
lim
4. Slant asymptote: y = mx + b
f (x)
x2
= lim x = +∞
m = lim
x→−∞ x
x→−∞ e
⇒ The function does not have slant asymptote.
Example 104
Sketch the graph of the function y =
p
3
1 − x 3.
SOLUTION
Domain D = R
y′ = − p
3
x2
(1 − x 3 )2
⇒ y ′ = 0 ⇔ x = 0.
So the critical numbers are x = 0, x = 1
2x
y ′′ = − p
⇒ y ′′ = 0 ⇔ x = 0
3
3
5
(1 − x )
113
113
Chapter 3. Derivatives and Differentials
1. The function does not have vertical asymptote because its domain is R.
2. lim
x→∞
p
3
1 − x 3 = ∞ ⇒ The function does not have horizontal asymptote.
3. Slant asymptote: y = mx + b
p
3
1 − x3
= −1
x
p
3
b = lim [ f (x) − mx] = lim ( 1 − x 3 + x) =
f (x)
= lim
m = lim
x→∞
x→∞ x
x→∞
= lim p
3
x→∞
x→∞
1
= 0.
p
3
(1 − x 3 )2 − x 1 − x 3 + x 2
⇒ y = −x is slant asymptote.
114
114
VII. Taylor - Maclaurin approximations
VII
Taylor - Maclaurin approximations
Remark 12
Function y = e x − 1 is approximately equal to polynomials when x is near 0.
Degree of polynomials is higher, the approximation is better.
Application: Approximate a function which is not a polynomial by a polynomial when x is near x 0 .
Question: How to find the polynomial approximation to f (x) when x is near
x0 ?
If f has derivatives up to the nth order, then an nth-degree polynomial given by
P (x) = c 0 + c 1 (x − x 0 ) + c 2 (x − x 0 )2 + · · · + c n (x − x 0 )n
will satisfy the conditions
f (x 0 ) = P (x 0 ), f ′ (x 0 ) = P ′ (x 0 ), f ′′ (x 0 ) = P ′′ (x 0 ), . . . , f (n) (x 0 ) = P (n) (x 0 )
if we take
c 0 = f (x 0 ), c 1 = f ′ (x 0 ), c 2 =
f ′′ (x 0 )
f (n) (x 0 )
, · · · , cn =
2!
n!
115
115
Chapter 3. Derivatives and Differentials
1
Taylor- Maclaurin approximations
Theorem VII.1 (Taylor’s Theorem)
Let f (x) be such that f ′ , f ′′ , . . . , f (n−1) exist where x is near x 0 , and f (n) (x 0 ) exists.
Then
f ′′ (x 0 )
f ′ (x 0 )
(x − x 0 ) +
(x − x 0 )2 + . . .
f (x) = f (x 0 ) +
1!
2!
...+
f (n) (x 0 )
(x − x 0 )n + o((x − x 0 )n ), as x → x 0 .
n!
o((x − x 0 )n )
= 0.
x→x 0 (x − x 0 )n
where lim
Corollary 7.0
Substituting x 0 = 0, we get formula Maclaurin approximation
f (x) =
n f (k) (0)
X
x k + o(x n ), x → 0.
k!
k=0
Corollary 7.0
1. If f (x) is even then f (x) =
n f (2k) (0)
X
x 2k + o(x 2n+1 )
(2k)!
k=0
2. If f (x) is odd then
n f (2k+1) (0)
X
f (x) =
x 2k+1 + o(x 2n+2 )
k=0 (2k + 1)!
Corollary 7.0
If x → x 0 , u(x) → 0 then
f (u(x)) =
2
n f (k) (0)
X
(u(x))k + o((u(x))n ).
k!
k=0
Some basic Maclaurin approximations
Corollary 7.0
1. e x = 1 + x +
xn
x2
+...+
+ o(x n ).
2!
n!
116
116
VII. Taylor - Maclaurin approximations
2. ln(1 + x) = x−
x2 x3
(−1)n−1 x n
+ −...+
+ o(x n )
2
3
n
3. sin x = x−
(−1)n x 2n+1
x3 x5
+ −...+
+ o(x 2n+2 )
3! 5!
(2n + 1)!
4. cos x = 1−
x2 x4
x 2n
+ − . . . + (−1)n
+ o(x 2n+1 )
2! 4!
(2n)!
5. (1 + x)α = 1 + αx +
α(α − 1) 2
α(α − 1) . . . (α − (n − 1)) n
x +...+
x + o(x n )
2!
n!
6.
1
= 1−x + x 2 − . . . + (−1)n x n + o(x n )
1+x
7.
1
= 1 + x + x 2 + . . . + x n + o(x n )
1−x
8. sinh x = x +
x3 x5
x 2n+1
+
+...+
+ o(x 2n+2 )
3! 5!
(2n + 1)!
9. cosh x = 1 +
x2 x4
x 2n
+
+...+
+ o(x 2n+1 )
2! 4!
(2n)!
Example 105
³
π´
.
Write down the Maclaurin polynomial of degree n for f (x) = sin 2x +
4
SOLUTION
³
π
π´
f (k) (x) = 2k sin 2x + + k
4
2
i
hπ
⇒ f (k) (0) = 2k sin (2k + 1)
4
So
hπ
i
³
n 2k
π´ X
=
sin (2k + 1) .x k + o(x n ).
sin 2x +
4
4
k=0 k!
Example 106
Write down the Maclaurin polynomial of degree n for f (x) = e x/2+2 .
SOLUTION
We have
e
x/2+2
2
= e .e
x/2
µ
¶
(x/2)2
(x/2)n
n
= e 1 + x/2 +
+...+
+ o((x/2) )
2!
n!
n (x/2)k
X
= e2
+ o(x n )
k!
k=0
2
Example 107
Write down the Maclaurin polynomial of degree n for f (x) =
117
117
1
·
2x + 3
Chapter 3. Derivatives and Differentials
SOLUTION
We have
1
1
=
=
2x + 3 3(1 + 2x/3)
=
¢
1¡
1−(2x/3) + (2x/3)2 − . . . + (−1)n (2x/3)n + o((2x/3)n ) =
3
µ ¶k
n
1X
k 2
=
x k + o(x n )
(−1)
3 k=0
3
Example 108
Write down the Maclaurin polynomial of degree 4 for f (x) = e x . ln(1 + x)
SOLUTION
We have
¶ µ
¶
µ
x2 x3 x4
x2 x3 x4
4
4
+
+
+ o(x ) × x −
+
−
+ o(x ) =
f (x) = 1 + x +
2! 3! 4!
2
3
4
µ
¶
µ
¶
µ
¶
1
1 1 1 3
1 1 1 1 4
2
= x + − + 1 x + − + x + − + − + x + o(x 4 ) =
2
3 2 2
4 3 4 6
1
1
= x + x 2 + x 3 + o(x 4 )
2
3
Example 109
Write down the Maclaurin polynomial of degree 6 for f (x) =
SOLUTION
The Maclaurin polynomial of degree 4 for
x2
·
1 + sin x
1
is
1 + sin x
µ
¶ µ
¶2 µ
¶3 µ
¶4
x3
1
x3
x3
x3
= 1− x −
+ x−
− x−
+ x−
+ o(x 4 ) =
1 + sin x
3!
3!
3!
3!
¶ µ
¶
x3
1 4
2
= 1− x −
+ x − x − x 3 + x 4 + o(x 4 ) =
3!
3
µ
5
2
= 1 − x + x 2 − x 3 + x 4 + o(x 4 )
6
3
5
2
So f (x) = x 2 − x 3 + x 4 − x 5 + x 6 + o(x 6 ).
6
3
Example 110
Find y (100) (1) if y(x) = ln x.
SOLUTION
Let u = x − 1 ⇒ x = u + 1. Then
y(x) = f (u) = ln(1 + u) = u −
u2
u 100
+ . . . + (−1)99
+....
2
100
118
118
VIII. Curves defined by Parametric Equations
According to Maclaurin approximation, we have
−
f (100) (0)
1
1
=
⇒ f (100) (0) = −
· 100! = −99!
100
100!
100
and
y ′ (x) = f ′ (u).u ′ (x) = f ′ (u) ⇒ y (100) (x) = f (100) (u)
Substituting x = 1 ⇒ u = 0, we have
y (100) (1) = f (100) (0) = −99!
VIII
1
Curves defined by Parametric Equations
Definition
Definition 72
Suppose that x and y are both given as functions of a third variable t (called a
parameter) by the equations
x = f (t ),
y = g (t )
(3.23)
(called parametric equations). Each value of t determines a point (x, y) in a coordinate plane.
2
Graphing Devices
119
119
Chapter 3. Derivatives and Differentials

1
1

 x = sin t + cos 5t + sin 13t
2
4
1
1

 y = cos t + sin 5t + cos 13t
2
4
3
Cycloid
Definition 73
The curve traced out by a point P on the circumference of a circle as the circle rolls
along a straight line is called a cycloid.
Ù
OT = P
T =rθ
½
⇒
x = OT − PQ = r θ − r sin θ = r (θ − sin θ)
,θ ∈ R
y = T C −QC = r − r cos θ = r (1 − cos θ)
120
120
IX. Calculus with Parametric Curves
IX
1
Calculus with Parametric Curves
Tangents
Theorem IX.1
Suppose that x and y are both given as functions of a third variable t by the equations
x = f (t ),
Then the Chain Rule gives
dy
=
dx
dy
dt
dx
dt
dy dy dx
=
·
and therefore
dt dx dt
′
=
y = g (t )
2
g (t ) d y
d dy
;
=
=
f ′ (t ) d x 2 d x d x
µ
¶
d
dt
³
dy
dx
dx
dt
´
i f f ′ (t ) ̸= 0
(3.24)
Example 111
A curve C is defined by the parametric equations x = t 2 , y = t 3 − 3t .
1. Show that C has two tangents at the point (3, 0) and find their equations.
2. Find the points on C where the tangent is horizontal or vertical.
3. Determine where the curve is concave upward or downward.
½
SOLUTION
x = t2 = 3
⇔
y = t 3 − 3t = 0
·
p
t= 3
p ,
t =− 3
d y d y/d t 3t 2 − 3
=
=
d x d x/d t
2t
p
p
1. The slope of the tangentpwhen t = ± 3 is dpy/d x = ± 3, so the equations of the
tangents at (3, 0) are y = 3(x − 3) and y = − 3(x − 3)
2. C has a horizontal tangent when d y/d x = 0 ⇒ d y/d t = 0, d x/d t ̸= 0 ⇒ 3t 2 − 3 = 0 ⇒
t = ±1. The corresponding points on C are (1, −2); (1, 2).
C has a vertical tangent when d x/d t = 0, d y/d t ̸= 0 ⇒ t = 0. The corresponding
point on C is (0, 0).
2
3.
d
dt
³
dy
dx
´
3
2
³
´
1 + t12
d y
3(t 2 + 1)
=
=
=
· Thus the curve is concave upward when t > 0
dx
d x2
2t
4t 3
dt
and concave downward when t < 0.
121
121
Chapter 3. Derivatives and Differentials
X
1
Exercises
Essay Questions
The derivative of the function y = f (x)
22 Find the derivative of the function y = f (x) = x 2 − 4|x| + 5.
SOLUTION
We have
 2
 x − 4x + 5, if x > 0
y = f (x) = x 2 + 4x + 5, if x < 0

5, if x = 0
f +′ (0) = lim+
x→0
x 2 − 4x
f (x) − f (0)
= lim+
= lim+ (x − 4) = −4.
x→0
x→0
x −0
x
f (x) − f (0)
x 2 + 4x
lim
= lim−
= lim− (x + 4) = 4.
x→0−
x→0
x→0
x −0
x
The derivative of the function y = f (x) is

 2x − 4, if x > 0
′
f (x) = 2x + 4, if x < 0

Ø, if x = 0
f −′ (0) =
............................................................................................□
23 Find the derivative of the following functions
³
´
p
1. y = ln x 2 + x 4 + 1 .
p
2. y =
3. y = x arctan x − ln
s¯
¯
¯x −4¯
¯
¯
4. y = ln ¯
x −2¯
x 2 − 2x + 3
2x + 1
122
122
p
x2 + 1
X. Exercises
SOLUTION
1. We have
p
4
′
2x + (xp +1)
(x +
2 x 4 +1
′
y =
=
=
p
p
2
4
2
x + x +1
x + x4 + 1
3
p
2x + p2x4
2x( x 4 + 1 + x 2 )
2x
x +1
=
=
=p
p
p
p
x 2 + x 4 + 1 (x 2 + x 4 + 1) x 4 + 1
x4 + 1
2
2. We have
′
y =
=
x 4 + 1)′
2
′
(x
p −2x+3) .(2x + 1) −
2
2 x −2x+3
p
x 2 − 2x + 3.(2x + 1)′
=
(2x + 1)2
(x − 1)(2x + 1) − 2(x 2 − 2x + 3)
3x − 7
=
p
p
(2x + 1)2 . x 2 − 2x + 3
(2x + 1)2 . x 2 − 2x + 3
3. We have
1
y = f (x) = x arctan x − ln(x 2 + 1)
2
⇒ y ′ = arctan x + x.(arctan x)′ −
= arctan x +
4. We have
1 (x 2 + 1)′
·
=
2 x2 + 1
x
x
= arctan x.
− 2
2
1+x
x +1
¯
¯
1 ¯¯ x − 4 ¯¯ 1
y = f (x) = ln ¯
= (ln |x − 4| − ln |x − 2|)
2
x −2¯ 2
µ
¶
1
1
1
′
⇒y =
−
=
2 x −4 x −2
=
2
1
1
·
=
2 (x − 4)(x − 2) (x − 4)(x − 2)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
( 1
24
e x , x ̸= 0
If y = f (x) =
, then find f ′ (2), f ′ (0).
0, x = 0
SOLUTION
p
µ
µ
¶
¶
1
1
1
1 1
e
′
′
2
2
If x ̸= 0 then f (x) = e . − 2 . So f (2) = e . − 2 = − · e = −
.
x
2
4
4
1
x
1
f +′ (0) =
f (x) − f (0)
ex
e +∞ +∞
lim+
= lim+
= + = + = +∞
x→0
x→0 x
x −0
0
0
Therefore, f ′ (0) does not exist.
......................................................... □
25 If y = f (x) = x 2 |x|, then find f ′ (0).
SOLUTION
123
123
Chapter 3. Derivatives and Differentials
We have
x 3 , if x > 0
y = f (x) = −x 3 , if x < 0

0, if x = 0


We also have
f +′ (0) = lim+
x→0
f −′ (0) =
So f ′ (0) = 0.
f (x) − f (0)
x3
= lim+
= lim+ x 2 = 0.
x→0 x
x→0
x −0
f (x) − f (0)
−x 3
lim
= lim−
= lim− (−x 2 ) = 0.
x→0−
x→0
x→0
x −0
x
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
p
3
26
If f (x) =
p
5
x 2 + 1. x + 2
, then find f ′ (0).
p
7
3
x +2
SOLUTION
We have
¯p
¯
5
¯ 3 x 2 + 1. p
x + 2 ¯¯
¯
ln | f (x)| = ln ¯
¯=
p
7
¯
¯
x3 + 2
= ln |
=
p
3
p
p
7
5
x 2 + 1| + ln | x + 2| − ln | x 3 + 2| =
1
1
1
ln |x 2 + 1| + ln |x + 2| − ln |x 3 + 2|
3
5
7
Differentiating both sides with respect to x, we have
f ′ (x) 1 2x
1
1
1 3x 2
= · 2
+ ·
− · 3
f (x) 3 x + 1 5 x + 2 7 x + 2
Substituting x = 0, we have
p
¶
5
1 2×0 1
1
1 3 × 02
2
f (0) = f (0) · 2
+ ·
− · 3
= p
7
3 0 +1 5 0+2 7 0 +2
10 2
µ
′
............................................................................................□
The second derivative of the function y = f (x)
27 Find the second derivative of the following functions
1. f (x) = (2x + 3)e −x
x2
1 + 2x
x
3. f (x) =
2 + 3e x
2. f (x) =
SOLUTION
124
124
X. Exercises
1. We have
f ′ (x) = 2e −x + (2x + 3)e −x .(−1) = (−2x − 1)e −x
⇒ f ′′ (x) = −2e −x + (−2x − 1)e −x .(−1) = (2x − 1)e −x
2. We have
f ′ (x) =
⇒ f ′′ (x) =
=
2x(1 + 2x) − x 2 .2 2x 2 + 2x
=
(1 + 2x)2
(1 + 2x)2
(4x + 2)(1 + 2x)2 − (2x 2 + 2x).2(1 + 2x).2
=
(1 + 2x)4
(4x + 2)(1 + 2x) − (2x 2 + 2x).4
2
=
(1 + 2x)3
(1 + 2x)3
3. We have
f ′ (x) =
f ′′ (x) =
2 + 3e x − x.3e x
(2 + 3e x )2
(3e x − 3e x − x.3e x )(2 + 3e x )2 − (2 + 3e x − x.3e x ).2(2 + 3e x ).3e x
=
(2 + 3e x )4
=
(−x.3e x )(2 + 3e x ) − (2 + 3e x − x.3e x ).2.3e x
=
(2 + 3e x )3
=
9xe 2x − 6xe x − 18e 2x − 12e x
(2 + 3e x )3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
The 1-st order differentials
28 Find the first order differential d f (1) of the following functions
1. f (x) = ln(1 + x 2 )
2. f (x) = (3x)x
SOLUTION
1. We have
f ′ (x) =
2x
2×1
′
⇒
f
(1)
=
=1
1 + x2
1 + 12
⇒ d f (1) = f ′ (1)d x = d x
2. We have
f ′ (x)
3
ln | f (x)| = x ln |3x| ⇒
= ln |3x| + x ·
= ln |3x| + 1
f (x)
3x
⇒ f ′ (1) = f (1)(ln |3 × 1| + 1) = 3(ln 3 + 1)
⇒ d f (1) = f ′ (1)d x = 3(ln 3 + 1)d x
125
125
Chapter 3. Derivatives and Differentials
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
29 Find the first order differential of the following functions
1. f (x) = e −x + ln |x|
2. f (x) = arctan
1
x
3. f (x) = (x 2 + 2)e 3x
SOLUTION
1. We have
1
1
= −e −x +
x
x
µ
¶
1
⇒ d f (x) = f ′ (x)d x = −e −x +
dx
x
f ′ (x) = e −x .(−1) +
2. We have
f ′ (x) =
− x12
1 + x12
=−
1
x2 + 1
⇒ d f (x) = f ′ (x)d x = −
dx
x2 + 1
3. We have
f ′ (x) = 2xe 3x + (x 2 + 2)e 3x .3 = (3x 2 + 2x + 6)e 3x
⇒ d f (x) = f ′ (x)d x = (3x 2 + 2x + 6)e 3x d x
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
The second order differentials
30 Find the second order differential d 2 f (0) of the following functions
1. f (x) = cos2 (2x)
2. f (x) = x3x
SOLUTION
1. We have
f ′ (x) = 2 cos(2x).(− sin(2x)).2 = −2 sin(4x)
⇒ f ′′ (x) = −2 × 4 cos(4x) = −8 cos 4x
⇒ f ′′ (0) = −8 cos 0 = −8
⇒ d 2 f (0) = f ′′ (0)d x 2 = −8d x 2
126
126
X. Exercises
2. We have
f ′ (x) = 3x + x.3x ln 3
⇒ f ′′ (x) = 3x ln 3 + 3x ln 3 + x.3x ln2 3
⇒ f ′′ (0) = 2 ln 3
⇒ d 2 f (0) = f ′′ (0)d x 2 = 2 ln 3d x 2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
Rates of change
31 Find the slope and an equation of the tangent line to the parabola y = 4x − x 2 at
the point (1, 3).
SOLUTION
We have y ′ = 4 − 2x.
1. The slope of the tangent line to the parabola y = 4x − x 2 at the point (1, 3) is
m = y ′ (1) = 4 − 2 × 1 = 2.
2. The equation of the tangent line to the parabola y = 4x − x 2 at the point (1, 3) is
y − 3 = m(x − 1) ⇔ y = 2x + 1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
32 If a ball is thrown into the air with a velocity of 40m/s, its height after t seconds
is given by y = 40t − 16t 2 .
1. Find the velocity after t = 2s.
2. When will the ball hit the ground? With what velocity will the ball hit the ground?
SOLUTION
′
We have y = 40 − 32t
1. The velocity after t = 2s is y ′ (2) = 40 − 32 × 2 = −24m/s
2. The ball will hit the ground when y = 0
40
⇔ 40t − 16t 2 = 0 ⇒ t =
= 2.5s ⇒ y ′ (2.5) = −40m/s
16
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
127
127
Chapter 3. Derivatives and Differentials
33 The cost (in dollars) of producing x units of a certain commodity is C (x) =
5000 + 10x + 0.05x 2 .
1. Find the average rate of change of C with respect to x when the production level
is changed
a) from x = 100 to x = 105
b) from x = 100 to x = 101
2. Find the instantaneous rate of change of C with respect to x when x = 100.
SOLUTION
If the number of items produced is increased from x 1 to x 2 , then the average rate of change
of the cost is
∆C C (x 2 ) −C (x 1 ) C (x 1 + ∆x) −C (x 1 )
=
=
∆x
x2 − x1
∆x
1. The average rate of change of C with respect to x when the production level is
changed
a) from x = 100 to x = 105 is
∆C C (105) −C (100)
=
=
∆x
105 − 100
(5000 + 10 × 105 + 0.05 × 1052 ) − (5000 + 10 × 100 + 0.05 × 1002 )
= 20.25
5
b) from x = 100 to x = 101 is
=
∆C C (101) −C (100)
=
=
∆x
101 − 100
=
(5000 + 10 × 101 + 0.05 × 1012 ) − (5000 + 10 × 100 + 0.05 × 1002 )
= 20.05
1
2. The instantaneous rate of change of cost with respect to the number of items produced, is call the marginal cost
∆C
= C ′ (x)
∆x→0 ∆x
marginal cost = lim
C ′ (x) = 10 + 0.1x
The instantaneous rate of change of C with respect to x when x = 100 is
C ′ (100) = 10 + 0.1 × 100 = 20.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
34 If a rod or piece of wire is homogeneous, then its linear density is uniform and is
defined as the mass per unit length (ρ = m/ℓ) and measured in kilograms per meter.
Suppose, however, that the rod is not homogeneous but that its mass measured from
p
its left end to a point is m = f (x) = x. Find the average density of the part of the rod
given by 1 É x É 1.2 and the density right at x = 1.
128
128
X. Exercises
SOLUTION
The mass of the part of the rod that lies between x = x 1 and x = x 2 is given by ∆m =
f (x 2 ) − f (x 1 ), so the average density of that part of the rod is
average density =
∆m f (x 2 ) − f (x 1 )
=
∆x
x2 − x1
The linear density is the rate of change of mass with respect to length
∆m d m
=
∆x→0 ∆x
dx
ρ = lim
The average density of the part of the rod given by 1 É x É 1.2 is
p
1.2 − 1
∆m f (1.2) − f (1)
=
=
≈ 0.48kg /m.
∆x
1.2 − 1
0.2
while the linear density right at x = 1 is
d m ¯¯
1 ¯¯
ρ=
= p ¯
= 0.5kg /m.
¯
d x x=1 2 x x=1
............................................................................................□
35 A current exists whenever electric charges move. The figure below shows part of a
wire and electrons moving through a plane surface, shaded red. If ∆Q is the net charge
that passes through this surface during a time period ∆t , then the average current
during this time interval is defined as
average current =
∆Q Q 2 −Q 1
=
∆t
t2 − t1
The current I at a given time t 1 is
∆Q dQ
=
∆t →0 ∆t
dt
I = lim
Thus the current is the rate at which charge flows through a surface. It is measured in
units of charge per unit time (coulombs per second, amperes). The quantity of charge
Q in coulombs (C) that has passed through a point in a wire up to time t (measured in
seconds) is given by
Q(t ) = t 3 − 2t 2 + 6t + 2.
Find the current when t = 0.5s and t = 1s.
SOLUTION
We have
I=
dQ
= 3t 2 − 4t + 6
dt
129
129
Chapter 3. Derivatives and Differentials
1. When t = 0.5s ⇒ I = 4.75(C /s) or I = 4.75A
2. When t = 1s ⇒ I = 5(C /s) or I = 5A
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
Related rates
36 Car A is travelling west at 50km/h and car B is travelling north at 60km/h. Both
are headed for the intersection of the two roads. At what rate are the cars approaching
each other when car A is 300km and car B is 400km from the intersection?
SOLUTION
1. At a given time t , let x be the distance from car A to the intersection of the roads C ,
let y be the distance from car B to C , let z be the distance between the cars A and B .
2. the given information:
3. the unknown:
dx
dy
= −50km/h and
= −60km/h (x and y are decreasing).
dt
dt
dz
=? when x = 300km, y = 400km
dt
4. The equation that relates x, y and z is given by Pythagorean theorem:
z2 = x2 + y 2
5. Differentiating each side with respect to t , we have
µ
¶
dz
dx
dy
dz 1 dx
dy
2z
= 2x
+ 2y
⇒
=
x
+y
dt
dt
dt
dt z
dt
dt
When x = 300, y = 400 ⇒ z =
q
x 2 + y 2 = 500, so
´
dz
1 ³
=
300 × (−50) + 400 × (−60) = −78km/h
d t 500
The cars are approaching each other at a rate of 78km/h.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
Indeterminate forms and L’ Hospital’s rule
130
130
X. Exercises
37 Evaluate the following limits
ln x
1. lim
x→1 x − 1
·
1
1
5. lim
−
x→0 ln(x + 1)
x
1 − cos x
x→0 sin x
1
2. lim
6. lim+ x x−1
x→1
x
3.
4.
lim
x→+∞
¸
e
x
7. lim+ (sin x)x
x→0
x2
x→+∞ e x
lim
8.
2
lim (x + 1) x .
x→+∞
SOLUTION
1. Since lim ln x = ln 1 = 0 and lim (x − 1) = 0, we can apply L’Hospital’s Rule:
x→1
x→1
ln x
(ln x)′
1/x
= lim
= lim
= 1.
′
x→1 x − 1
x→1 (x − 1)
x→1 1
I = lim
2. Since lim 1 − cos x = 0 and lim sin x = 0, we can apply L’Hospital’s Rule:
x→0
x→0
1 − cos x
(1 − cos x)′
sin x 0
= lim
= = 0.
= lim
′
x→0 sin x
x→0 (sin x)
x→0 cos x
1
I = lim
3. We have lim e x = +∞ and lim x = +∞. Applying L’Hospital’s Rule confirms our
x→+∞
x→+∞
suspicions, as
ex
(e x )′
e x +∞
= lim
=
= +∞.
=
lim
x→+∞ (x)′
x→+∞ x
x→+∞ 1
1
I = lim
∞
since lim e x = +∞ and lim x 2 = +∞. Applyx→+∞
x→+∞
∞
ing L’Hospital’s Rule twice, we get
4. Note that this limit has the form
x2
(x 2 )′
2x ∞
=
lim
=
= lim x =
x
x
′
x→+∞ e
x→+∞ (e )
x→+∞ e
∞
I = lim
2
2
(2x)′
= lim x =
= 0.
x
′
x→+∞ e
x→+∞ (e )
+∞
= lim
5. Note that this limit has the form ∞ − ∞. If we add the fractions, we get a form to
which we can apply L’Hospital’s Rule. We have
·
¸
1
1
x − ln(x + 1)
I = lim
−
= lim
=
x→0 ln(x + 1)
x→0 x ln(x + 1)
x
x
x − ln(x + 1)
x − ln(x + 1) 0
· lim
= lim
=
2
x→0 ln(x + 1) x→0
x→0
x
x2
0
lim
Applying L’Hospital’s Rule, we get
(x − ln(x + 1))′
= lim
I = lim
x→0
x→0
(x 2 )′
131
131
1
x +1 =
2x
1−
Chapter 3. Derivatives and Differentials
x +1−1
1
1
=
= lim x + 1 = lim
x→0 2(x + 1)
x→0
2x
2
6. Note that this limit has the form 1∞ .
1
ln y = ln x x−1 =
1
· ln x.
x −1
We now consider the limit
lim+ ln y = lim+
x→1
x→1
1
ln x
0
· ln x = ∞ · 0 = lim+
=
x→1 x − 1
x −1
0
1
(ln x)
1
= lim+
= lim+ x = = 1.
′
x→1 (x − 1)
x→1 1
1
′
⇒ I = lim+ y = lim+ e ln y = e 1 = e.
x→1
x→1
7. Note that this limit has the form 00 . We let y = (sin x)x , so that
ln y = ln(sin x)x = x ln(sin x).
We now consider the limit
lim+ ln y = lim+ x ln(sin x) = 0 · ∞ = lim+
x→0
x→0
x→0
ln(sin x) ∞
=
1
∞
x
Applying L’Hospital’s Rule, we get
cos x
(ln(sin x))′
x =
lim ln y = lim+
= lim+ sin−2
x→0+
x→0
x→0 −x
(x −1 )′
= lim+
x→0
−x 2 cos x
x
= lim+
· lim (−x cos x) =
x→0 sin x x→0+
sin x
= 1 × 0 × 1 = 0.
⇒ I = lim+ y = lim+ e ln y = e 0 = 1.
x→0
x→0
2
8. Note that this limit has the form ∞0 . We let y = (x + 1) x , so that
2
ln y = ln(x + 1) x =
2
ln(x + 1).
x
We now consider the limit
2
ln(x + 1) = 0 · ∞ =
x→+∞ x
lim ln y = lim
x→+∞
= lim
x→+∞
2 ln(x + 1) ∞
=
x
∞
132
132
X. Exercises
Applying L’Hospital’s Rule, we get
′
lim ln y = lim
x→+∞
x→+∞
(2 ln(x + 1))
= lim
x→+∞
(x)′
2·
1
x +1 =
1
2
= 0.
x→+∞ x + 1
= lim
We now have that
⇒ I = lim y = lim e ln y = e 0 = 1.
x→+∞
x→+∞
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
38 Find the mistake in the string of equalities
x2
2x
2
2
=
lim
=
lim
=
= 2.
x→0 e x − 1
x→0 e x
x→0 e x
1
lim
SOLUTION
x2
0
x2
has
the
form
·
Therefore,
the
first
equality,
lim
=
x→0 e x − 1
x→0 e x − 1
0
1. The first limit lim
2x
, holds.
x→0 e x
lim
2x 0
= = 0 and L’Hospital’s Rule does not apply here.
x→0 e x
1
2. Notice that lim
x2
= 0.
x→0 e x − 1
3. The correct evaluation is then lim
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
Applications of Differentiation
39 The Hubble Space Telescope was deployed on April 24, 1990, by the space shuttle
Discovery. A model for the velocity of the shuttle during this mission, from lift-off at
t = 0 until the solid rocket boosters were jettisoned at t = 126s, is given by
v(t ) = 0.001302t 3 − 0.09029t 2 + 23.61t − 3.083 (feet/second) .
Using this model, estimate the absolute maximum and minimum values of the acceleration of the shuttle between lift-off and the jettisoning of the boosters.
SOLUTION
∀t ∈ [0, 126], we have
a(t ) = v ′ (t ) = 3 × 0.001302t 2 − 2 × 0.09029t + 23.61,
⇒ a ′ (t ) = 6 × 0.001302t − 2 × 0.09029
⇒ a ′ (t ) = 0 ⇔ t 0 =
2 × 0.09029
≈ 23.12
6 × 0.001302
133
133
Chapter 3. Derivatives and Differentials
Critical number is t 0 ≈ 23.12. Evaluating a(t ) at the critical number and at the endpoints,
we have
a(0) = 23.61, a(t 0 ) ≈ 21.52, a(126) = 62.87
So the maximum acceleration is 62.87 f t /s 2 and the minimum acceleration is about
21.52 f t /s 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
40 A farmer has 2400 ft of fencing and wants to fence off a rectangular field that
borders a straight river. He needs no fence along the river. What are the dimensions of
the field that has the largest area?
SOLUTION
We wish to maximize the area A of the rectangle. Let x and y be the depth and width of the
rectangle (in feet). Then we express A in terms of x and y
A = xy
The total length of the fencing is 2400 ft. Thus
2x + y = 2400 ⇒ y = 2400 − 2x ⇒ A = x(2400 − 2x)
Note that x Ê 0 and y = 2400 − 2x Ê 0 ⇒ x É 1200. So the function that we wish to maximize
is
A(x) = 2400x − 2x 2 , ∀x ∈ [0, 1200]
A ′ (x) = 2400 − 4x ⇒ A ′ (x) = 0 ⇔ x = 600
The maximum value of A must occur either at the critical number x = 600 or at the
endpoints of interval [0, 1200]. Since
A(0) = 0, A(600) = 720000, A(1200) = 0,
the maximum value is A(600) = 720000. Thus the rectangular field should be 600 ft deep
and 1200 ft wide. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
41 A cylindrical can is to be made to hold 1ℓ of oil. Find the dimensions that will
minimize the cost of the metal to manufacture the can.
134
134
X. Exercises
SOLUTION
Let r be the radius and h be the height of the can (in meters). In order to minimize the cost
of the metal, we minimize the total surface area of the cylinder (top, bottom, and sides). So
the surface area is
A = 2πr 2 + 2πr h
We use the fact that the volume is given as 1ℓ = 10−3 m 3
V = πr 2 h = 10−3 ⇒ h =
10−3
.
πr 2
Substitution of h into the expression for A gives
µ −3 ¶
10
2.10−3
2
2
A = 2πr + 2πr
=
2πr
+
πr 2
r
Therefore the function that we want to minimize is
A(r ) = 2πr 2 +
2.10−3
,
r
r ∈ (0, +∞)
2.10−3 4πr 3 − 2.10−3
⇒ A (r ) = 4πr −
=
r2
r2
p
3
⇒ A ′ (r ) = 0 ⇔ 4πr 3 − 2.10−3 = 0 ⇔ r = 10−1 0.5/π
′
Substitution of r into the expression for h gives
10−3
10−3
10−1
=
=
=
p
3
πr 2
10−2 π(0.5/π)2/3
0.52 π
p
3
p
10−1 0.5
−1 3
=
0.5/π = 2r
=
2.10
p
0.5 3 π
p
3
Thus, to minimize the cost of the can, the radius should be 10−1 0.5/π(m) and the height
should be equal to twice the radius. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
h=
135
135
Chapter 3. Derivatives and Differentials
2
Multiple-choice Questions
Tangent problems
Question 1 (L.O.1): A point M (a, b) is on the graph of the function f (x) = 1.55e x − x.
Given that the tangent line to the graph at M is parallel to the line y = 2.47x +2.82. Find
the value of a.
A 0.8059
B 0.4466
C 1.7944
D 1.4552
E 1.2394
SOLUTION
Since the tangent line to the graph at M is parallel to the line y = 2.47x + 2.82, then
f ′ (a) = 2.47 ⇒ 1.55e a − 1 = 2.47 ⇒ a = ln
µ
¶
2.47 + 1
= 0.805899663027613 ≈ 0.8059.
1.55
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 2 (L.O.1): Find ALL points on the graph of f (x) = 4x 3 − 192x where the tangent line is parallel to the x−axis.
A (4; 512)
B (4; −512) and (−4; 512)
C (4; −512)
D (4; 512) and (−4; −512)
E (−4; 512)
SOLUTION
The tangent line is parallel to the x−axis when its slope is equal to 0. It means that
f ′ (x) = 12x 2 − 192 = 0 ⇔ x = ±4.
Therefore, all points on the graph of f (x) where the tangent line is parallel to the x−axis
are (4; −512) and (−4; 512).
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 3 (L.O.1): Find ALL points on the graph of f (x) =
is perpendicular to the line y = 16x − 4.
A (−12; 3/4)
B (12; 3/4)
C (12; −3/4) and (−12; 3/4)
D (−12; −3/4)
E (12; 3/4) and (−12; −3/4)
9
where the tangent line
x
SOLUTION
The tangent line is perpendicular to the line y = 16x − 4 when its slope is equal to −
means that
f ′ (x) = −
9
1
= − ⇔ x = ±12.
2
x
16
136
136
1
· It
16
X. Exercises
Therefore, all points on the graph of f (x) where the tangent line is perpendicular to the
line y = 16x − 4, are (12; 3/4) and (−12; −3/4) .
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 4 (L.O.1): Find the real value of a such that the tangent line to the graph of
p
21
the function y = 3 x at x = a has y−intercept
.
2
C 49
A 46
B 52
D 44
E 48
SOLUTION
p
3
The tangent line at x = a has an equation of the form y = 3 a + p (x − a). This tangent
2 a
p
3 a 21
=
⇒ a = 49.
line has y−intercept 21/2 when
2
2
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 5 (L.O.1): Let the curve be given by paramtric equations
½
x(t ) = (3 − 4t ) e 3t
¡
¢
y(t ) = 3t 2 − 5t − 4 e 3t
Find the set of all values of t such that the tangent at M (x(t ), y(t )) has the slope of
37
− ?
31 ½
¾
¾
¾
½
½
38
503
410
A − ;3
B −
; −1
;7
C −
93
93
½ 93
¾
½
¾
131
148
D −
;4
E
;1
93
93
SOLUTION
¡
¢
y ′ (t ) (6t − 5) e 3t + 3 · 3t 2 − 5t − 4 e 3t
′
The slope of the tangent at M (x(t ), y(t )) is y (x) = ′
=
·
x (t )
3 · (3 − 4t ) e 3t − 4e 3t
Therefore,

38
2
37
−9t
+
9t
+
17
37
t = −
′

y (x) = − ⇔
=− ⇔
93
31
12t − 5
31
t = 3
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Differentiable functions
Question 6 (L.O.2): Determine ALL values of x such that f (x) =
entiable.
A x ̸= 5
B x = −5
C x =5
SOLUTION
x 2 − 11
The domain of f (x) = 2
is D = R\{±5}.
x − 25
⇒ y′ =
−28x
(x 2 − 25)2
137
137
D x = ±5
x 2 − 11
is not differx 2 − 25
E x ̸= ±5
Chapter 3. Derivatives and Differentials
So the function f is not differentiable when x = ±5.
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
p
3
Question 7 (L.O.2): Determine ALL values of x such that f (x) = (x 2 − 25)2 is not
differentiable.
A x = ±5
B x ̸= ±5
C x ̸= 5
D x = −5
E x =5
The domain of f (x) =
p
3
SOLUTION
(x 2 − 25)2 is D = R.
⇒ y′ =
2
4x
× (x 2 − 25)−1/3 × 2x = p
3
3
3 (x + 5)(x − 5)
So the function f is not differentiable when x = ±5.
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 8 (L.O.2): Find S = a + b such that function f is differentiable at x = 0, if
½
f (x) =
A 7e 5
2x 2 + 7x + 5, x É 0
ln(ax + b), x > 0
C 8e 5
B 8
D 9
E 6e 5
SOLUTION
In order to be differentiable, function has to be continuous
lim f (x) = lim− f (x) ⇒ ln b = 5 ⇒ b = e 5 .
x→0+
x→0
Function f is differentiable at x = 0 if and only if
f +′ (0) = f −′ (0) ⇒
a
= 7 ⇒ a = 7b = 7e 5 .
b
Therefore, S = a + b = 8e 5
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Derivatives of composite functions
Question 9 (L.O.1): Let f (x) and g (x) be functions satisfying the conditions: f (−4) =
f ′ (−4) = 9 and g (9) = g ′ (9) = −4. Consider the function h(x) = ( f ◦g ◦ f )(x) = f (g ( f (x))).
Find the value of h ′ (−4).
C −322
A −328
B −325
D −324
E −329
SOLUTION
Using the chain rule, we have
h ′ (x) = f ′ (g ( f (x))).g ′ ( f (x)). f ′ (x) ⇒ h ′ (−4) = f ′ (g ( f (−4))).g ′ ( f (−4)). f ′ (−4) =
= f ′ (−4).g ′ (9). f ′ (−4) = 9 × (−4) × 9 = −324.
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
138
138
X. Exercises
Question 10 (L.O.2): An importer of Rwandan coffee estimates that local consumers
4346
will buy approximately D(p) = 2 pounds of the coffee per week when the price is p
p
dollars per pound. It is estimated that t weeks from now, the price of Rwandan coffee
will be p(t ) = 0.02t 2 + 0.08t + 9 dollars per pound. At what rate will the weekly demand
for the coffee be changing with respect to time 6 weeks from now?
C −3.4126
A −3.1978
B −2.3627
D −3.2408
E −2.621
SOLUTION
µ
¶
2
Using the chain rule, we have D(t ) = D(p(t )) ⇒ D (t ) = D (p).p (t ) = 4346× − 3 ×(0.04t +
p
0.08)
When t = 6 ⇒ p(6) = 0.02 × 62 + 0.08 × 6 + 9 = 10.2 and
′
′
′
¶
2
D (6) = 4346 × −
× (0.04 × 6 + 0.08) = −2.62101303420253 ≈ −2.621.
10.23
′
µ
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 11 (L.O.2): When electric blenders are sold for p dollars apiece, local con7928
sumers will buy approximately D(p) =
blenders per month. It is estimated that t
p
p
months from now, the price of the blenders will be p(t ) = 0.03 t 3 +4 dollars. Compute
the rate at which the monthly demand for the blenders will be changing with respect
to time 16 months from now.
A −40.9061
B −40.7186
C −40.9638
D −40.9057
E −41.3092
SOLUTION
µ
¶
1
Using the chain rule, we have D(t ) = D(p(t )) ⇒ D ′ (t ) = D ′ (p).p ′ (t ) = 7928 × − 2 ×
p
p
(0.045 t )
p
When t = 16 ⇒ p(16) = 0.03 × 163 + 4 ≈ 5.92 and
¶
p
1
D (16) ≈ 7928 × −
× (0.045 × 16) = −40.7185902118335 ≈ −40.7186.
2
5.92
′
µ
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Derivatives of inverse functions
Question 12 (L.O.1): If f (x) = 2x + 2 sinh (x − 2), find ( f −1 )′ (4).
9
13
19
15
A
B −
C
D −
4
4
4
4
E
1
4
SOLUTION
1
1
Let y = f (x) = 2x + 2 sinh (x − 2) ⇒ x (y) = ′
=
. When y = 4 we have
y (x) 2 cosh (x − 2) + 2
x = 2. Therefore,
1
1
( f −1 )′ (4) = x ′ (4) = ′
=
y (2) 4
′
139
139
Chapter 3. Derivatives and Differentials
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Rate of change
Question 13 (L.O.2): The management fee for running services in an area is calculated
by the formula f (x) = 0.0682x 2 + 0.1082x (in dollars), where x is the number of homes
being present in the area. Suppose that at the moment when 60 homes are present in
this area, the number of homes is increasing at a rate of 2 homes per week. What is the
rate of change (in dollars/week) in the management fee at this moment?
C 16.6788
A 16.664
B 16.1436
D 16.5844
E 17.1895
SOLUTION
The rate of change (in dollars/week) in the management fee at the moment when x = 60
homes are present in the area and x ′ (t ) = 2 homes per week is
f ′ (t ) = f ′ (x).x ′ (t ) = f ′ (60) × 2 = 16.5844 ≈ 16.5844.
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Related rates
Question 14 (L.O.2): Recall that volume of a right cylinder is V = π.r 2 .h, where r is
the radius of the base and h is the height. Given a right cylinder with the radius of the
base of 9 cm and the height of 5 cm. The radius of the base is increasing at a rate of
0.8 cm/s, and its height is increasing at a rate of 0.21 cm/s. How fast is the volume of
cylinder increasing?
A 278.9661
B 280.2004
C 279.6332
D 279.1033
E 279.8283
SOLUTION
The formula for the volume of a right cylinder is
V = π.r 2 .h,
(3.25)
where r is the radius of the base and h is the height.
In this problem, V, r and h are functions of the time t in seconds. Taking the derivative of
both sides of equation (3.25) with respect to time yields
µ
¶
dV
dr
2 dh
=π r
+ h.2r
(3.26)
dt
dt
dt
Since the radius of the base is increasing at a rate of 0.8 cm/s and the height is increasing
at a rate of 0.21 cm/s,
dr
dh
= 0.8,
= 0.21.
dt
dt
Substituting these, as well as r = 9 and h = 5, into equations (3.26) yields
£
¤
dV
= π 92 × 0.21 + 5 × 2 × 9 × 0.8 = 89.01π ≈ 279.6332
dt
140
140
X. Exercises
Because the sign of
dV
is positive, the volume of the right cylinder is increasing at a rate of
dt
279.6332 cm 3 /s.
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 15 (L.O.2): Water pours into a conical tank of height 10(m) and radius 4(m)
at a rate of 7(m 3 /mi n).
4
r
10
h
At what rate is the water level rising when the level is 5(m) high?
C 0.1141
A 0.5570
B −0.3594
D 0.0975
E 0.7804
SOLUTION
Let V and h be the volume and height of the water in the tank at time t . Our problem is:
dh
dV
Compute
at h = 5 given that
= 7m 3 /mi n. When the water level is h, the volume of
dt
dt
1
2
water in the cone is V = πhr , where r is the radius of the cone at height h, but we cannot
3
use this relation unless we eliminate the variable r . Using similar triangles, we see that
dV
4
1
dV
dh
dh
r
dt
=
⇒ r = 2/5 × h ⇒ V = πh(2/5 × h)2 ⇒
= 4/25πh 2
⇒
=
h 10
3
dt
dt
dt
4/25πh 2
dh
7
7
=
≈ 0.5570(m/mi n).
=
2
dt
4/25π5
4π
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
When h = 5, the level is rising at a rate of
Question 16 (L.O.2): A leaky water tank is in the shape of an inverted right circular
cone with depth of 11(m) and top radius 3(m). When the water in the tank is 3(m) deep,
it is leaking out at a rate of 1/6(m 3 /mi n).
3
r
11
h
How fast is the water level in the tank dropping at that time?
D 0.1284
A 0.0792
B 0.3362
C −0.6292
E −0.5812
SOLUTION
Let V and h be the volume and depth of the water in the tank at time t . Our problem is:
141
141
Chapter 3. Derivatives and Differentials
dh
dV
at h = 3 given that
= 1/6m 3 /mi n. When the water level is h, the volume
dt
dt
1
of water in the cone is V = πhr 2 , where r is the radius of the cone at height h, but we
3
cannot use this relation unless we eliminate the variable r . Using similar triangles, we see
that
Compute
dV
3
1
dV
dh
r
dh
dt
=
⇒ r = 3/11 × h ⇒ V = πh(3/11 × h)2 ⇒
= 9/121πh 2
⇒
=
h 11
3
dt
dt
dt
9/121πh 2
When h = 3;
−1/6
121
dV
dh
= −1/6 ⇒
=
=−
≈ −0.0792(m/mi n). Because the sign
2
dt
dt
9/121π3
486π
dh
is negative, the water level in the tank is dropping a rate of 0.0792(m/mi n).
dt
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
of
Question 17 (L.O.2): Boyle’s law states that when a sample of gas is compressed at a
constant temperature, the pressure P and volume V satisfy equation PV = C = const .
Suppose that at a certain instant the volume is 647cm 3 , the pressure is 184kP A, and the
pressure is increasing at a rate of 28kP A/mi n. At what rate is the volume decreasing at
this instant?
A 98.1919
B 99.3825
D 97.6455
E 98.362
C 98.4565
SOLUTION
Since the pressure P and volume V satisfy equation
PV = C = const ,
(3.27)
where P and V are functions of the time t in minutes, so we can differentiate both sides of
equation (3.27) with respect to time and receive
dP
dV
·V +P ·
= 0.
dt
dt
(3.28)
Since the pressure P is increasing at a rate of 28 kPA/min, so
dP
= 28
dt
Substituting these, as well as V = 647cm 3 and P = 184kP A, into equations (3.28) yields
28 × 647 + 184 ·
⇒
Because the sign of
dV
=0
dt
dV
28 × 647
=−
= 98.4565217391304 ≈ −98.4565.
dt
184
dV
is negative, the volume is decreasing at a rate of 98.4565 (cm 3 /mi n).
dt
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Undetermined coefficients
142
142
X. Exercises
Question 18 (L.O.2): Let y = Ax 3 +B x +C , where A, B,C ∈ R, satisfy the equation y ′′′ +
2y ′′ − 5y ′ + 9y = 27x 3 − 45x 2 + 54x + 89. Find the value of S = A + B +C .
C 16
A 11
B 9
D 15
E 14
SOLUTION

′
y
=
3Ax 2 + B

y ′′ = 6Ax
We have y = Ax 3 + B x +C ⇒
 ′′′
y
= 6A
Substituting these derivatives into the equation y ′′′ +2y ′′ −5y ′ +9y = 27x 3 −45x 2 +54x +89,
we receive
6A + 2 × 6Ax − 5 × (3Ax 2 + B ) + 9 × (Ax 3 + B x +C ) = 27x 3 − 45x 2 + 54x + 89
⇔ 9Ax 3 − 15Ax 2 + (12A + 9B )x + (6A − 5B + 9C ) = 27x 3 − 45x 2 + 54x + 89


9A = 27



 A=3
−15A = −45
⇔
⇔ B =2


12A + 9B = 54


C =9
6A − 5B + 9C = 89
Therefore, S = A + B +C = 14
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 19 (L.O.2): Let y = Ax cos x + B x sin x, where A, B ∈ R, satisfy the equation
y ′′ + y = −4 sin x + 10 cos x. Find the value of S = A + B.
A 6
B 9
C 8
D 5
E 7
SOLUTION
We have y = Ax cos x + B x sin x
½
⇒
y ′ = A cos x − Ax sin x + B sin x + B x cos x
y ′′ = −A sin x − A sin x − Ax cos x + B cos x + B cos x − B x sin x
Substituting these derivatives into the equation y ′′ + y = −4 sin x + 10 cos x, we receive
−2A × sin x + 2B × cos x = −4 sin x + 10 cos x
½
½
−2A = −4
A=2
⇔
⇔
2B = 10
B =5
Therefore, S = A + B = 7
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 20 ([L.O.2): Let y = Ax p + B x + C , where A, B,C , p ∈ R, satisfy the equation
30
y ′′ = 7 and y ′ (1) = −3.0, y(1) = 5.0. Find the value of S = (A + B +C ).p.
x
A 11
B 7
D 8
E 12
C 9
SOLUTION
If y = Ax p + B x +C
½
⇒
y ′ = Apx p−1 + B
y ′′ = Ap(p − 1)x p−2
143
143
Chapter 3. Derivatives and Differentials
Substituting these derivatives into the equation y ′′ =
30
, we receive
x7
Ap(p − 1)x p−2 = 30x −7
½
½
Ap(p − 1) = 30
A=1
⇒
⇒
p − 2 = −7
p = −5
Also,
½
A × 1p + 1B +C = 5.0
⇒
Ap × 1p−1 + B = −3.0
½
B =2
C =2
Therefore, S = (A + B +C ).p = 7
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Linear approximation
Question 21 (L.O.2): A spherical balloon is inflated so that its radius increases from
30 cm to 30.53 cm in 2 seconds. By approximately how much has its volume increased
(using linear approximation) (in cm 3 /s)?
A 2998.0140
B 2997.0794
C 2998.0033
D 2996.8578
E 2996.1505
SOLUTION
4
The volume of the spherical balloon is V = πR 3 . So the change of volume in 1 second is
3
∆V ≈ V ′ (R).R ′ (t ) = 4πR 2 .R ′ (t ) ≈ 4π × 302 ×
30.53 − 30
= 954.000000000002π ≈ 2997.0794.
2
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 22 (L.O.1): Use the linear approximation formula to find an approximate
value of f (x) at x = 7.68, given that f (7) = 4 and f ′ (7) = 0.41.
A 4.2788
B 4.1361
D 4.5952
E 3.6586
C 3.6549
SOLUTION
The approximate value of f (x) at x = 7.68, is
f (7.68) ≈ f (7) + f ′ (7)∆x = f (7) + f ′ (7) × (7.68 − 7) = 4.2788 ≈ 4.2788.
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 23 (L.O.2): A company estimates that volume sales S(x) of a type of commodity depends on the amount of money x (in thousands dollars) spent on advertising,
that is given by:
S(x) = −0.0014x 3 + 0.4468x 2 + 1.9363x + 325.
Use the differential (also known as: the linear approximation formula) to estimate
the change in the total sales of this commodity if the advertising expenditures are
increased from 28 to 28.8689 (thousands dollars).
A 21.4195
B 19.7626
D 20.7240
E 21.1726
C 20.5619
144
144
X. Exercises
SOLUTION
The change in the total sales of this commodity if the advertising expenditures are increased
from 28 to 28.8689 (thousands dollars) is
∆S ≈ S ′ (28)∆x = (3 × (−0.0014) × 282 + 2 × 0.4468 × 28 + 1.9363) × (28.8689 − 28) =
= 20.5619173473358 ≈ 20.5619.
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 24 (L.O.2): A ball of ice melts so that its radius decreases from 14 cm
to 13.2376 cm. By approximately how much does the volume of the ball decrease?
A −1877.80
B −1877.7054
C −1878.4110
D −1877.5370
E −1877.1782
SOLUTION
4 3
The volume of the ball of ice is V = πR
3
⇒ ∆V ≈ dV = V ′ (R)d R = V ′ (R)∆R = 4πR 2 ∆R
Since ∆R = 13.2376 − 14 = −0.7624 cm and R = 14 cm, so
∆V ≈ 4π × 142 × (−0.7624) = −597.721π ≈ −1877.80 cm 3 .
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 25 (L.O.1): Let f be an even function and f is differentiable in R. Given that
f ′ (1) = 2.81, f (1) = −3. Use differential formula to approximate f (−0.22).
C −5.673
A −6.0772
B −5.1918
D −6.0849
E −6.0731
SOLUTION
Using property of even function and differential formula, we have
f (−0.22) = f (0.22) ≈ f (1) + f ′ (1)(0.22 − 1) = −5.1918 ≈ −5.1918
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
The second order differential
Question 26 (L.O.1): Find d 2 y(0) if y = cos6 (7x).
A −292d x 2
B −289d x 2
C −298d x 2
D −299d x 2
E −294d x 2
SOLUTION
We have
y ′ = 6 cos5 (7x)(− sin(7x)) × 7 = −42 cos5 (7x) sin(7x)
⇒ y ′′ = −42 × 5 cos4 (7x)(− sin(7x)) × 7 × sin(7x) − 42 × 7 cos5 (7x) cos(7x).
145
145
Chapter 3. Derivatives and Differentials
Thus
y ′′ (0) = −42 × 7 = −294 ⇒ d 2 y(0) = −294d x 2 .
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Monotonicity
ln(3x + 4)2
, then which statement is always true?
3x + 4 µ
¶
4 e 4 e
is decreasing on − − , − +
¶
µ 3 3 3¶ 3 µ
4 4 e
4 e 4
is increasing on − − , − and − , − +
3 3 3
µ 3 3 3
¶
4 e 4 e
is increasing on − − , − +
3 3 3 3
is increasing on R
µ
¶ µ
¶
4 e
4 e
is increasing on − − , 0 ∪ 0, − +
3 3
3 3
Question 27 (L.O.1): If y =
A The function y
B The function y
C The function y
D The function y
E The function y
SOLUTION
We have
¡
¡
¢¢
2
2
−
log
+
4)
3
·
(3x
2(3
−
3
ln
|3x
+
4|)
ln(3x + 4)2 2 ln |3x + 4|
=
⇒ y′ =
y=
=
3x + 4
3x + 4
(3x + 4)2
(3x + 4)2
Thus
−4 − e
4 e
x=
=− −

3
3 3
y ′ = 0 ⇔ |3x + 4| = e ⇔ 
−4 + e
4 e
x=
=− +
3
3 3

x
−∞
f ′ (x)
4 e
− −
3 3
−
+
0
−4/3
+∞
0
4 e
− +
3 3
−
+
0
µ
¶
4 e
f − +
3 3
+∞
f (x)
4 e
f − −
3 3
µ
¶
−∞
0
¶
µ
¶
4 e
4 e
Therefore, y is increasing on − − , −4/3 and −4/3, − + .
3 3
3 3
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
µ
146
146
X. Exercises
tan x − 3
Question 28 (L.O.1): Find all real numbers m such that y =
is increasing on
tan x − m
³ π´
the interval 0; .
4
C mÊ0
A 1Ém<3
B m>3
D m É 0 or 1 É m < 3
E mÉ0
SOLUTION
1
> 0, ∀x ∈ R and y ′ (x) = y ′ (t ).t ′ (x). So if y ′ (x) >
Let t = tan x ⇒ t ∈ (0; 1). Then t ′ (x) =
2x
cos
³ π´
0, ∀x ∈ 0;
⇒ y ′ (t ) > 0, ∀t ∈ (0; 1). We have
4
y(t ) =
t −3
3−m
⇒ y ′ (t ) =
t −m
(t − m)2
Thus y is increasing on (0; 1) when
½
y ′ (t ) > 0
⇔
m ∉ (0; 1)
½
m<3
⇔
m ∉ (0; 1)
½
1Ém<3
mÉ0
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Local extreme values
Question
Studying the local extreme values of the function f (x) =
µ 29 (L.O.1):
¶
3x − 9
arctan 2
, which statement is always true?
x + 72
A f has local minimum when x = −6, and no local maximum
B f has local minimum when x = −6, and local maximum when x = 12
C f has local maximum when x = −6, and local minimum when x = 12
D f has no local minimum, and local maximum when x = 12
E f has no local minimum, and no local maximum
SOLUTION
µ
Since f (x) = arctan
3x − 9
x 2 + 72
¶
−3x 2 + 18x + 216
¡
¢
3 −x 2 + 6x + 72
−3x 2 + 18x + 216
(x 2 + 72)2
′
⇒ f (x) =
= 2
=
¶
µ
(x + 72)2 + (3x − 9)2 x 4 + 153x 2 − 54x + 5265
3x − 9 2
1+ 2
x + 72
⇒ f ′ (x) = 0 ⇔ x = −6 ∨ x = 12
147
147
Chapter 3. Derivatives and Differentials
x
−∞
f ′ (x)
−
+∞
12
−6
+
0
0
0
−
f (12)
f (x)
f (−6)
0
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Maximum and minimum values
Question 30 (L.O.2): Find the maximum value of
½
f (x) =
A 23
B 31
e 2−x + x 2 + 9x + 5, if 0 É x É 2
x 2 − 24x + 72, if 2 < x É 12
C 25
D 30
E 28
SOLUTION
If 0 É x É 2 then
f (x) = e 2−x + x 2 + 9x + 5 ⇒ f ′ (x) = 2x − e 2−x + 9
and f ′′ (x) = e 2−x + 2 > 0, ∀x ∈ (0, 2). Therefore,
f ′ (x) Ê f ′ (0) = −e 2 + 9 > 0, ∀x ∈ (0, 2) ⇒ f max = f (2) = 28, x ∈ [0, 2].
If 2 É x É 12 then
f (x) = x 2 − 24x + 72 ⇒ f ′ (x) = 2x − 24 < 0, ∀x ∈ (2, 12) ⇒ f max = f (2) = 28, ∀x ∈ [2, 12].
The function f is continuous at x = 2. Thus, f max = f (2) = 28, ∀x ∈ [0, 12].
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Concavity
Question 31 (L.O.1): If y = x 3 e (12/11)x , then the number of point(s) of inflection of the
graph of y is
A 4
B 1
C 0
D 2
E 3
SOLUTION
Domain: D = R
y ′ = 3x 2 e 12/11x +
′′
⇒ y = 6xe
12/11x
12
× x 3 e 12/11x .
11
µ ¶2
12 2 12/11x
12
12 2 12/11x
+3× x e
+3× x e
+
x 3 e 12/11x =
11
11
11
148
148
X. Exercises
= xe
12/11x
h µ 12 ¶2
11
x2 + 6 ×
i
12
x +6 .
11
y ′′ = 0 ⇔ x = 0 ∨ x = −3/2 ∨ x = −4.
x
−∞
−4
+∞
0
−3/2
f ′′ (x)
−
0
+
0
−
0
+
f (x)
CD
PI
CU
PI
CD
PI
CU
Therefore, the graph of y has 3 points of inflection.
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Applications of differentiation in Phyics
Question 32 (L.O.1): A particle moving on the x−axis has position s(t ) = t 3 − 39t 2 +
432t + 12 (m) after an elapsed time of t seconds. What is the total distance travelled by
the particle during the first 31 seconds?
A 6699
B 6707
C 6704
D 6709
E 6702
SOLUTION
We have s(t ) = t − 39t + 432t + 12 ⇒ v(t ) = s ′ (t ) = 3t 2 − 78t + 432.
3
2
·
v(t ) > 0 ⇔
0<t <8
18 < t < 31
and v(t ) < 0 ⇔ 8 < t < 18
So the total distance travelled by the particle during the first 31 seconds is
|s(8) − s(0)| + |s(18) − s(8)| + |s(31) − s(18)| = 6704.
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 33 (L.O.1): A particle moving on the x−axis has position s(t ) = t 3 − 45t 2 +
600t + 9 (m) after an elapsed time of t seconds. What is the displacement travelled by
the particle during the first 37 seconds?
A 11243
B 11250
C 11248
D 11251
E 11253
SOLUTION
We have s(t ) = t − 45t + 600t + 9. So the displacement travelled by the particle during the
first 37 seconds is
s(37) − s(0) = 11248.
3
2
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Applications of differentiation in Economics
149
149
Chapter 3. Derivatives and Differentials
Question 34 (L.O.2): For Luce Landscaping, the total revenue from the yard maintenance of x homes is given by R(x) = 1868x − 4x 2 (dollars) and the total cost is given
by C (x) = 2938 + 18x (dollars). Suppose that Luce is adding 19 homes per day at the
moment when the 458th customer is signed. At that moment, what is the rate of change
of total profit P in dollars per day, if the total profit P (x) = R(x) −C (x)?
A −34469
B −34465
C −34471
D −34466
E −34463
SOLUTION
We have
h
i
P (x(t )) = R(x(t )) −C (x(t )) ⇒ P ′ (t ) = P ′ (x).x ′ (t ) = R ′ (x) −C ′ (x) x ′ (t ).
From the given information, we have x = 458 and x ′ (t ) = 19. Therefore,
h
i
h
i
P ′ (t ) = 1868 − 2 × 4x − 18 x ′ (t ) = 1868 − 2 × 4 × 458 − 18 × 19 = −34466.
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
150
150
Chapter 4
Integration
Learning Objectives
Study the Fundamental Theorem of Calculus.
Study some techniques of integration
Study some applications of integrals.
I
Anti-derivatives and indefinite integrals
Example 112
Data on the growth of world population provided by the U.S. Census Bureau can be
used to create a model of Earth’s population growth. According to this model, the
rate of change of the world’s population since 1950 is given by p(t ) = −0, 012.t 2 +
48.t − 47925, where t is the calendar year and p(t ) is in millions of people per year.
1. Given that the population in 2000 was about 6000 million people, find an
equation for P (t ), the total population as a function of the calendar year.
2. Use the equation P (t ) to predict the world population in 2050.
SOLUTION
1. P (t ) is the antiderivative of p(t )
p(t ) = −0, 012.t 2 + 48.t − 47925
t3
t2
+ 48. − 47925.t +C
3
2
To find C , substitute 2000 for t and 6000 for P (t ). We receive C = 31856000 and
⇒ P (t ) = −0, 012.
P (t ) = −0, 004.t 3 + 24.t 2 − 47925.t + 31856000
2. Substitute 2050 for t , according to the model the world population in 2050 should be
about P (2050) = 9250 million people.
151
Chapter 4. Integration
1
Definition
Definition 74
A function F is called an anti-derivative of f on an interval X , if F (x) is continuous
and differentiable on X and F ′ (x) = f (x), or d F (x) = f (x)d x for all x ∈ X .
Theorem I.1
If F is an anti-derivative of f on an interval X ⊂ R then the most general antiderivative of f on X is Φ(x) = F (x) +C , where C is an arbitrary constant
Example 113
1
The general anti-derivative of f (x) = x 2 is x 3 +C .
3
Definition 75
Let F be any anti-derivative of f on an interval X ⊂ R. The indefinite integral of f (x)
is defined by
Φ(x) = F (x) +C ,
where C is an arbitrary constant.
Remark 13
Z
Indefinite integral is denoted by
f (x)d x.
The process of computing an integral is called integration.
Here, f (x) is called the integrand and the term d x identifies x as the variable
of integration.
152
152
I. Anti-derivatives and indefinite integrals
2
Some basic formulas of indefinite integrals
Z
Some basic formulas of indefinite integrals
Z
0.d x = C .
cos xd x = sin x +C .
Z
Z
1.d x = x +C .
x α+1
x dx =
+C , α ̸= −1.
α+1
Z
Z
dx
= ln |x| +C .
x
Z
Z
ax
+C , a > 0, a ̸= 1.
a dx =
ln a
Z
Z
α
x
cosh xd x = sinh x +C .
Z
x
dx
cosh2 x
Z
dx
= arcsin x +C , x ̸= ±1.
p
1 − x2
Z
= − cot x +C .
sinh xd x = cosh x +C .
e d x = e +C .
Z
dx
sin2 x
x
Z
dx
= tan x +C .
cos2 x
dx
sinh2 x
= tanh x +C .
= − coth x +C .
¯x −a¯
dx
1
¯
¯
ln
=
¯
¯ +C .
x 2 − a 2 2a
x +a
Z
¯
¯
p
dx
¯
¯
= ln ¯x + x 2 + a ¯ +C .
p
2
x +a
Z
dx
= arctan x +C .
1 + x2
Z
sin xd x = − cos x +C .
Some simple rules of indefinite integrals
1. Rule I. If a ̸= 0 then
Z
Z
a f (x)d x = a
f (x)d x.
2. Rule II.
Z h
Z
i
f (x) ± g (x) d x =
Z
f (x)d x ±
g (x)d x.
Z
f (t )d t = F (t ) +C then
3. Rule III. If
Z
f (ax + b)d x =
1
F (ax + b) +C , (a ̸= 0).
a
Example 114
Z
Find
p
dx
a2 − x2
·
SOLUTION
153
153
Chapter 4. Integration
Substitute t =
x
from the formula
a
Z
Z
1
x
dt
dx
=
arcsin
=
arcsin
t
+C
⇒
+C
p
q
1
¡ x ¢2
a
1− t2
a
1− a
dx
Z
⇒
p
Z
a2 − x2
=
1
x
x
dx
q
¡ x ¢2 = a · a. arcsin a +C = arcsin a +C .
a. 1 − a
Example 115
dx
Z
Find
x2 + a2
·
SOLUTION
x
Substitute t = from the formula
a
Z
Z
x
dt
dx
1
= arctan t +C ⇒ ¡ ¢2
= 1 arctan +C
2
x
t +1
a
+1 a
a
Z
⇒
II
1
dx
=
2
x + a2
Z
a2
dx
h¡ ¢
x 2
a
+1
i=
1
x
1
x
· a. arctan +C = arctan +C .
2
a
a
a
a
Techniques of indefinite integration
The substitution rule
Theorem II.1 (The substitution rule)
Let composite function f (u(x)) define on interval X , and let function t = u(x) be
differentiable on interval X . If f (t ) has anti-derivative F (t ) on an interval T ⊇ u(X )
then
Z
(4.1)
f (u(x))d u(x) = F (u(x)) +C .
Remark 14 (Case I)
Z
If we can not compute the integral
g (x)d x directly, we often look for a new variable
u and function f (u) for which
Z
Z
g (x)d x =
′
µZ
Z
f (u(x)).u (x)d x =
f (u(x))d u(x) =
Z
where the integral
¶¯
¯
f (t )d t ¯¯
Z
f (t )d t is easier to evaluate than
154
154
g (x)d x.
t =u(x)
II. Techniques of indefinite integration
Example 116
Z
Find
sin3 x cos xd x.
SOLUTION
Let t = sin x, d t = cos xd x. This gives us
Z
Z
t4
sin4 (x)
3
sin x cos xd x = t 3 d t = +C =
+C .
4
4
Remark 15 (Case II)
Z
In some cases, the integral
f (x)d x will be easier to evaluate if we change x by a
new function x = ϕ(t ) with a new variable t . At this time, we have
f (x)d x = f (ϕ(t )).ϕ′ (t )d t .
So
Z
Z
f (x)d x =
f (ϕ(t ))ϕ′ (t )d t
In the result, we substitute t = ϕ−1 (x), where ϕ−1 is the inverse function of ϕ.
Example 117
Evaluate I =
Z p
a 2 − x 2 d x,
a > 0 is a constant.
π
π
x
É t É · Then t = arcsin , d x = a cos t d t .
2
2
a
p
p
p
a 2 − x 2 = a 2 − a 2 sin2 t = a 2 cos2 t = a| cos t | =
π
π
= a cos t . Note that cos t Ê 0 because − É t É · Thus the Substitution Rule gives
2
2
Z p
Z
a 2 − x 2 d x = a 2 cos2 t d t =
SOLUTION Let x = a sin t , where −
¶
µ
Z
a2
a2
1
(1 + cos 2t )d t =
=
t + sin 2t +C =
2
2
2
x
Substituting t = arcsin , we have
a
³
³
x
x´
x ´i
a2 h
I=
arcsin + sin arcsin cos arcsin
+C =
2
a
a
a
r
·
¸
³
³
a2
x
x´
x´
2
=
arcsin + sin arcsin
1 − sin arcsin
+C
2
a
a
a
Ã
!
r
³ x ´2
a2
x x
1−
=
arcsin +
+C .
2
a a
a
Z p
1 p
a2
x
a2 − x 2d x = x a2 − x 2 +
arcsin +C
2
2
a
155
155
(4.2)
Chapter 4. Integration
2
Integration by Parts
If u and v are differentiable functions, then
d
[u(x).v(x)] = u(x).v ′ (x) + u ′ (x).v(x)
dx
Z
⇒ [u(x).v ′ (x) + u ′ (x).v(x)]d x = u(x).v(x)
Z
Z
′
⇒ u(x).v (x)d x + u ′ (x).v(x)d x = u(x).v(x)
Theorem II.2
If functions u = u(x) and v = v(x) are differentiable on interval X ⊂ R, then
Z
Z
ud v = uv −
vd u
Example 118
Z
Find I =
x sin xd x.
SOLUTION
Let
u = x,
d v = sin xd x.
Then
d u = d x,
v = − cos x.
Thus, using formula for integration by parts, we have
Z
Z
I = x sin xd x = x(− cos x) − (− cos x)d x =
Z
= −x cos x +
cos xd x = −x cos x + sin x +C .
Example 119
Z
Find I =
ln xd x.
SOLUTION
Let
u = ln x,
Then
du =
d v = d x.
1
d x,
x
v = x.
Integrating by parts, we get
Z
Z
Z
dx
I = ln xd x = x ln x − x ·
= −x ln x − d x = x ln x − x +C .
x
156
156
(4.3)
II. Techniques of indefinite integration
Example 120
Find I =
Z p
x 2 + a.d x,
where a is any constant.
SOLUTION
Suppose that we choose u =
p
x 2 + a,
d v = d x. Then
xd x
du = p
,
x2 + a
v = x.
Thus, using formula for integration by parts, we have
I=
Z p
=x
I =x
x 2 + a.d x = x
p
p
x2 + a −
x2 + a −
=x
p
Z
p
Z
x 2 + a.d x + a
x2 + a − I
p
xd x
=
x·p
x2 + a
Z
x2 + a
dx +
p
x2 + a
Z p
⇒ 2I = x
x2 + a −
Z
+a
x2 + a + a
p
ad x
p
x2 + a
Z
p
dx
x2 + a
=
dx
x2 + a
Z
p
dx
x2 + a
Therefore,
Z p
p
¯
p
x
x 2 + a a ¯¯
¯
2
2
x + a.d x =
+ ln ¯x + x + a ¯ +C
2
2
Remark 16
We use method of integration by parts in the following cases:
Z
1. x k lnm xd x (k, m ∈ Z0 )
Z
2.
x k e ax d x
Z
Z
k
x sin axd x,
3.
Z
4.
(k ∈ Z0 )
e
ax
Z
sin bxd x,
x k cos bxd x
(k ∈ Z0 )
e ax cos bxd x.
157
157
(4.4)
Chapter 4. Integration
Example 121
Prove the reduction formula
I n+1 =
1
x
2n − 1
· 2
+
In
2
2
n
2na (x + a )
2na 2
(4.5)
where
dx
Z
In =
(x 2 + a 2 )n
SOLUTION
Let
u=
1
(x 2 + a 2 )n
Then
du =
(n ∈ N)
(4.6)
, d v = d x.
−2nxd x
, v = x.
(x 2 + a 2 )n+1
So integration by parts gives
x
In = 2
+ 2n
(x + a 2 )n
=
x
+ 2n
(x 2 + a 2 )n
=
Therefore
x
(x 2 + a 2 )n
Z
Z
x 2d x
=
(x 2 + a 2 )n+1
(x 2 + a 2 ) − a 2
dx =
(x 2 + a 2 )n+1
+ 2nI n − 2na 2 I n+1
·
¸
1
x
I n+1 =
+ (2n − 1)I n .
2na 2 (x 2 + a 2 )n
Since
Z
I1 =
dx
x2 + a2
=
x
1
arctan +C ,
a
a
so when n = 1 we can calculate I 2 , and then I 3 , etc.
Example 122
Z
Find
dx
(x 2 + 4)2
·
SOLUTION
By formula (4.5) we have a = 2, n = 1. Therefore
Z
dx
1
x
1
= · 2
+
2
2
(x + 4)
8 x +4 8
=
Z
dx
x2 + 4
1
x
1
x
· 2
+
· arctan +C .
8 x + 4 16
2
158
158
=
III. Integration of rational functions by partial fractions
III Integration of rational functions by partial fractions
1
Partial fractions
Definition 76
The fractions of the forms
1)
3)
A
;
x −a
Mx +N
x 2 + px + q
2)
;
4)
A
(x − a)k
(k = 2, 3, . . .);
Mx +N
(x 2 + px + q)m
where A, a, M , N , p, q are constants and q −
(m = 2, 3, ...)
p2
> 0, are called partial fractions.
4
The
Z indefinite integrals of these partial fractions are:
A
1.
d x = A ln |x − a| +C .
Z x −a
A
A
2.
dx =
+C , (n ̸= 1).
n
n−1
(1 − n)(x
µ
¶Z
Z − a)
Z (x − a)
M
(2x + p)d x
Mp
dx
Mx +N
dx =
+ N−
=
3.
2
2
x + px + q
2
x + px + q
2
(x + p/2)2 + q − p 2 /4
µ
¶
M
Mp
1
x + p/2
2
ln |x + px + q| + N −
·p
arctan p
+C
=
2
2
q − p 2 /4
q − p 2 /4
µ
¶Z
Z
Z
Mx +N
2x + p
dx
M
Mp
4.
d
x
=
d
x
+
N
−
=
(x 2 + px + q)m
2
(x 2 + px + q)m
2
(x 2 + px + q)m
µ
¶Z
Mp
dx
M (x 2 + px + q)−m+1
·
+ N−
,
=
2
2
−m + 1
2
(x + px + q)m
where
Z
dx
=
2
(x + px + q)m
can be found by putting t = x +
2
Z
·³
x+
p ´2
2
dx
µ
¶¸m
p2
+ q−
4
p2
p
,q −
= a 2 and applying reduction formula.
2
4
Integration of rational functions by partial fractions
Z
Integration of rational functions I =
with degrees n and m respectively.
Method of partial fractions
P n (x)
d x, where P n (x),Q m (x) are polynomials
Q m (x)
1. If n Ê m then we divide P n (x) into Q m (x)
P n (x)
R(x)
= S(x) +
Q m (x)
Q m (x)
159
159
Chapter 4. Integration
2. If n < m then we factorize the denominator as
Q m (x) = (x − a)k . . . (x 2 + px + q)ℓ ,
where k + . . . + 2ℓ = m,
p2
−q <0
4
The partial fraction decomposition of the integrand
P n (x)
has the form
Q m (x)
M 1 x + N1
P n (x)
A1
A2
Ak
M ℓ x + Nℓ
+...+ 2
=
+
+...+
+...+ 2
2
k
Q m (x) x − a (x − a)
x + px + q
(x − a)
(x + px + q)ℓ
In order to find the coefficients A 1 , A 2 , . . . , A k , M 1 , . . . , M ℓ , N1 , . . . , Nℓ , we can choose values
x that simplify the equation.
Example 123
Z
Find
x3 + x
d x.
x −1
SOLUTION Since the degree of the numerator is greater than the degree of the denominator,
we perform the division.
Z
x3 + x
dx =
x −1
=
Z µ
¶
2
x +x +2+
dx =
x −1
2
x3 x2
+
+ 2x + 2 ln |x − 1| +C .
3
2
Example 124
Z
Evaluate I =
x 2 + 2x + 6
d x.
(x − 1)(x − 2)(x − 4)
SOLUTION The method of partial fractions gives
A
B
C
x 2 + 2x + 6
=
+
+
(x − 1)(x − 2)(x − 4) x − 1 x − 2 x − 4
⇒ x 2 + 2x + 6 ≡ A(x − 2)(x − 4) + B (x − 1)(x − 4) +C (x − 1)(x − 2), ∀x ∈ R
We put x = 1 and get: 9 = A(1 − 2)(1 − 4) ⇒ A = 3
We put x = 2, we get: 14 = B (2 − 1)(2 − 4) ⇒ B = −7
We put x = 4, we get: 30 = C (4 − 1)(4 − 2) ⇒ C = 5
Z
I =3
dx
−7
x −1
Z
dx
+5
x −2
Z
dx
=
x −4
= 3 ln |x − 1| − 7 ln |x − 2| + 5 ln |x − 4| +C =
¯
¯
¯ (x − 1)3 (x − 4)5 ¯
¯
¯ +C
= ln ¯
¯
(x − 2)7
160
160
III. Integration of rational functions by partial fractions
Example 125
Z
Evaluate I =
x2 + 1
d x.
(x − 1)3 (x + 3)
SOLUTION The partial fraction decomposition is
x2 + 1
A
B
C
D
=
+
+
+
3
3
2
(x − 1) (x + 3) (x − 1)
(x − 1)
x −1 x +3
⇒ x 2 + 1 ≡ A(x + 3) + B (x − 1)(x + 3) +C (x − 1)2 (x + 3) + D(x − 1)3 , ∀x ∈ R
1
We put x = 1, we get: 2 = 4A ⇒ A =
2
5
We put x = −3, we get: 10 = −64D ⇒ D = −
32
5
3
Now we equate coefficients of x in both sides, we get C + D = 0 ⇒ C =
32
3
We put x = 0, we get: 1 = 3A − 3B + 3C − D ⇒ B = . So
8
Z
Z
Z
Z
1
dx
3
dx
5
dx
5
dx
I=
+
+
−
=
3
2
2 (x − 1)
8 (x − 1)
32 x − 1 32 x + 3
¯
¯
1
5 ¯¯ x − 1 ¯¯
3
=−
+
ln
−
+C
4(x − 1)2 8(x − 1) 32 ¯ x + 3 ¯
Example 126
Z
Evaluate I =
dx
x5 − x2
·
SOLUTION We factorize the denominator
x 5 − x 2 = x 2 (x − 1)(x 2 + x + 1).
Then the partial fraction decomposition is
1
x5 − x2
=
C
Dx +E
A B
+ 2
+ +
2
x
x x −1 x +x +1
⇒ 1 ≡ A(x − 1)(x 2 + x + 1) + B x(x − 1)(x 2 + x + 1)+
+C x 2 (x 2 + x + 1) + (D x + E )x 2 (x − 1), ∀x ∈ R.
We put x = 0, we get: 1 = −A ⇒ A = −1
1
We put x = 1, we get: 1 = 3C ⇒ C =
3
Now we equate coefficients of x 4 , x 3 , x 2 in both sides
1 ≡ A(x 3 − 1) + B (x 4 − x) +C (x 4 + x 3 + x 2 ) + D x 4 + E x 3 − D x 3 − E x 2 ,
we have





B =0
B +C + D = 0
1
A +C + E − D = 0 ⇔ D = − 3




C −E = 0
 E=1
3


161
161
Chapter 4. Integration
Therefore
Z
1
dx
x −1
−
dx =
I =−
2
x −1 3 x +x +1
Z
1 1
1 2x + 1 − 3
= + ln |x − 1| −
dx =
x 3
6 x2 + x + 1
Z
1 1
1
1
dx
2
= + ln |x − 1| − ln(x + x + 1) +
=
2
x 3
6
2 x +x +1
¢
¡
Z
1
d
x
+
1 1
1
1
2
= + ln |x − 1| − ln(x 2 + x + 1) +
=
¡
¢
1 2
x 3
6
2
x + 2 + 43
Z
=
IV
Z
1 1
(x − 1)2
1
2x + 1
+ ln 2
+ p arctan p +C .
x 6 x +x +1
3
3
Integration of non-rational functions
Z
1
dx 1
+
x2 3
Type 1:
¶ µ
¶
µ
¶ ¶
µ µ
ax + b p n
ax + b p 1 ax + b p 2
,
,...,
dx
R x,
cx + d
cx + d
cx + d
where
p 1 , p 2 , . . . , p n are rational numbers,
a, b, c, d are real numbers.
SOLUTION Let
ax + b
= tm,
cx + d
where m is the lowest common multiple of denominators of rational numbers p 1 , p 2 , . . . , p n
Example 127
dx
Z
Evaluate I =
p
3
p
(2x + 1)2 − 2x + 1
SOLUTION
Let 2x + 1 = t 6 ⇒ x =
t6 −1
, d x = 3t 5 d t .
2
Z
I=
3t 5 d t
=3
t4 − t3
Z
t 2d t
=3
t −1
Z µ
¶
1
t +1+
dt =
t −1
3
= t 2 + 3t + 3 ln |t − 1| +C .
2
Substitute t = (2x + 1)1/6 , we have
3
I = (2x + 1)1/3 + 3(2x + 1)1/6 + 3 ln |(2x + 1)1/6 − 1| +C .
2
162
162
V. Trigonometric Integrals
2
dx
Z
Type 2:
p
ax 2 + bx + c
Example 128
dx
Z
Evaluate I =
p
x 2 + 2x + 5
SOLUTION
dx
Z
I=
p
d (x + 1)
Z
=
p
x 2 + 2x + 1 + 4
(x + 1)2 + 4
¯
¯
p
¯
¯
= ln ¯x + 1 + x 2 + 2x + 5¯ +C .
=
Example 129
Z
Evaluate I =
dx
·
p
2
−3x + 4x − 1
SOLUTION
µ
¶
2
d x−
Z
Z
1
dx
3
I= s ·
µ
¶¸ = p3 s
µ
¶2 =
1
2
4
2
1
3
− x2 − 2 · · x +
− x−
9
3
9
9
3
1
1
x − 2/3
+C = p arcsin(3x − 2) +C .
= p arcsin
1/3
3
3
V
1
Trigonometric Integrals
Z
Type 1:
R(sin x, cos x)d x
Where R(sin x, cos x) is a rational function with respect to variables sin x, cos x.
SOLUTION
x
Let t = tan · Then
2
2d t
x = 2 arctan t , d x =
,
1+ t2
2t
1− t2
sin x =
,
cos
x
=
1+ t2
1+ t2
Example 130
Z
Evaluate I =
dx
sin x
SOLUTION
x
Let t = tan · Then
2
sin x =
2t
2d t
,dx =
·
2
1+t
1+ t2
163
163
Chapter 4. Integration
So
2d t
1+t 2
2t
1+t 2
Z
I=
Z
=
¯
dt
x ¯¯
¯
= ln |t | +C = ln ¯tan ¯ +C .
t
2
Example 131
Z
Evaluate I =
dx
·
cos x
x
SOLUTION Let t = tan · Then
2
cos x =
1− t2
2d t
,dx =
·
2
1+t
1+ t2
So
2d t
1+t 2
1−t 2
1+t 2
Z
I=
h
Z
=
2d t
=
1− t2
Z
i
(1 + t ) + (1 − t ) d t
(1 − t )(1 + t )
=
¯
¯
¯ 1 + tan x2 ¯
¯
¯ +C .
= − ln |1 − t | + ln |1 + t | +C = ln ¯
1 − tan x2 ¯
Example 132
Z
Evaluate I =
dx
4 sin x + 3 cos x + 5
x
SOLUTION Let t = tan · Then
2
sin x =
2t
1− t2
2d t
,
cos
x
=
,dx =
·
2
2
1+t
1+t
1+ t2
So
Z
I=
Z
=
VI
1
2d t
1+t 2
2t
1−t 2
4. 1+t
2 + 3. 1+t 2
Z
+5
=2
dt
=
2t 2 + 8t + 8
dt
1
1
=−
+C = −
+C .
2
(t + 2)
t +2
tan x2 + 2
Definite integrals
Area under a curve
One way to estimate the area between the graph of a function f (x) and the x−axis for an
interval from x = a to x = b is by filling the region with rectangles, whose areas we know
how to compute.
164
164
VI. Definite integrals
Let f (x) be defined on the interval [a, b](a < b). We subdivide [a, b] into n equal subintervals [x i −1 , x i ] (i = 1, . . . , n). We take the sample points x i∗ ∈ [x i −1 , x i ]
The total area A n of the n rectangles is given by the sum of the areas
A n = f (x 1∗ )∆x + f (x 2∗ )∆x + . . . + f (x n∗ )∆x, where ∆x =
b−a
n
To make the width of the rectangles approach 0, we let the number of rectangles approach
∞. Therefore, the exact area of the region under the graph of the function is lim A n . This
n→∞
Z b
limit is called a definite integral and is denoted by
f (x)d x.
a
Z
b
a
f (x)d x = lim
n→∞
n
X
i =1
f (x i∗ ).∆x
(4.7)
Definition 77
1. A n = f (x 1∗ )∆x + f (x 2∗ )∆x + . . . + f (x n∗ )∆x is called Riemann sum.
2. If lim A n exists, we say that f is integrable on [a, b].
n→∞
Z
3. The symbol
is called an integral sign.
4. f (x) is called the integrand
5. a and b are called the limits of integration, a is the lower limit and b is the
165
165
Chapter 4. Integration
upper limit.
6. The procedure of calculating an integral is called integration.
2
Principle of Mathematical Induction
1. 1 + 2 + 3 + . . . + n =
n(n + 1)
2
2. 12 + 22 + 32 + . . . + n 2 =
n(n + 1)(2n + 1)
6
3. 13 + 23 + 33 + . . . + n 3 =
n 2 (n + 1)2
4
Example 133
Use the limits to find the area of the region between the graph of y = x 2 and the
x−axis from x = 0 to x = 1.
SOLUTION f (x) = x 2 , a = 0, b = 1. We subdivide [a, b] into n equal sub-intervals ∆x =
b−a 1
i
1
i
= · Then find x i = a + i ∆x = 0 + i · = . Let x i∗ = x i = · Now we can calculate
n
n
n n
n
the integral that gives the area
Z
0
1
2
x d x = lim
n→∞
n
X
i =1
(x i∗ )2 ∆x
n µ i ¶2 µ 1 ¶
X
= lim
=
n→∞
n
i =1 n
n(n + 1)(2n + 1) 1
12 + 22 + . . . + n 2
= lim
=
3
n→∞
n→∞
n
6n 3
3
= lim
166
166
VI. Definite integrals
3
Geometric meaning
Remark 17
Z
If f (x) Ê 0 on the interval [a, b] then the integral
b
f (x)d x is the area of the region
a
between the graph of y = f (x) and the x−axis from x = a to x = b.
4
Applications of integral in Construction
The contractors have been hired to clean the Gateway Arch in St. Louis. The Arch is very
close to a parabola in shape, 630 m high and 630 m across at the bottom. The equation of
x2
· The idea for approaching the project is to build
the Arch is approximately y = 630 −
157.5
scaffolding in the entire space under the Arch, so that the cleaning crew can easily climb
up and down to any point on the Arch.
To settle the matter, the contractors want to find out how much area there is under the
Arch.
167
167
Chapter 4. Integration
The area is given by
315
Z
µ
−315
5
¶
x2
630 −
d x = 264600(m 2 )
157.5
Properties of the definite integrals
Properties of the definite integrals
a
Z
1.
b
a
a
a
b
5.
a
VII
1
Z
f (x)d x =
Z bh
Z
f (x)d x
a
f (x)d x = 0
b
3.
4.
b
a
Z
2.
Z
Z
f (x)d x = −
c
a
b
Z
f (x)d x +
Z
i
f (x) ± g (x) d x =
Z
α. f (x)d x = α
c
f (x)d x, ∀c ∈ [a, b].
b
a
Z
f (x)d x ±
b
g (x)d x
a
b
a
f (x)d x, ∀α = const ant
Techniques of definite integration
The Fundamental Theorem of Calculus
Theorem VII.1 (The Fundamental Theorem of Calculus)
If f is continuous on [a, b], then the function g defined by
Z x
g (x) =
f (t )d t , a É x É b,
a
is continuous on [a, b] and differentiable on (a, b), and g ′ (x) = f (x).
168
168
VII. Techniques of definite integration
2
Newton-Leibniz’s formula
Function g is an anti-derivative of f on the interval [a, b]. If F is any other anti-derivative
of f on [a, b] then
Zx
F (x) = g (x) +C = f (t )d t +C .
a
If we put x = a, we get
Za
F (a) =
f (t )d t +C = C .
a
If we put x = b we get
Zb
F (b) =
Zb
f (t )d t +C =
a
f (t )d t + F (a).
a
Remark 18
Z
b
Z
b
f (x)d x is a number; it does not depend on x. So
The integral
Z b
f (t )d t
a
a
f (x)d x =
a
Theorem VII.2 (Newton-Leibniz’s formula)
If f is continuous on [a, b], then
Z
b
a
¯b
¯
f (x)d x = F (x)¯ = F (b) − F (a),
a
where F is any anti-derivative of f .
Example 134
Z
Evaluate I =
SOLUTION
π/4
π/6
dx
cos2 x
p
¯π/4
π
π
3
¯
I = tan x ¯ = tan − tan = 1 −
π/6
4
6
3
169
169
(4.8)
Chapter 4. Integration
3
Integration by parts for definite integrals
Theorem VII.3
If functions u ′ and v ′ are continuous on interval [a, b], then
Zb
¯b Zb
¯
ud v = uv ¯ − vd u,
(4.9)
a
a
a
Example 135
Z
Calculate I =
1
xe −x d x
0
SOLUTION Let
u = x, d v = e −x d x
⇒ d u = d x, v = −e −x .
Using formula for integration by parts, we have
¯1 Z 1
¯1
¯
−x ¯
I = −xe ¯ +
e −x d x = −e −1 − e −x ¯ =
0
0
0
= −2e −1 + 1.
4
The substitution rule for definite integrals
Theorem VII.4 (Case I)
If u ′ is continuous on [a, b] and f is continuous on the range of t = u(x), then
Zb
Zβ
′
f (u(x)).u (x)d x =
f (t )d t ,
(4.10)
α
a
The limits of integration change from x = a and x = b to the corresponding limits for
u : α = u(a), β = u(b).
Theorem VII.5 (Case II)
If ϕ(t ) and ϕ′ (t ) are continuous on the interval [α, β], f [ϕ(t )] is continuous on the
interval [a, b], then
Zβ
Zb
f (x)d x =
a
f [ϕ(t )]ϕ′ (t )d t ,
α
where a = ϕ(α), b = ϕ(β).
170
170
(4.11)
VII. Techniques of definite integration
Example 136
Ze
ln2 x
dx
x
Calculate I =
1
dx
x
· Then
t
x
SOLUTION If we let t = ln x ⇒ d t =
Z
I=
0
1
1
0
e
Therefore
1
¯1
t 3 ¯¯
1
1
t d t = ¯ = (13 − 03 ) = ·
3 0 3
3
2
Example 137
Calculate I =
Z 2p
0
4 − x 2 d x.
SOLUTION If we let x = 2 sin t ⇒ d x = 2 cos t d t . Then
x
0
t
0
2
π Therefore
2
Zπ/2p
Zπ/2
2
I=
4 − 4 sin t .2 cos t d t = 4. cos2 t d t =
0
1
= ·4
2
5
0
·
¸
Zπ/2
sin 2t π/2
(1 + cos 2t )d t = 2. t +
= π.
2
0
0
Integral of Symmetric functions
Theorem VII.6
Suppose that f is continuous on the interval [−a, a].
1. If f is odd function: f (−x) = − f (x), then
Za
f (x)d x = 0
−a
2. If f is even function: f (−x) = f (x), then
Za
Za
f (x)d x = 2
−a
0
Example 138
π/3
x sin x
dx
2
−π/3 cos x
Z
Calculate I =
f (x)d x
171
171
Chapter 4. Integration
SOLUTION
Since f (x) =
h π πi
x sin x
satisfies
f
(−x)
=
f
(x),
it
is
even
on
the
interval
− ,
and so
cos2 x
3 3
Zπ/3
I =2
0
x sin x
d x.
cos2 x
Let
sin xd x
1
⇒ d u = d x, v =
2
cos x
cos x
Using formula for integration by parts, we have


Zπ/3
¯
π/3
dx 
x ¯
=
I = 2
¯ −
cos x 0
cos x
u = x, d v =
0
¶
µ
¯
³ x π ´¯¯π/3 ¶
π
5π
2π
¯¯
¯
=2
− ln ¯tan + ¯¯
− ln tan
.
=2
3 cos(π/3)
2 4 0
3
12
µ
Example 139
x 2 arctan x
dx
p
−1
1 + x2
Z
Calculate I =
1
SOLUTION
x 2 arctan x
Since f (x) = p
satisfies f (−x) = − f (x), it is odd on the interval [−1, 1] and so
1 + x2
Z 1 2
x arctan x
d x = 0.
I=
p
−1
1 + x2
VIII
1
Improper integral of Type 1: Infinite intervals
+∞
Z
Definition of an improper integral of type 1
f (x)d x
a
Definition 78
Let f (x) be defined for every number x Ê a and be integrable on every interval [a, b].
Z b
Then Φ(b) =
f (x)d x is defined on the interval [a, +∞). The limit
a
I = lim Φ(b) = lim
b→+∞
Z
b
b→+∞ a
f (x)d x
(4.12)
is called an improper
integral of type 1 of function f (x) on the interval [a, +∞) and
Z
+∞
f (x)d x.
denoted by
a
Z
1. If the limit I = lim
b
b→+∞ a
f (x)d x exists (as a finite number) then the improper
172
172
VIII. Improper integral of Type 1: Infinite intervals
Z
+∞
f (x)d x are called convergent.
integrals
a
Z b
f (x)d x does not exist or is equal to ∞ then the im2. If the limit I = lim
b→+∞
Z +∞ a
proper integrals
f (x)d x are called divergent.
a
2
Geometric meaning
Geometric meaning
Z +∞
If f (x) Ê 0, ∀x ∈ [a, +∞) and the integral
f (x)d x is convergent then the ima
Z +∞
proper integrals
f (x)d x can be interpreted as an area of the region S =
a
{(x, y)|x Ê a, 0 É y É f (x)}.
Z
+∞
f (x)d x, if
According to geometric meaning of an improper integral of type 1:
a
lim f (x) = A ̸= 0
x→+∞
and
Z +∞ f (x) is integrable on every interval [a, b] ⊂ [a, +∞), then the improper integrals
f (x)d x are divergent.
a
173
173
Chapter 4. Integration
3
Newton-Leibniz’s Formula
Theorem VIII.1 (Newton-Leibniz’s Formula)
Suppose that f (x) has anti-derivative F (x) on the interval
Z [a, +∞) and is intergrable
+∞
f (x)d x is convergent if
on every interval [a, b]. The improper integral of type 1
a
and only if lim F (b) = F (+∞) exists as a finite number. Then
b→+∞
+∞
Z
¯+∞
¯
f (x)d x = F (+∞) − F (a) = F (x)¯
a
a
(4.13)
Example 140
+∞
Z
Evaluate I =
cos xd x.
0
SOLUTION
¯+∞
¯
= lim sin b − sin 0 = lim sin b.
I = sin x ¯
0
b→+∞
b→+∞
The limit lim sin b does not exist. Therefore the improper integral I is divergent.
b→+∞
Example 141
+∞
Z
Evaluate I =
2
xe −x d x
0
SOLUTION
1
I =−
2
Z
+∞
e
0
−x 2
¯
2
1 −x 2 ¯¯+∞
1
1 1
d (−x ) = − e ¯
= lim − e −b + =
b→+∞ 2
2
2 2
0
2
So the given integral I is convergent.
Example 142
For what values of α is the integral
+∞
Z
I=
1
dx
xα
convergent?
SOLUTION If α ̸= 1 then
If α > 1, then lim
1
b→+∞ b α−1
If α < 1, then lim
1
b→+∞ b α−1
1
I =−
α−1
µ
lim
1
b→+∞ b α−1
= 0. Therefore I =
−
1
¶
1α−1
1
and so the integral I converges.
α−1
= +∞ and so the integral I diverges.
If α = 1, then I = lim ln |b| − ln 1 = +∞ and so the integral I diverges.
b→+∞
174
174
VIII. Improper integral of Type 1: Infinite intervals
Summary
4
1. If α > 1 then I =
Z
2. If α É 1 then I =
Z
1
1
+∞ d x
xα
+∞ d x
xα
converges.
diverges.
A comparison test for improper integrals of type 1
Theorem VIII.2
Suppose that f and g are continuous functions on every interval [a, b] ⊂ [a, +∞)
with 0 É g (x) É f (x), ∀x Ê a.
Z +∞
Z +∞
1. If
f (x)d x is convergent, then
g (x)d x is convergent.
a
Z
a
+∞
Z
+∞
g (x)d x is divergent then
2. If
a
f (x)d x is divergent.
a
Remark 19
1. If the area under the top curve y = f (x) is finite, then so is the area under the
bottom curve y = g (x).
2. If the area under y = g (x) is infinite, then so is the area under y = f (x).
Example 143
Z
Determine whether the integral is convergent or divergent
1
SOLUTION
+∞ 1 + e −x
x
1 + e −x 1
>
x
x
Z +∞
Z +∞
1
1 + e −x
Since
d x is divergent, so the integral
d x is divergent.
x
x
1
1
175
175
dx
Chapter 4. Integration
Theorem VIII.3
Suppose f (x), g (x) be integrable on every [a, b] ⊂ [a, +∞) and f (x), g (x) Ê 0, ∀x Ê a.
f (x)
=λ
Evaluate lim
x→∞ g (x)
Z ∞
Z ∞
1. If λ = 0 and
g (x)d x converges then
f (x)d x converges.
a
2. If λ > 0 then
a
∞
Z
Z
∞
g (x)d x and
a
3. If λ = +∞ and
Z
f (x)d x either converge or diverge.
a
∞
∞
Z
g (x)d x diverges then
f (x)d x diverges.
a
a
Example 144
Z
Determine whether the integral is convergent or divergent I =
+∞
p
1
dx
x 2 − 2x + 3
SOLUTION
We have p
1
x 2 − 2x + 3
> 0, ∀x Ê 1 and
p
+∞ d x
Z
where
x
1
1
x→+∞
x 2 − 2x + 3
∼
1
x
diverges, therefore I diverges.
Example 145
Z
Determine whether the integral is convergent or divergent I =
SOLUTION
We have p
3
x +1
x 7 − 3x − 2
> 0, ∀x Ê 2 and
p
3
+∞
Z
where
2
dx
x 4/3
x +1
x 7 − 3x − 2
x
x→+∞
∼
x 7/3
converges, therefore I converges.
176
176
=
1
x 4/3
2
+∞
(x + 1)d x
p
3
x 7 − 3x − 2
IX. Improper integral of Type 2: Infinity discontinuous integrands
IX Improper integral of Type 2: Infinity discontinuous integrands
b
Z
1
Definition of an improper integral of type 2
f (x)d x on
a
[a, b)
−
Suppose that f is defined on a finite interval [a, b) but has a vertical
Z asymptote as x → b
and f is integrable on every interval [a, η] ⊂ [a, b). Then Φ(η) =
the interval [a, b).
η
f (x)d x is defined on
a
Definition 79
The limit of function Φ(η) as η → b − is called an improper integral of type 2 on the
interval [a, b)
Z b
Z η
f (x)d x = lim− Φ(η) = lim−
f (x)d x
(4.14)
η→b
a
η→b
a
η
Z
f (x)d x exists (as a finite number) then the improper
1. If lim− Φ(η) = lim−
η→b
η→b a
Z b
integral of type 2
f (x)d x converges.
a
2. If lim− Φ(η) = lim−
η→b
η→b
Z
integral of type 2
Z
η
a
b
f (x)d x does not exist or is equal to ∞ then the improper
f (x)d x diverges.
a
2
Geometric meaning
Geometric meaning
Z b
If f (x) Ê 0, ∀x ∈ [a, b) and the integral
f (x)d x is convergent then the improper
a
Z b
integrals
f (x)d x can be interpreted as an area of the region S = {(x, y)|a É x <
a
b, 0 É y É f (x)}, where x = b is the vertical asymptote of the graph of function f (x)
177
177
Chapter 4. Integration
3
Newton-Leibniz’s formula
Theorem IX.1 (Newton-Leibniz’s Formula)
Suppose that f (x) has anti-derivative F (x) on every intervals [a, η] ⊂ [a, b) and
Z b
f (x)d x is convergent if and only if
lim− f (x) = ∞. The improper integral of type 2
x→b
a
lim− F (η) = F (b − 0) exists as a finite number. Then
η→b
b
Z
a
¯b −
¯
f (x)d x = F (b − 0) − F (a) = F (x)¯ .
a
(4.15)
Example 146
1dx
Z
Evaluate I =
x
0
SOLUTION
1
= +∞. Therefore x = 0 is vertical asymptote. Since
x→0+ x
¯1
¯
I = ln |x|¯ = ln 1 − lim ln |a| = +∞.
We have lim
0
a→0+
so improper integral I is divergent.
Example 147
Z
Evaluate I =
1
arccos x
dx
p
−1 1 − x 2
SOLUTION
arccos x
We have lim p
= +∞. Therefore x = −1 is vertical asymptote. In other hand, x = 1
x→−1+
1 − x2
arccos x
is not vertical asymptote because lim p
= 1. Since
x→1−
1 − x2
Z 1
¯1
1
¯
I =−
arccos xd (arccos x) = − · (arccos2 x)¯
−1
2
−1
1
π2
= − (arccos2 1 − lim arccos2 a) =
a→−1+
2
2
so improper integral I is convergent.
Example 148
Z
Evaluate I =
b
a
dx
, (a < b)
(b − x)α
SOLUTION
Z
I = lim−
η→b
η
a
¯η
dx
1
−α+1 ¯
=
−
lim
(b
−
x)
¯ =
a
(b − x)α
−α + 1 η→b −
178
178
IX. Improper integral of Type 2: Infinity discontinuous integrands
=
1
1
lim− (b − η)−α+1 +
(b − a)−α+1
α − 1 η→b
−α + 1
If α < 1 then lim− (b − η)−α+1 = 0.
η→b
If α > 1 then lim− (b − η)−α+1 = ∞.
If α = 1 then
η→b
Zη
I = lim−
η→b
a
¯η
dx
¯
= − lim− ln |b − x|¯ =
a
η→b
b−x
= − lim− ln |b − η| + ln(b − a) = ∞.
η→b
Summary
4
1. If α < 1, then improper integral
Z
2. If α Ê 1, then improper integral
Z
b
a
b
a
dx
converges.
(b − x)α
dx
diverges.
(b − x)α
A comparison test for improper integrals of type 2
Theorem IX.2
Suppose f (x), g (x) be integrable on every [a, η] ⊂ [a, b) and not be bounded as
f (x)
x → b − . Moreover, for every x ∈ [a, b), we have 0 É f (x), 0 É g (x), lim−
=λ
x→b g (x)
Z b
Z b
1. If λ = 0 and
g (x)d x converges then
f (x)d x converges.
a
2. If λ > 0 then
a
b
Z
Z
b
g (x)d x and
a
3. If λ = +∞ and
Z
f (x)d x either converge or diverge.
a
b
Z
b
g (x)d x diverges then
a
f (x)d x diverges.
a
Example 149
Z
Determine whether the integral is convergent or divergent I =
SOLUTION
cos2 x
dx
p
3
1 − x2
cos2 x
cos2 x
1
lim p
= lim− p
.
=∞
x→1− 3 1 − x 2
x→1 3 1 + x (1 − x)1/3
cos2 x
cos2 x
1
·
=
p
p
3
3
1 + x (1 − x)1/3
1 − x2
We have α =
0
1
1
< 1, therefore I converges.
3
179
179
x→1−
∼
cos2 1
1
·
.
p
3
2 (1 − x)1/3
Chapter 4. Integration
Example 150
Z
Determine whether the integral is convergent or divergent I =
SOLUTION
lim+
x→0
ln(1 +
p
3
x)
= lim+
e sin x − 1
ln(1 +
x→0
p
3
x)
Due to α =
X
e sin x
p
3
x)
−1
dx
x 1/3
1
= lim+ 2/3 = ∞
x→0 x
x
x→0+
∼
e sin x − 1
0
1 ln(1 +
x 1/3
1
= 2/3
x
x
2
< 1 so I converges.
3
Application of integration
1 Area between the graph of a function y = f (x) and the
x−axis
Theorem X.1
Suppose f (x) is integrable on the interval [a, b]. The area between the graph of a
function y = f (x) and the x−axis for an interval from x = a to x = b is
Z
A=
2
b¯
a
¯
¯ f (x)¯ d x
(4.16)
Area between curves
Theorem X.2
The area between the curves y = f (x) and y = g (x) and between x = a and x = b is
Z
A=
b¯
a
¯
¯ f (x) − g (x)¯ d x
180
180
(4.17)
X. Application of integration
Example 151
Find the area of the region bounded by the curve y = 4x − x 2 and the x−axis.
SOLUTION
The points of intersection of y = 4x − x 2 and the x−axis are defined by the equation
·
2
4x − x = 0 ⇔
x =0
x =4
·
¸
Z 4³
´
1 3 4 32
2
2
A=
4x − x d x = 2x − x
=
3
3
0
0
Example 152
Find the area enclosed by the curves y = 3 − x and y = x 2 − 9
SOLUTION
181
181
Chapter 4. Integration
The points of intersection of y = 3 − x and y = x 2 − 9 are defined by the equation
3 − x = x 2 − 9 ⇔ x 2 + x − 12 = 0
·
x = −4
⇔ (x − 3)(x + 4) = 0 ⇔
x =3
Z 3
Z 3h
i
2
(3 − x) − (x − 9) d x =
A=
(−x 2 − x + 12)d x =
−4
−4
· 3
¸3
¸
· 3
x
x2
32
3
= − −
+ 12x
= − − + 12 × 3 −
3
2
3
2
−4
¸
·
343
(−4)3 (−4)2
−
+ 12 × (−4) =
− −
3
2
6
3
Volume problem
When designing a building, architects must perform numerous detailed calculations. For
instance, in order to analyze a building’s heating and cooling systems, engineers must
calculate the volume of air being processed.
We subdivide [a, b] into n sub-intervals, each of width ∆x =
b−a
· We denote x i = a + i ∆x,
n
for i = 0, 1, . . . , n.
For any point x i∗ ∈ [x i −1 , x i ], the area of the cross sections is A(x i∗ ). The volume Vi of the
i −th slice is approximately equal
Vi ≈ A(x i∗ ).∆x
The total volume V of the solid is approximately equal
V≈
n
X
i =1
A(x i∗ ).∆x
182
182
X. Application of integration
Notice that as the number of slices increases, the volume approximation should improve
and we get the exact volume by computing
V = lim
n
X
n→∞
i =1
A(x i∗ ).∆x
The volume of a solid with cross-sectional area A(x) is
b
Z
V=
A(x)d x
(4.18)
a
Example 153
Find volume of a sphere of radius r .
SOLUTION
The cross-sectional area is
A(x) = π.y 2 = π(r 2 − x 2 )
Using the definition of volume with a = −r and b = r , we have
Z r
Z r
Z r
V=
A(x)d x =
π(r 2 − x 2 )d x = 2π (r 2 − x 2 )d x =
−r
0
−r
µ
¸r
¶
·
r3
4
x3
3
2
= 2π r −
= πr 3 .
= 2π r .x −
3 0
3
3
4
The volume of a solid of revolution
Theorem X.3
Suppose that f (x) Ê 0 and f is continuous on the interval [a, b]. The volume of the
solid resulting from revolving the region under the curve y = f (x) and the x−axis,
for a É x É b, about the x−axis, is
Vx = π
Z
b
f 2 (x)d x
a
183
183
(4.19)
Chapter 4. Integration
We can find the volume of solid by slicing it perpendicular to the x−axis and recognizing
that each cross section is a circular disk of radius r = f (x).
Z b
Z b
Z b
2
V=
A(x)d x =
π f (x)d x = π
f 2 (x)d x
a
a
a
Example 154
p
Revolve the region under the curve y = x on the interval [0, 4] about the x−axis
and find the volume of the resulting solid of revolution.
SOLUTION
p
The radius of each cross section is given by r = x. Then we get the volume
¯4
Z 4 ³ ´
Z 4
p 2
x 2 ¯¯
= 8π.
V=
π x dx = π
xd x = π ·
2 ¯0
0
0
Theorem X.4
Suppose that g (y) Ê 0 and g is continuous on the interval [c, d ]. Then, revolving the
region bounded by the curve x = g (y) and the y−axis, for c É y É d , about the y−axis
generates a solid. The volume of this solid is
Vy = π
Z
d
g 2 (y)d y
c
184
184
(4.20)
X. Application of integration
Example 155
Find the volume of the solid resulting fromprevolving the region bounded by the
curves y = 4 − x 2 and y = 1 from x = 0 to x = 3 about the y−axis.
SOLUTION
p
The radius of any of the circular cross section is r = x = 4 − y by solving the equation
p
y = 4 − x 2 ⇔ x = 4 − y.
Since the surface extends from y = 1 to y = 4, the volume is
Z 4 ³
Z 4
´2
p
V=
π 4− y dy =
π(4 − y)d y =
1
1
·
µ
·
¸4
¶¸
1
y2
9π
= π (16 − 8) − 4 −
·
= π 4y −
=
2 1
2
2
5
The volume by cylindrical shells
Let R denote the region bounded by the graph of y = f (x) and the x−axis on the interval
[a, b] and f (x) Ê 0 on [a, b].
185
185
Chapter 4. Integration
b−a
· On
n
each sub-interval [x i −1 , x i ], pick a point x i∗ and construct the rectangle of height f (x i∗ ).
Revolving this rectangle about the y−axis forms a thin cylindrical shell.
We first partition the interval [a, b] into n sub-intervals of equal width ∆x =
To find the volume of this thin cylindrical shell, imagine cutting the cylinder from top to
bottom and then flattening out the shell.
Notice that the length of such a thin sheet corresponds to the circumference of the cylindrical shell, which is 2π × radius = 2πx i∗ .
The volume Vi of the i −th cylindrical shell is approximately equal
Vi = length × width × height =
= (2π × radius) × thickness × height = (2πx i∗ )∆x f (x i∗ ).
The total volume V of the solid can be approximated by the sum of the volumes of the n
cylindrical shells
n
X
V≈
2πx i∗ f (x i∗ )∆x
i =1
186
186
X. Application of integration
Theorem X.5
The volume of the solid, obtained by rotating about the y−axis the region: 0 É y É
f (x), a É x É b, is
Z b
¯
¯
¯x f (x)¯ d x
V y = 2π
(4.21)
a
Example 156
Find the volume of the solid obtained by rotating about the y−axis the region
bounded by y = x 2 , x = 1, x = 2, y = 0
SOLUTION 1
Z
V = 2π
1
2
Z
x. f (x)d x = 2π
1
2
x4
x.x d x = 2π
4
SOLUTION 2
187
187
2
·
¸2
=
1
15π
2
Chapter 4. Integration
· 2 ¸4
Z 4³ ´
y
15π
p 2
V = π.2 .4 − π.1 .1 − π
y d y = 15π − π.
=
2 1
2
1
2
6
2
Arc length
Assume that f is continuous on [a, b] and differentiable on (a, b). We begin by partitioning
the interval [a, b] into n equal pieces: A = M 0 , M 1 , . . . , M i −1 , M i , . . . , M n = B, where x i −
b−a
x i −1 = ∆x =
· We approximate the arc length M i −1 M i by the straight-line distance
n
between the two points
q
L i ≈ (x i − x i −1 )2 + (y i − y i −1 )2
Since f is continuous on [a, b] and differentiable on (a, b), f is also continuous on the
sub-interval [x i −1 , x i ] and is differentiable on (x i −1 , x i ). By the Mean Value Theorem, we
have
y i − y i −1 = f (x i ) − f (x i −1 ) = f ′ (x i∗ )(x i − x i −1 ) = f ′ (x i∗ ).∆x,
for some number x i∗ ∈ (x i −1 , x i ). This gives us the approximation
q
L i ≈ (x i − x i −1 )2 + (y i − y i −1 )2 =
r
=
r
=
h
i2
(x i − x i −1 )2 + f ′ (x i∗ )(x i − x i −1 ) =
h
1+ f
i2
′ (x ∗ )
i
r
(x i − x i −1 ) =
h
i2
1 + f ′ (x i∗ ) ∆x
Adding together the lengths of these n line segments, we get an approximation of the total
arc length
r
h
i2
n
X
∗
′
L≈
1 + f (x i ) ∆x.
i =1
188
188
X. Application of integration
Notice that as n gets larger, this approximation should approach the exact arc length Ù
AB
L = lim
n
X
n→∞
L i = lim
n
X
n→∞
i =1
b
Z
=
r
h
i2
1 + f ′ (x i∗ ) ∆x i =
i =1
r
h
i2
1 + f ′ (x) d x
a
Theorem X.6 (The Arc Length Formula)
If f ′ is continuous on [a, b], then the length of the curve Ù
AB : y = f (x), a É x É b, is
b
Z
L=
r
a
h
i2
1 + f ′ (x) d x
(4.22)
Example 157
Find the arc length of the portion of the curve y =
x 2 ln x
−
, with 1 É x É 3.
2
4
1
· Using the arc length formula, we get
4x
s
µ
¶2
Z 3r
Z 3
h
i2
1
L=
1 + f ′ (x) d x =
1+ x −
dx =
4x
1
1
SOLUTION We have f ′ (x) = y ′ = x −
Z 3r
=
Z
1
3
1
s
1
1
x2 +
+ dx =
2
16x
2
(4x 2 + 1)2
dx =
16x 2
3
Z
1
1
s
16x 4 + 1 + 8x 2
dx =
16x 2
4x 2 + 1
dx =
4x
x2 1
=
+ ln |x|
2 4
·
3
Z
¸3
189
189
Z 3µ
1
1
= 4 + ln 3
4
1
¶
1
dx =
x+
4x
Chapter 4. Integration
7
Area of a surface of Revolution
The surface area of a circular cylinder with radius r and height h is
A = 2πr h
(4.23)
because we can imagine cutting the cylinder and unrolling it to obtain a rectangle with
dimensions 2πr and h.
The surface area of a circular cone with base radius r and slant height ℓ is
A = πr ℓ
(4.24)
because we can imagine cutting the cone along the dashed line and flattening it to form a
2πr
sector of a circle with radius ℓ and central angle θ =
·
ℓ
1
1 2πr
A = ℓ2 θ = ℓ2
= πr ℓ.
2
2
ℓ
The surface area of a frustum of a circular cone with slant height ℓ and upper and lower
radii r 1 , r 2 is found
A = πℓ(r 1 + r 2 )
by subtracting the areas of two cones:
A = πr 2 (ℓ1 + ℓ) − πr 1 ℓ1 = π[(r 2 − r 1 )ℓ1 + r 2 ℓ].
190
190
(4.25)
X. Application of integration
From similar triangles we have
ℓ1 ℓ1 + ℓ
=
⇒ r 2 ℓ1 = r 1 ℓ1 + r 1 ℓ ⇒ (r 2 − r 1 )ℓ1 = r 1 ℓ
r1
r2
⇒ A = π[(r 2 − r 1 )ℓ1 + r 2 ℓ] = π(r 1 ℓ + r 2 ℓ) = πℓ(r 1 + r 2 ).
Suppose that y = f (x) Ê 0, ∀x ∈ [a, b] and f has a continuous derivative on [a, b]. Consider
the surface, which is obtained by rotating the curve y = f (x), a É x É b, about the x−axis.
We divide the interval [a, b] into n subintervals with endpoints a = x 0 , x 1 , . . . , x n = b and
b−a
· The part of the surface between x i −1 and x i is approximated by
equal width ∆x =
n
taking the line segment P i −1 P i and rotating it about the x−axis.
The surface area S i of part of the surface on the interval [x i −1 , x i ] is approximately the
surface area of the frustum of the cone
r
h
i
h
i2
S i ≈ π(y i + y i −1 )ℓ = π f (x i ) + f (x i −1 ) 1 + f ′ (x i∗ ) ∆x,
where x i∗ is some number in (x i −1 , x i ). When ∆x is small, we have f (x i ) ≈ f (x i∗ ) and
f (x i −1 ) ≈ f (x i∗ ), since f is continuous. Therefore
r
S i ≈ 2π f
The total surface area
S≈
n
X
i =1
i2
h
1 + f ′ (x i∗ ) ∆x.
(x i∗ )
r
2π f
(x i∗ )
h
1+ f
191
191
′ (x ∗ )
i
i2
∆x.
Chapter 4. Integration
As n gets larger, this approximation approaches the actual surface area
r
h
i2
n
X
S = lim
2π f (x i∗ ) 1 + f ′ (x i∗ ) ∆x =
n→∞
Z
=
i =1
b
a
r
h
i2
2π f (x) 1 + f ′ (x) d x.
Theorem X.7
Suppose that f is positive and has a continuous derivative on [a, b]. The surface
area of the surface obtained by rotating the curve y = f (x), a É x É b, about x−axis is
Z
S = 2π
r
b
f (x)
a
h
1+ f
′ (x)
i2
dx
(4.26)
Example 158
Find the area of the surface obtained by rotating the arc y = sin 2x, 0 É x É
the x−axis.
π
, about
2
SOLUTION We have y ′ = 2 cos 2x. Using the area of surface of revolution formula, we get
Zb
S = 2π
r
f (x)
h
1+ f
′ (x)
i2
Zπ/2
p
d x = 2π sin 2x 1 + 4 cos2 2xd x
a
0
π
1
x 0
Let t = 2 cos 2x ⇒ d t = −4 sin 2xd x ⇒ sin 2xd x = − d t ,
2 . So
4
t 2 −2
µ ¶
Z −2 p
Z
Z 2p
1
π 2p
2
2
S = 2π
1+t − dt =
1+t dt = π
1 + t 2d t
4
2
2
−2
0
192
192
XI. Exercises
· p
· p
·
¸
´¸2
´¸
p
p
p ´
2
t
1 ³ p
1 ³
1 ³
2
2
2
2
=π
=π
1 + t + ln t + 1 + t
1 + 2 + ln 2 + 1 + 2
= π 5 + ln 2 + 5
2
2
2
2
2
0
XI
1
Exercises
Essay Questions
Antiderivatives
42 If an object’s downward acceleration is given by a(t ) = s ′′ (t ) = −32m/s 2 , find the
position function s(t ). Assume that the initial velocity is v(0) = s ′ (0) = −100m/s and
the initial position is s(0) = 100000m
SOLUTION
We have to undo 2 derivatives, so we compute 2 antiderivatives. First, we have
Z
Z
′
′′
v(t ) = s (t ) = s (t )d t = (−32)d t = −32t +C 1 .
Since v(0) = s ′ (0) = −100, we must have
−100 = v(0) = −32 × 0 +C 1 ⇒ C 1 = −100.
Thus, the velocity is v(t ) = s ′ (t ) = −32t − 100. Next, we have
Z
Z
′
s(t ) = s (t )d t = (−32t − 100)d t = −16t 2 − 100t +C 2 .
Using the initial position s(0) = 100000m, we have
100000 = s(0) = −16 × 02 − 100 × 0 +C 2 ⇒ C 2 = 100000
and
s(t ) = −16t 2 − 100t + 100000
............................................................................................□
43 Find a function f (x) such that the point (1, 2) is on the graph of y = f (x), the
slope of the tangent line at (1, 2) is 3 and f ′′ (x) = x − 1.
SOLUTION
We have to undo 2 derivatives, so we compute 2 antiderivatives. First, we have
Z
Z
x2
f ′ (x) = f ′′ (x)d x = (x − 1)d x =
− x +C 1 .
2
Since f ′ (1) = 3, we must have
3 = f ′ (1) =
12
7
− 1 +C 1 ⇒ C 1 = ·
2
2
193
193
Chapter 4. Integration
x2
7
−x + ·
2
2
Next, we have
¶
Z
Z µ 2
7
x 3 x 2 7x
x
′
−x + dx =
−
+
+C 2
f (x) = f (x)d x =
2
2
6
2
2
Thus, f ′ (x) =
Using the condition f (1) = 2, we have
2 = f (1) =
7
13 12 7 × 1
− +
+C 2 ⇒ C 2 = − ·
6
2
2
6
and
x 3 x 2 7x 7
f (x) =
−
+
− ·
6
2
2
6
............................................................................................□
Indefinite integrals
44
¶
Z µ
p
1
Find the general antiderivative
3 x − 4 dx
x
SOLUTION
We have
p
¶
Z
¡ 1/2
¢
1
3 x − 4 dx =
3x − x −4 d x =
x
p
1
x 3/2 x −3
−
+C = 2 x 3 + 3 +C =
= 3·
3/2
−3
3x
p
1
= 2x x + 3 +C .
3x
............................................................................................□
Z 1/3
45
x −3
dx
Find the general antiderivative
x 2/3
Z µ
SOLUTION
We have
Z
x 1/3 − 3
x 2/3
Z
dx =
¡
¢
x −1/3 − 3x −2/3 d x =
p
x 2/3
x 1/3
3 p
3
−3·
+C = · x 2 − 9. 3 x +C .
2/3
1/3
2
............................................................................................□
=
Techniques of integration
46
Z
Find
3
x 2 e −x d x.
194
194
XI. Exercises
SOLUTION
dt
Let t = −x , d t = −3x d x ⇒ x d x = − · This gives us
3
µ
¶
Z
Z
dt
1
1
t
e · −
= − · e t d t = − · e t +C =
3
3
3
3
2
2
3
1
= − · e −x +C .
3
............................................................................................□
Z
47
x
Find
d x.
4
x +1
SOLUTION
dt
· This gives us
Let t = x , d t = 2xd x ⇒ xd x =
2
Z
Z
1
1
dt
dt
=
·
=
· arctan t +C =
2(t 2 + 1) 2
t2 +1 2
2
1
· arctan x 2 +C .
2
............................................................................................□
Z
48
Find I = e x sin xd x
=
SOLUTION
Let
u = ex,
d v = sin xd x.
Then
d u = e x d x,
v = − cos x.
Thus, using formula for integration by parts, we have
Z
Z
x
x
I = e sin xd x = e (− cos x) + e x cos xd x
Let
u = ex,
d v = cos xd x.
Then
d u = e x d x,
v = sin x.
Thus, using formula for integration by parts again, we have
Z
Z
x
x
x
x
I = e (− cos x) + e cos xd x = −e cos x + e sin x − e x sin xd x
= −e x cos x + e x sin x − I ⇒ 2I = −e x cos x + e x sin x
Dividing by 2 and adding the constant of integration, we get
1
I = e x (sin x − cos x) +C
2
195
195
Chapter 4. Integration
............................................................................................□
Z
49
Find I = arctan xd x
SOLUTION
Let
u = arctan x,
Then
du =
d v = d x.
1
d x,
1 + x2
v = x.
Integrating by parts, we get
Z
I=
Z
arctan xd x = x. arctan x −
x·
dx
=
1 + x2
Z
1
1 d (x 2 + 1)
= x. arctan x −
= x. arctan x − · ln(1 + x 2 ) +C .
2
2
x +1
2
............................................................................................□
Z
Z
50
1
n −1
n
n−1
x+
sinn−2 xd x, where n Ê 2 is an
Prove that sin xd x = − cos x. sin
n
n
integer.
SOLUTION
Let
u = sinn−1 x,
d v = sin xd x.
Then
d u = (n − 1) sinn−2 x. cos xd x,
v = − cos x.
Integrating by parts, we get
Z
Z
n
n−1
I = sin xd x = − cos x. sin
x + (n − 1) sinn−2 x cos2 xd x
n−1
I = − cos x. sin
n−1
= − cos x. sin
Z
x + (n − 1)
Z
x + (n − 1)
sinn−2 x(1 − sin2 x)d x =
n−2
sin
= − cos x. sinn−1 x + (n − 1)
Z
Z
xd x − (n − 1)
sinn xd x =
sinn−2 xd x − (n − 1)I
⇒ nI = − cos x. sinn−1 x + (n − 1)
Z
sinn−2 xd x
Z
n −1
1
sinn−2 xd x.
⇒ I = − cos x. sinn−1 x +
n
n
............................................................................................□
Integration of rational functions by partial fractions
196
196
XI. Exercises
51
1
Z
Find I =
x2 − x
d x.
SOLUTION
The method of partial fractions gives
1
A
B
1
=
=
+
x 2 − x x(x − 1 x x − 1
⇒ 1 ≡ A(x − 1) + B x, ∀x ∈ R
We put x = 0 and get: 1 = A(0 − 1) ⇒ A = −1
We put x = 1, we get: 1 = B × 1 ⇒ B = 1
Z
I =−
dx
+
x
Z
dx
=
x −1
= − ln |x| + ln |x − 1| +C .
............................................................................................□
Integration of non-rational functions
Z p
52
Find
Let t =
p
x +4
d x.
x
SOLUTION
x + 4 ⇒ t = x + 4 ⇒ x = t − 4 ⇒ d x = 2t d t . Therefore
Z p
Z
Z
x +4
t
t2
dx =
·
2t
d
t
=
2
dt =
x
t2 −4
t2 −4
¯
¯
¶
Z µ
¯t −2¯
4
1
¯
¯ +C =
ln
= 2 1+ 2
d t = 2t + 8 ·
t −4
2×2 ¯t +2¯
¯p
¯
¯ x +4−2¯
p
¯
¯
= 2 x + 4 + 2 ln ¯ p
¯ +C .
¯ x +4+2¯
2
2
............................................................................................□
Trigonometric Integrals
53
Z
Find
cos3 xd x.
SOLUTION
Let t = sin x, d t = cos xd x. This gives us
Z
Z
Z
cos3 xd x = cos2 x. cos xd x = (1 − sin2 x) cos xd x =
197
197
Chapter 4. Integration
1 3
1
· t +C = sin x − · sin3 x +C .
3
3
............................................................................................□
Z
54
Find sin4 xd x.
Z
(1 − t 2 )d t = t −
=
SOLUTION
We have
Z
Z
4
sin xd x =
2
Z µ
2
(sin x) d x =
1 − cos 2x
2
¶2
dx =
Z
1
=
(1 − 2 cos 2x + cos2 2x)d x =
4
¸
Z ·
1
1
=
1 − 2 cos 2x + (1 + cos 4x) d x =
4
2
µ
¶
1 3
1
=
· x − sin 2x + sin 4x +C .
4 2
8
............................................................................................□
Z
55
Find tan3 xd x.
SOLUTION
We have
¶
1
−1 dx =
tan xd x = tan x. tan xd x = tan x
cos2 x
Z
Z
Z
Z
dx
d (− cos x)
= tan x ·
− tan xd x = tan xd (tan x) −
=
2
cos x
cos x
Z
Z
3
Z
2
µ
tan2 x
+ ln | cos x| +C .
2
............................................................................................□
=
Definite integrals
56
Z4
Calculate I =
¯ 2
¯
¯x + 2x − 3¯ d x.
−1
SOLUTION
We have
Z1
I=
¯ 2
¯
¯x + 2x − 3¯ d x +
Z4
1
−1
Z1
=
−1
¯ 2
¯
¯x + 2x − 3¯ d x =
2
Z4
(−x − 2x + 3)d x +
1
198
198
(x 2 + 2x − 3)d x =
97
·
3
XI. Exercises
............................................................................................□
57
Zln 8
dx
Calculate I = p
·
ex + 1
0
SOLUTION
p
x
2t d t
x
2
x
· Then
Let t = e x + 1 ⇒ e = t − 1 ⇒ e d x = 2t d t ⇒ d x = 2
t
t −1
0
p
2
ln 8
and
3
¯
¯¯
Z3
¯ t − 1 ¯¯3
2t d t
2d t
¯
¯¯ =
=
= ln ¯
(t 2 − 1)t p (t 2 − 1)
t + 1 ¯¯p2
Z3
I=
p
2
2
p
p
1
2−1
2+1
= ln − ln p
= ln p
·
2
2+1
2 2−2
............................................................................................□
58
Zπ/6
Calculate I = (e sin x + 2 cos x) cos xd x.
0
SOLUTION
We have
Z
I=
=e
e
sin x
0
π/6
Z
=
π/6
e
sin x
0
π/6
Z
cos xd x +
2 cos2 xd x =
π/6
Z
d (sin x) +
0
(1 + cos 2x)d x =
0
p
¶
¯π/6 µ
π
3
sin 2x ¯¯π/6 p
·
¯ = e −1+ +
¯ + x+
0
0
2
6
4
sin x ¯
............................................................................................□
59
¶
Ze µ
2
Calculate I =
4x +
ln xd x.
x
1
SOLUTION
We have
e
Z
I=
1
e 2 ln xd x
Z
4x ln xd x +
x
1
=
Z e
¯e Z e
¯
= 2x ln x ¯ − 2xd x + 2 ln xd (ln x) =
2
1
1
1
¯e
¯e
¯e
¯
¯
¯
= 2x 2 ln x ¯ − x 2 ¯ + ln2 x ¯ = e 2 + 2.
1
1
1
............................................................................................□
199
199
Chapter 4. Integration
60
2
Z
Calculate I =
1
(3x 2 + 4x − 2) ln xd x.
SOLUTION
dx
Let u = ln x, d v = (3x 2 + 4x − 2)d x ⇒ d u =
, v = x 3 + 2x 2 − 2x. Integrating by parts, we
x
have
¯2 Z 2 x 3 + 2x 2 − 2x
¯
I = (x 3 + 2x 2 − 2x) ln x ¯ −
dx =
1
x
1
¯2 Z 2
¯
3
2
= (x + 2x − 2x) ln x ¯ − (x 2 + 2x − 2)d x =
1
1
¶¯
¯2
10
x3
¯2
¯
2
+ x − 2x ¯ = 12 ln 2 − ·
= (x + 2x − 2x) ln x ¯ −
1
1
3
3
3
µ
2
............................................................................................□
Z π/3
61
Calculate I =
(3x + 2) cos 2xd x.
0
SOLUTION
sin 2x
Let u = 3x + 2, d v = cos 2xd x ⇒ d u = 3d x, v =
· Integrating by parts, we have
2
Z π/3
sin 2x ¯¯π/3
3 sin 2x
I = (3x + 2) ·
dx =
¯ −
2 0
2
0
= (3x + 2) ·
sin 2x ¯¯π/3 3 (− cos 2x) ¯¯π/3
¯ − ·
¯ =
0
2 0
2
2
p
(π + 2) 3 9
− ·
=
4
8
............................................................................................□
Z 1
62
Calculate I =
x arctan xd x.
0
SOLUTION
dx
x2 + 1
Let u = arctan x, d v = xd x ⇒ d u =
,
v
=
· Integrating by parts, we have
1 + x2
2
¯1 Z 1 x 2 + 1 d x
x2 + 1
¯
I=
· arctan x ¯ −
·
=
0
2
2
1 + x2
0
¯1 x ¯1
x2 + 1
¯
¯
=
· arctan x ¯ − ¯ =
0
2
2 0
π 1
− ·
4 2
............................................................................................□
=
200
200
XI. Exercises
63
π/2
Z
Calculate I =
0
¶
1 + sin x
ln
d x.
1 + cos x
µ
SOLUTION
π
π
x
0
π/2
and
Let t = − x ⇒ x = − t ⇒ d x = −d t . Then
t π/2
0
2
2
³π
´

¶
Z π/2 µ
Z 0
1
+
sin
−
t
1 + sin x


I=
ln
dx =
ln 
³2
´  (−d t ) =
π
1 + cos x
0
π/2
1 + cos
−t
2
¶
¶
Z π/2 µ
Z π/2 µ
1 + cos t
1 + sin t
=
ln
dt = −
ln
d t = −I
1 + sin t
1 + cos t
0
0
⇒ 2I = 0 ⇒ I = 0.
............................................................................................□
Improper integrals
64
Z+∞
Evaluate the improper integral I =
dx
p
1
x x2 + x + 1
.
SOLUTION
Let
t=
1
dt
1
⇒ x = ⇒ dx = − 2 ·
x
t
t
Then
So
Z0
I=
Z1
=
0
1
1 t.
q
x
t
1 +∞
1
0
− dt 2t
Z1
1
t2
+ 1t + 1
=
0
dt
=
p
1+ t + t2
¡
¢
¯¸1
· ¯
p
¯
¯
d t + 12
1
2
¯ t + + t + t + 1¯
=
ln
q¡
¯
¯
¢2
2
0
t + 12 + 43
¶
µ
¶
µ ¶
µ
3 p
3
2
= ln + 3 − ln
= ln 1 + p .
2
2
3
............................................................................................□
65
Z+∞
arctan x
Evaluate the improper integral I =
d x.
x2
1
SOLUTION
201
201
Chapter 4. Integration
Using integration by parts, we have


dx

 du =
 u = arctan x
1 + x2
⇒
dx

 dv =
 v = −1
x2
x
So
·
¸+∞ Z+∞
1
dx
I = − · arctan x
+
=
x
x(1 + x 2 )
1
1
Z+∞
Z+∞
·
¸+∞
dx
xd x
π
1
2
−
= + ln |x| − ln(1 + x )
=
x
1 + x2 4
2
1
1
1
¯¸+∞
· ¯
¯
¯
1
π ln 2
π
π
x
¯
¯
= + ln ¯ p
= + ln 1 − ln p = +
¯
2
4
4
2
2 4
1+x 1
............................................................................................□
π
= +
4
66
Z3
Evaluate the improper integral I =
1
dx
p
4x − x 2 − 3
SOLUTION
We have
Z3
d (x − 2)
I=
p
1
h
1 − (x − 2)2
= arcsin(x − 2)
i3
1
=
= lim− arcsin(x − 2) − lim+ arcsin(x − 2)
x→3
x→1
π π
+ = π.
2 2
............................................................................................□
= arcsin 1 − arcsin(−1) =
67
Z1
Evaluate the improper integral I =
0
dx
p
1−x
SOLUTION
We have
Z1
i1
d (1 − x) h p
I =− p
= −2 1 − x =
0
1−x
0
³ p
´
= lim− −2 1 − x − (−2) = 2
x→1
............................................................................................□
Applications of integration
68
Find the area enclosed by the curves y = (x − 1)2 and x 2 −
202
202
y2
=1
2
XI. Exercises
SOLUTION
y2
2
2
The points of intersection of y = (x − 1) and x −
= 1 are defined by the equation
2
(x − 1)4
= 1 ⇔ x 4 − 4x 3 + 4x 2 − 4x + 3 = 0
2
·
x =1
2
⇔ (x − 1)(x − 3)(x + 1) = 0 ⇔
x =3
Z 3p
A=
[ 2(x 2 − 1) − (x − 1)2 ]d x =
x2 −
1
p h
i3 10 p2
i3 1 h
p
p
2 p 2
3
2
x x − 1 − ln |x + x − 1 − (x − 1)
−
ln(3 + 8)
=
=
1
1
2
3
3
2
............................................................................................□
69 Find the area enclosed by the curves y = x − x 2 and y = x p1 − x.
SOLUTION
p
The points of intersection of y = x − x 2 and y = x 1 − x are defined by the equation
·
p
p
p
x =0
2
x − x = x 1 − x ⇔ x 1 − x( 1 − x − 1) = 0 ⇔
x =1
1
Z
A=
Let t =
0
Z 1 p
Z 1
p
2
[x 1 − x − (x − x )]d x =
x 1 − xd x − (x − x 2 )d x = I 1 − I 2
0
0
p
x
1 − x ⇒ t 2 = 1 − x ⇒ x = 1 − t 2 ⇒ d x = −2t d t . Then
t
Z
I1 =
0
1
0
1
1
and therefore
0
¸1
· 3
Z 0
Z 1
p
t
t5
4
2
2
4
x 1 − xd x =
(1 − t )t (−2t d t ) = 2 (t − t )d t = 2
−
= ·
3
5 0 15
1
0
Z
I2 =
So A = I 1 − I 2 =
4 1
1
− = ·
15 6 10
0
1
x2 x3
(x − x )d x =
−
2
3
·
2
¸1
1
= ·
6
0
............................................................ □
70 Revolve the region under the curve y 2 = (x − 1)3 on the interval [1, 2] about the
x−axis and find the volume of the resulting solid of revolution.
203
203
Chapter 4. Integration
SOLUTION
We have
¯2 π
1
(x − 1)3 d x = π (x − 1)4 ¯1 =
4
4
1
1
............................................................................................□
V =π
Z
2
2
y dx = π
Z
2
71 Revolve the region bounded by the curves y = x 2 , y = 0, x + y = 2 about the x−axis
and find the volume of the resulting solid of revolution.
SOLUTION
We have
V =π
Z
0
1
2 2
(x ) d x + π
2
Z
1
x5
(2 − x) d x = π
5
2
·
¸1
·
¸2
(2 − x)3
π π 8π
+π −
= + =
3
5 3 15
0
1
............................................................................................□
72
Find the arc length of the portion of the curve y =
x3 1
+ , with 1 É x É 4.
12 x
SOLUTION
2
x
1
We have f ′ (x) = y ′ =
− · Therefore
4 x2
s
¶2
µ 2
Z 4q
Z 4
x
1
′
2
L=
1 + [ f (x)] d x =
1+
−
dx =
4 x2
1
1
s
s
¶2
Z 4
Z 4 µ 2
4
x
1 1
x
1
=
+
+ dx =
+
dx =
42 x 4 2
4 x2
1
1
¶
· 3
¸4
Z 4µ 2
x
1
x
1
=
+
dx =
−
= 6.
4 x2
12 x 1
1
204
204
XI. Exercises
............................................................................................□
73
1
1
Find the arc length of the portion of the curve y = ln(1 − x 2 ), with − É x É .
2
2
SOLUTION
−2x
We have f (x) = y =
· Using the arc length formula, we get
1 − x2
s
µ
¶
Z 1/2
Z 1/2 q
−2x 2
′
2
dx =
1 + [ f (x)] d x =
1+
L=
1 − x2
−1/2
−1/2
′
′
Z
=
1/2
s
−1/2
s
(1 − x 2 )2 + 4x 2
dx =
(1 − x 2 )2
1/2
Z
−1/2
s
1 + x 4 − 2x 2 + 4x 2
dx =
(1 − x 2 )2
¶
Z 1/2 µ
x2 + 1
2
dx =
−1 +
dx =
=
2
1 − x2
−1/2 1 − x
−1/2
−1/2
¸
¶
Z 1/2 ·
Z 1/2 µ
1−x +1+x
1
1
=
−1 +
+
dx =
dx =
−1 +
(1 − x)(1 + x)
1+x 1−x
−1/2
−1/2
¯
¯¸
·
¯ 1 + x ¯ 1/2
¯
¯
= −x + ln ¯
= 2 ln 3 − 1.
1 − x ¯ −1/2
Z
1/2
(1 + x 2 )2
dx =
(1 − x 2 )2
Z
1/2
............................................................................................□
p
74
Find the area of the surface obtained by rotating the arc y = x 2 + 4, 0 É x É 1,
about the x−axis.
We have f ′ (x) = y ′ = p
Zb
S = 2π
x
x2 + 4
SOLUTION
· Using the area of surface of revolution formula, we get
Z1 p
q
′2
f (x) 1 + f (x)d x = 2π
x2 + 4
a
= 2π
Z1 p
0
µ
1+ p
0
s
x2 + 4
s
x2 + 4 + x2
d x = 2π
x2 + 4
Z1 p
¶2
x
x2 + 4
dx =
1
2x 2 + 4d x
0
p Z p
= 2 2π
x 2 + 2d x =
0
Ã
p !
· p
¸1
p
p p
p
x
2
1+ 3
2
2
= 2 2π
x + 2 + ln(x + 2 + x ) = π 2 3 + 2 ln p
·
2
2
2
0
............................................................................................□
2
Multiple-choice Questions
Anti-derivatives
205
205
Chapter 4. Integration
Question 1 (L.O.1): Find the antiderivative
F (x) of the function f (x) = 8 sin x + 4 cos x
³π´
which satisfies the condition F
= 12.
2
A −8 cos x + 4 sin x + 8
B −8 cos x − 4 sin x + 8
C −8 cos x + 4 sin x − 8
D 8 cos x − 4 sin x − 8
E 8 cos x + 4 sin x + 8
SOLUTION
Z
We have F (x) =
f (x) dx = 4 sin (x) − 8 cos (x) +C . Moreover,
F
³π´
2
= 4 +C = 12 ⇒ C = 8.
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 2 (L.O.1): Find the antiderivative F (x) of the function f (x) = 8ex + 4x which
satisfies the condition F (0) = 14.
C −8ex + 2x 2 − 6
A 8ex + 2x 2 − 6
B −8ex + x 2 + 6
x
2
x
2
D 8e + 2x + 6
E e + 2x + 6
Z
We have F (x) =
SOLUTION
x
2
f (x) dx = 8e + 2x +C . Moreover,
F (0) = 8e0 + 2 × 02 +C = 14 ⇒ C = 6
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 3 (L.O.1): Suppose that f ′′ (x) = −18x − 100 sin(5x) + 9e 3x and f ′ (0) = 23,
f (0) = −379. Evaluate f (2).
A −2.2641
B −2.7473
C −2.0485
D −3.6915
E −3.5396
SOLUTION
We have f (x) = −18x − 100 sin(5x) + 9e 3x ⇒ f ′ (x) = −9x 2 + 20 cos(5x) + 3e 3x + C 1 . Since
f ′ (0) = 23, so C 1 = 0. Thus, f (x) = −3x 3 + 4 sin(5x) + e 3x + 0x + C 2 . Since f (0) = −379, nên
C 2 = −380. Therefore, f (2) ≈ −2.7473.
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
′′
Fundamental theorem of calculus
Z
Question 4 (L.O.2): Find the maximum and minimum values of f (x) =
22)d t on the interval [−9, 9].
A f max ≈ 433.8311; f mi n ≈ −0.1873
B f max ≈ 433.4807; f mi n ≈ 0.638
C f max ≈ 432.9731; f mi n ≈ 0.1885
D f max ≈ 433.7281; f mi n ≈ 0.0568
E f max ≈ 432.8333; f mi n ≈ 0
206
206
x
−9
(t 2 − 9t −
XI. Exercises
SOLUTION
We have
Z
f (x) =
x
−9
2
·
2
′
x = −2
x = 11
(t − 9t − 22)d t ⇒ f (x) = x − 9x − 22 = 0 ⇔
−9
Z
f (−9) =
Z
f (−2) =
−9
−2
−9
Z
f (9) =
(t 2 − 9t − 22)d t = 0;
(t 2 − 9t − 22)d t ≈ 432.8333;
9
−9
(t 2 − 9t − 22)d t ≈ 90.
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Riemann sum
Question 5 (L.O.2): Using the midpoint rule with 5 equal interval on [3; 11] to estimate
Z 11
the following integral I =
f (x)d x where f (x) is given by the following table
3
3 3.8 4.6 5.4 6.2 7.0 7.8 8.6 9.4 10.2 11
x
f (x) 6.8 4.3 6.6 5.0 3.4 5.8 5.1 5.5 4.2 5.8 6.0
A 41.8
B 41.88
C 42.24
D 41.6
E 42.26
SOLUTION
11 − 3
Each subinterval has length of ∆x =
= 1.6. Therefore, the subintervals consist of
5
[3, 4.6]; [4.6, 6.2]; [6.2, 7.8]; [7.8, 9.4]; [9.4, 11.0]. The midpoints of these subintervals are
3.8; 5.4; 7.0; 8.6; 10.2. Thus, the approximation of the integral using midpoint rule is
h
i
I ≈ ∆x × 4.3 + 5.0 + 5.8 + 5.5 + 5.8 ≈ 42.24
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Definite integrals
Z
Question 6 (L.O.1): Let
π
2
0
2.63 sin x] dx.
A 63.8129
B 64.3499
Z
f (x) dx = 9.51.
Evaluate
C 64.0289
I
=
D 64.5836
0
π
2
[6.49 f (x) +
E 64.7637
SOLUTION
We have
Z
I=
0
π
2
Z
[6.49 f (x) + 2.63 sin x] dx = 6.49
207
207
0
π
2
Z
f (x) dx + 2.63
0
π
2
sin x dx =
Chapter 4. Integration
¯ π2
¯
= 61.7199 − 2.63 cos x ¯ = 61.7199 + 2.63 = 64.3499(≈ 64.3499).
0
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Z 6.4
Z 6.4
Question 7 (L.O.1): Let
f (x) dx = 6.88 and
g (x) dx = −2.47. Evaluate I =
−2.5
−2.5
Z 6.4 h
i
18.86x + 5.55 f (x) − 7.76g (x) dx.
−2.5
A 384.0257
B 384.9296
C 385.2162
D 384.6665
E 385.0722
SOLUTION
We have
Z 6.4
Z
£
¤
18.86x + 5.55 f (x) − 7.76g (x) dx = 18.86
−2.5
6.4
Z
x dx+5.55
−2.5
= 18.86
6.4
Z
6.4
f (x) dx−7.76
−2.5
−2.5
g (x) dx =
x 2 ¯¯6.4
+ 5.55 × 6.88 − 7.76 × (−2.47) = 384.6665(≈ 384.6665).
¯
2 −2.5
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 8 (L.O.2): Let g (x) = f −1 (x) be the inverse function of function y = f (x) =
Z 11.01
x/2
e + 5.34. Evaluate
g (x)d x.
6.34
A 9.6536
B 10.3370
C 10.9418
D 10.7950
E 9.4292
SOLUTION
Since y = f (x) = e +5.34 ⇒ x = 2 ln(y −5.34). Thus g (x) = f −1 (x) = 2 ln(x −5.34). Therefore,
Z 11.01
g (x)d x ≈ 10.3370.
x
2
6.34
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
The average value of a function
Question 9 (L.O.2): The interest rates charged by Madison Finance on auto loans for
used cars over a certain 12−month period in 2022 are approximated by the function
r (t ) = −
1 3 1 2
t + t − 2t + 27,
21
3
(0 É t É 12)
where t is measured in months and r (t ) is the annual percentage rate. What is the
average rate on auto loans extended by Madison over the 12−month period?
A 11.1063
B 11.2527
C 10.2379
D 10.4286
E 10.6659
SOLUTION
The average rate on auto loans extended by Madison over the 12−month period is
¶
Z 12
Z µ
1
1 12
1 3 1 2
73
r ave =
r (t )d t =
− t + t − 2t + 27 d t =
≈ 10.4286.
12 − 0 0
12 0
21
3
7
208
208
XI. Exercises
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Work done
Question 10 (L.O.2): When a particle is located a distance x meter from the origin,
1
a force of given F (x) = 2
(newton) acts on it. How much work is done in
x + 4x
p + 53
moving it from x = 5 to x = 7 3 − 2.
C 0.0374
A −0.0604
B −0.0687
D 0.0634
E −0.4302
SOLUTION
We have
p
7 3−2
Z
W=
F (x)d x =
5
p
7 3−2
Z
5
1
=
2
x + 4x + 53
Z
p
7 3−2
5
p
d (x + 2)
1
(x + 2) ¯¯7 3−2
= arctan
=
¯
5
(x + 2)2 + 49 7
7
p
1
1 ³π π´ π
= (arctan 3 − arctan 1) =
−
=
≈ 0.0374.
7
7 3 4
84
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 11 (L.O.2): When a particle is located a distance x meter from the origin, a
1
force of given F (x) = p
(newton) acts on it. How much work is
−16x 2 + 128x + 1040
done in moving it from x = 17/2 to x = 13.
A 0.2618
B −0.7293
C 0.5967
D −0.3280
E −0.6007
SOLUTION
We have
13
13
1
W=
F (x)d x =
dx =
p
4
17/2
17/2 −16x 2 + 128x + 1040
Z
Z
1
Z
13
17/2
d (x − 4)
p
81 − (x − 4)2
=
µ
¶
1
(x − 4) ¯¯13
1
1
1 ³π π´ π
= arcsin
= arcsin 1 − arcsin =
−
=
≈ 0.2618.
¯
17/2
4
9
4
2
4 2 6
12
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 12 (L.O.2): When a particle is located a distance x meter from the origin, a
force of given F (x) = ln(6 + 5x) (newton) acts on it. How much work is done in moving
it from x = 10 to x = 17.
A 30.1241
B 30.4412
D 30.0137
E 29.2949
C 29.4318
SOLUTION
56 log (56)
91 log (91)
We have W =
F (x)d x =
ln(6 + 5x)d x = −
−7+
≈ 30.0137.
5
5
10
10
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Z
17
Z
17
209
209
Chapter 4. Integration
Question 13 (L.O.2): When a particle is located a distance x meter from the origin, a
60
force of given F (x) = p
(newton) acts on it. How much work is done in moving
196 − x 2
it from x = 1 to x = 6.
C 23.0572
A 22.2853
B 22.7360
D 21.5142
E 23.0559
6
6
SOLUTION
¶
µ ¶
1
3
We have W =
F (x)d x =
d x = −60 asin
+ 60 asin
≈ 22.2853.
p
2
14
7
1
1
196 − x
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Z
Z
60
µ
Applications of definite integrals in Physics
Question 14 (L.O.2): Evaluate the approximation of the displacement of one particle
with velocity v(t ) = 9 arcsin(t ) (m/s) from t = 0s to t = 0.7s.
C 2.5228
A 3.0163
B 3.2834
D 1.9853
E 2.3123
SOLUTION
The displacement of one particle is
Z
D=
0
0.7
Z
v(t )d t =
0.7
9 arcsin(t )d t .
0

dt
 du =
u = arcsin t
p
Let
⇒
1− t2

dv = dt
v = t
Therefore,
Z 0.7
¯0.7 9 Z 0.7
¯0.7
tdt
¯
¯
= 9t arcsin t ¯ +
D = 9t arcsin t ¯ − 9
(1 − t 2 )−1/2 d (1 − t 2 ) =
p
2
0
0
2 0
0
1−t
¯0.7
¯0.7
p
¯
¯
= 9t arcsin t ¯ + 9 1 − t 2 ¯ = 2.31228981433631 ≈ 2.3123.
½
0
0
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 15 (L.O.2): The velocity of a dragster t seconds after leaving the starting
line is v(t ) = 72t e −0.7t ft/sec. Find the approximation of the distance traveled by the
dragster during the first 9 seconds.
A 145.0934
B 144.0773
C 143.9867
D 144.9122
E 144.9691
SOLUTION
The distance is
Z 9
Z 9¯
Z 9
¯
¯
¯
72t e −0.7t d t = 144.969057569783 ≈ 144.9691
D=
|v(t )|d t =
¯72t e −0.7t ¯d t =
0
0
0
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Applications of definite integrals in Economics
210
210
XI. Exercises
Question 16 (L.O.1): The accumulated, or total, future value after T years of an income
stream of R(t ) thousand dollars per year, earning interest at the rate of r per year
Z T
rT
compounded continuously, is given by A = e
R(t )e −r t d t . A company recently
0
bought an automatic car-washing machine that is expected to generate 3 thousand
dollars in revenue per year, t years from now, for the next 7 years. If the income is
reinvested in a business earning interest at the rate of r = 11% per year compounded
continuously, find the total accumulated value of this income stream at the end of 7
years.
A 30.6382
B 31.6300
C 32.4296
D 32.2997
E 31.7604
SOLUTION
We have r = 11%, R(t ) = 3, T = 7. Therefore, the total accumulated value of this income
stream at the end of 7 years is
A = er T
T
Z
0
R(t )e −r t d t = 31.6299887395886 ≈ 31.6300
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Area problems
Question 17 (L.O.2): Find the area of the region bounded by the graphs of y =
1
; y = 0; x = 0; and x = 4.
p
9 16 − x 2
A 0.5049
B 0.3040
C 0.2790
D 0.2973
E 0.1745
SOLUTION
The required area is
¶
µ
Z 4
1h
x i4 1
π
dx
b
=
A=
arcsin
=
lim− arcsin − arcsin 0 =
≈ 0.1745
p
9
4 0 9 b→4
4
18
0 9 16 − x 2
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 18 (L.O.2): Find the area of the region bounded by the graphs of y = e 2.53x ;
y = e 4.75x and x = 3.03.
A 374034.3816
B 374034.2241
C 374034.3701
D 374034.9554
E 374034.1488
SOLUTION
The required area is
Z
A=
0
3.03
(e 4.75x − e 2.53x )d x =
h e 4.75x
4.75
−
e 2.53x i3.03
=
2.53 0
= 374034.381624145 ≈ 374034.3816
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
211
211
Chapter 4. Integration
Volume problems
p
Question 19 (L.O.2): Let the region D be bounded by the curve y = 15x 2 + 2, x−axis
and the lines x = 0, x = 5. Evaluate the volume V of the solid, revolving the region D
about x−axis.
A 1995.0119
B 1995.0890
C 1994.9113
D 1994.7434
E 1993.9313
SOLUTION
We have
V =π
Z 5 ³p
0
15x 2 + 2
´2
dx = π
Z
0
5¡
¢
£
¤5
15x 2 + 2 dx = π 5x 3 + 2x 0 = 635π ≈ 1994.9113
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 20 (L.O.2): Find the volume of the solid obtained by rotating the region
bounded by the curves x = 0, y = −16, y = x 2 − 8x about the y−axis.
C 133.9908
A 134.0465
B 134.0413
D 133.6074
E 134.2850
SOLUTION
We have
VO y
Z 4³
Z 4
´
128π
2
= 2π
|x(−16)| − |x(x − 8x)| d x = 2π x(16 − 8x + x 2 )d x =
≈ 134.0413
3
0
0
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Arc length
Question 21 (L.O.2): Find the length of the arc y = 3 ln x, where 6 É x É 9.
A 3.2399
B 2.8595
C 3.9077
D 2.5529
E 3.1169
SOLUTION
3
We have y = 3 ln x ⇒ y = · The arc length is
x
s
µ ¶2
µ ¶
µ ¶
Z 9q
Z 9
p
p
3
1
1
′
2
d x = −3 5−3 asinh
+3 asinh
+3 10 ≈ 3.2399
L=
1 + [y (x)] d x =
1+
x
3
2
6
6
′
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 22 (L.O.2): A steady wind blows a kite due west. The kite’s height above the
³p
´3
ground from horizontal position x = 0 to x = a > 0 meters is given by y = 3 + 2x .
Find the real number a such that the distance travelled by the kite is 20.66 meters.
A 3.6074
B 3.2822
D 2.8492
E 3.5988
C 2.5416
212
212
XI. Exercises
SOLUTION
³p
´3
p
′
We have y = 3 + 2x ⇒ y = 3 2x + 3· The distance travelled by the kite is
L=
Z aq
0
1 + [y ′ (x)]2 d x
Z ar
=
0
p
3
h p
i2
(18a + 28) 2 56 7
1 + 3 2x + 3 d x =
−
= 20.66
27
27
⇒ a ≈ 2.8492.
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 23 (L.O.2): Find the length of the arc y = P (x), where 3 É x É 9. Here P (x) is
a polynomial of degree 2 such that P (3) = 62, P (4) = 96, P (9) = 386.
C 323.3469
A 323.5810
B 323.6091
D 324.0597
E 324.9452
SOLUTION
Let P (x) = ax 2 + bx + c. Since P (3) = 62, P (4) = 96, P (9) = 386, so we have


9a + 3b + c = 62
16a + 4b + c = 96 ⇒ a = 4; b = 6; c = 8.

81a + 9b + c = 386
Thus, P (x) = 4x 2 + 6x + 8 ⇒ P ′ (x) = 8x + 6. The arc length is
L=
Z 9q
3
1 + [y ′ (x)]2 d x
Z 9r
=
3
h
i2
1 + 8x + 6 d x =
p
p
15 901 asinh (30) asinh (78) 39 6085
=−
−
+
+
≈ 324.0597.
8
16
16
8
¤ The correct answer is D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 24 (L.O.2): Find the length of the arc y =
8.
A 69.4857
′
We have y (x) =
B 69.7190
p
5x 2 + 7.
C 69.7552
Z xp
2
5t 2 + 7d t , where 2 É x É
D 69.0281
E 68.6696
SOLUTION
The arc length is
Z 8r
Z 8q
hp
i2
L=
1 + [y ′ (x)]2 d x =
1+
5x 2 + 7 d x =
2
p
= −2 7 −
2
³p ´
p
4 5 asinh 210
5
+
p
¡ p ¢
4 5 asinh 2 10
5
p
+ 8 82 ≈ 69.4857
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
213
213
Chapter 4. Integration
Area of surface of revolution
Question 25 (L.O.2): Find the area of the surface of revolution obtained by rotating
p
the curve y = 2 x + 6, where 1 É x É 8 about the x−axis.
C 502.8761
A 502.2993
B 502.5967
D 502.4058
E 502.0702
SOLUTION
1
We have y = 2 x + 6 ⇒ y = p · The surface area is
x
p
′
8
Z
A = 2π
1
|y|
q
1 + [y ′ (x)]2 d x
Z
= 2π
1
8
s
p
(2 x + 6)
·
1
1+ p
x
¸2
dx =
p !
³
³ p ´
p ´ 26 2
p ´
= −2π 6 log 1 + 2 +
+ 2π 6 asinh 2 2 + 36 + 36 2 ≈ 502.2993.
3
Ã
³
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
q
Question 26 (L.O.2): Rotating the curve x = − y 2 + 3, where 0 É y É 3 about the
y−axis, evaluate the surface area of the obtained surface of revolution.
A 53.5843
B 54.2794
C 54.0413
D 53.1674
E 54.2563
SOLUTION
q
y
′
2
· The surface area is
We have x = − y + 3 ⇒ x = − p
y2 + 3
Z
A = 2π
Z
= 2π
s
3
2π
0
0
3
Z 3 q
q
|x| 1 + [x ′ (y)]2 d y = 2π | − y 2 + 3|
s
0
2y 2 + 3
y2 + 3
q
y 2 + 3d y = 2π
à p
¡p ¢
3 2 asinh 6
4
1+
y2
dy =
y2 + 3
p !
3 21
+
≈ 54.0413.
2
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Improper integrals
Z+∞ 2
3x d x
Question 27 (L.O.2): Evaluate the integral
7 + x4
1
A 1.8614
B 2.4366
C 1.9137
D 1.5139
SOLUTION
We have
Z+∞ 2
Z+∞
3x d x
3/x 2 d x
I=
=
7 + x4
1 + 7/x 4
1
1
214
214
E 1.3924
XI. Exercises
Let t =
1
dx
⇒ dt = − 2 ·
x
x
Z0
I=
1
−3d t
=
1 + 7t 4
Z1
0
3d t
≈ 1.9137
1 + 7t 4
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
+∞
Z
Question 28 (L.O.2): Evaluate the integral
A 1.1263
Let t =
p
4
B 0.3787
x 2 + 625 ⇒ t 4
Z
=
6
C −0.6059
D 0.7021
E −0.0613
SOLUTION
= x + 625 ⇒ 2t d t = xd x. Then,
2
3
+∞
dx
=
p
p
4
671 x · x 2 + 1
Z
I=
dx
p
p
4
671 x · x 2 + 625
+∞
xd x
=
p
p
4
671 x 2 · x 2 + 625
Z
+∞
Z
6
2t 3 d t
=
t (t 4 − 625)
Z +∞ 2
Z +∞
Z +∞
2t 2 d t
(t + 25) + (t 2 − 25)d t
dt
dt
=
=
+
=
t 4 − 625
(t 2 − 25)(t 2 + 25)
t 2 − 25
t 2 + 25
6
6
6
¯
¯ ¯
¯
t ¯¯+∞ π 1
6 1 ¯¯ 1 ¯¯
1 ¯¯ t − 5 ¯¯ ¯¯+∞ 1
=
ln
− arctan −
ln
=
=
¯ + arctan ¯
10 ¯ t + 5 ¯ 6
5
5 6
10 5
5 10 ¯ 11 ¯
+∞
= −0.175211610119639 + 0.1 log (11) +
π
≈ 0.3787
10
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 29 (L.O.2): The total profit of a firm in dollars, from producing x units of
an item, is P (x). The firm is able to determine that its marginal profit is given by
P ′ (x) = 190e −1.05x . Suppose that it were possible for the firm to make infinitely many
units of this item. What would its total profit be?
A 180.8926
B 181.1377
C 180.4869
D 181.9074
E 180.9524
Z
The total profit is P (x) =
∞
0
SOLUTION
·
¸
Z ∞
190 −1.05x ∞
190
−1.05x
P (x)d x =
190e
dx =
e
=
≈
−1.05
1.05
0
0
′
180.9524.
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Z+∞
8d x
Question 30 (L.O.2): Evaluate the integral
.
p
x2 + x4
5
A 0.9952
B 0.9963
C 1.5895
D 1.1703
SOLUTION
We have
Z+∞
Z+∞
8d x
8/x 2 d x
I=
=
p
p
1 + 1/x 2
x2 + x4
5
5
215
215
E 1.7271
Chapter 4. Integration
Let t =
1
dx
⇒ dt = − 2 ·
x
x
Z0
I=
1/5
−8d t
=
p
1+ t2
Z1/5
8d t
p
1+ t2
0
= 1.58952088279393 ≈ 1.5895
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 31 (L.O.2): Let a be the real number such that lim
x→+∞
³p
´
x 2 + 8x + 2− x − a =
Z+∞
dx
·
(x + a)(x − 2)2
0. Evaluate a +
6
A 4.5249
B 4.0162
We have a = lim
³p
x→+∞
x 2 + 8x + 2 − x
C 3.4425
D 3.4233
E 4.8000
´ SOLUTION8x + 2
= lim p
= 4. Then,
x→+∞ x 2 + 8x + 2 + x
Z+∞
a+
6
µ
¶ µ
¶ µ ¶
dx
log (10)
log (4)
1
= 4+ −
+
+
≈ 4.0162
2
(x + a)(x − 2)
36
36
24
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
9 md x
Question 32 (L.O.2): Find the approximation of m which satisfies
=
p
0
81 − x 2
Z 5 2
x dx
p
1
x −1
A 18.2816
B 17.4858
C 17.6462
D 18.3127
E 18.3730
Z
SOLUTION
x i9 π
Let A =
= arcsin
= ·
p
9 0 2
− x2
Z 05 81
p
x 2d x
Let B =
· Using substitution method, let t = x − 1 ⇒ t 2 = x − 1 ⇒ 2t d t = d x, we
p
1
x −1
have
p
p
p
· 5
¸ 4
Z p4
3
p
2( 4)5 4( 4)3
2t
4t
2
2
B=
2(t + 1) d t =
+
+ 2t
=
+
+ 2 4.
5
3
5
3
0
0
Z
9
dx
h
B 824
=
≈ 17.4858
A 15π
¤ The correct answer is B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Therefore, m =
216
216
Chapter 5
Ordinary Differential Equations
Learning Objectives
Study the first order differential equations.
Study the second order differential equations
Study some applications of Ordinary Differential Equations.
I
Ordinary differential equations
1
Introduction
Perhaps the most important of all the applications of calculus is to differential equations. Many of the principles, or laws, underlying the behaviour of the natural world
are statements or relations involving rates at which things happen. When expressed in
mathematical terms, the relations are equations and the rates are derivatives. Equations
containing derivatives are differential equations. Therefore, to understand problems, we
study the motion of fluids, the flow of current in electric circuits, the increase-decrease of
populations...
2
Ordinary Differential Equations
217
Chapter 5. Ordinary Differential Equations
Definition 80
1. A differential equation is an equation that contains an unknown function and
one or more of its derivatives.
2. The order of a differential equation is the order of the highest derivative that
occurs in the equation. The equation F (x, y, y ′ , . . . , y (n) ) = 0 is an ordinary
differential equation of the nth order.
3. A function f is called a general solution of an ordinary differential equation
F (x, y, y ′ , . . . , y (n) ) = 0 if the equation is satisfied when y = f (x) and its derivatives are substituted into the equation.
3
Direction fields
Suppose we are asked to sketch the graph of the solution of the problem
y ′ = x + y,
y(0) = 1.
Drawing short line segments at a number of points (x, y) with slope x + y, we receive
direction field. The direction field allows us to visualize the general shape of the solution
curves.
Definition 81
Suppose we have a first-order differential equation of the form
y ′ = F (x, y),
218
218
II. Separable Differential Equations
where F (x, y) is some expression in x and y. If we draw short line segments with
slope F (x, y) at several points (x, y), the result is called a direction field.
These line segments indicate the direction in which a solution curve is heading, so the
direction field helps us visualize the general shape of these curves.
II
1
Separable Differential Equations
Definition of separable equations
Definition 82
A separable equation is a first-order differential equation, which has a form
P (x)d x +Q(y)d y = 0
(5.1)
Integrating both sides, we receive general solution
Z
Z
P (x)d x + Q(y)d y = C .
This equation defines y implicitly as a function of x.
Example 159
Solve the differential equation xd x + (y + 1)d y = 0
SOLUTION
Integrating both sides, we receive general solution
Z
Z
xd x +
2
(y + 1)d y = C ⇒
x2 y 2
+
+y =C
2
2
Another form of separable equations
Corollary 2.0
f 1 (x).g 1 (y)d x + f 2 (x).g 2 (y)d y = 0
can be reduced to a separable differential equation.
If f 1 (x), f 2 (x), g 1 (y), g 2 (y) ̸= 0, then we divide both sides by f 2 (x)g 1 (y). So
f 1 (x)
g 2 (y)
dx +
d y = 0.
f 2 (x)
g 1 (y)
Z
Z
f 1 (x)
g 2 (y)
⇒
dx +
d y = C.
f 2 (x)
g 1 (y)
219
219
(5.2)
Chapter 5. Ordinary Differential Equations
Example 160
Solve the differential equation x(y 2 − 4)d x + yd y = 0.
SOLUTION Dividing both sides by y 2 − 4 ̸= 0, we receive
xd x +
yd y
= 0.
y2 − 4
Integrating both sides of the equation, we have
x 2 + ln |y 2 − 4| = ln |C | ⇒ y 2 − 4 = C e −x
3
2
A cool problem
Physical information: Experiments show that when the time rate of change of the temperature T of a body B (which conducts heat well, for example, as a copper ball does)
is proportional to the difference between T and the temperature TS of the surrounding
medium.
Example 161
Suppose that a bread is removed from a 100o C oven and placed in a room with a
temperature of 25o C . In 20mi n the bread has a temperature of 60o C . We want to
determine the time required to cool the bread to a temperature of 30o C , when we
can finally enjoy eating it.
SOLUTION
1. The rate of change of the temperature T (t ) is the derivative
dT
.
dt
2. According to Newton’s law of cooling, the rate at which the temperature T (t ) changes
in a cooling body is proportional to the difference between the temperature of the
220
220
II. Separable Differential Equations
body and the constant temperature T s of the surrounding medium. Symbolically,
the statement is expressed as
dT
= k(T − T s ),
dt
where k is a constant of proportionality.
Separating the variables we have
dT
dT
= kd t ⇔
= kd t
T − Ts
T − 25
Z
Z
dT
⇒
= k d t ⇒ ln |T − 25| = kt + lnC
T − 25
⇒ T − 25 = e kt +lnC = e kt .e lnC = C e kt
We can apply the initial condition: if t = 0mi n then T = 100o C . So 100−25 = C e k.0 ⇒ C = 75
To find k we use the condition: if t = 20mi n then T = 60o C . So
µ ¶1
7 20
k.20
k
60 − 25 = 75e
⇒e =
15
Therefore
µ
7
T = 75
15
¶t
20
+ 25.
We want to know when T (t ) = 30o C . Thus we solve
µ ¶t
7 20
−20 ln 15
≈ 71mi n
30 = 75
+ 25 ⇒ t =
15
ln 7 − ln 15
Example 162
Suppose that in winter the daytime temperature in a certain office building is maintained at 70o F. The heating is shut off at 10pm and turned on again at 6am. On a
certain day the temperature inside the building at 2am was found to be 65o F. The
outside temperature was 50o F at 10pm and had dropped to 40o F by 6am. What was
the temperature inside the building when the heat was turned on at 6am?
Note. The temperature of surrounding media TS can replaced with the average of the two
known 50o F and 40o F. Thus TS = 45o F. By Newton’s law,
dT
dT
= k(T − TS ) ⇒
= kd t
dt
T − 45
⇒ ln |T − 45| = kt + ln |C | ⇒ T (t ) = 45 +C e kt .
We choose 10 pm to be t = 0. Then the given initial condition is T (0) = 70 and yields a
particular solution, called T p .
T (0) = 45 +C e 0 = 70 ⇒ C = 25 ⇒ T p (t ) = 45 + 25e kt .
In order to determine k, we use T (4) = 65, where t = 4 is 2 am.
T p (4) = 45 + 25e 4k = 65 ⇒ e 4k = 0.8 ⇒ k = −0.056
⇒ T p (t ) = 45 + 25e −0.056t ⇒ T p (8) = 45 + 25e −0.056×8 ≈ 61o F.
221
221
Chapter 5. Ordinary Differential Equations
4
Radiocarbon Dating
Example 163
In the atmosphere and in living organisms, the ratio of radioactive carbon 14
6 C (made
12
radioactive by cosmic rays) to ordinary carbon 6 C is constant. When an organism
dies, its absorption of 14
6 C by breathing and eating terminates. Hence one can
estimate the age of a fossil by comparing the radioactive carbon ratio in the fossil
with that in the atmosphere. To do this, one needs to know the half-life of 14
6 C , which
is 5715 years.
(CRC Handbook of Chemistry and Physics, 83rd ed., Boca Raton: CRC Press, 2002,
page 11-52, line 9).
In September 1991 the famous Iceman (Oetzi), a mummy from the Neolithic period
of the Stone Age found in the ice of the Oetztal Alps (hence the name "Oetzi") in
Southern Tyrolia near the Austrian-Italian border, caused a scientific sensation.
12
When did Oetzi approximately live and die if the ratio of carbon 14
6 C to 6 C in this
mummy is 52.5% of that of a living organism?
SOLUTION
Note.
Radioactive decay is governed by the ODE y ′ (t ) = k y, where t is time and y is the ratio of
14
12
6 C to 6 C .
Z
Z
dy
dy
dy
= ky ⇒
= kd t ⇒
= kd t
dt
y
y
⇒ ln |y| = kt + lnC ⇒ y = C e kt .
In order to determine k we use the fact that when t = 5715, haft of the original substance is
still present, it means that y(5715) = 0.5y 0 , where y 0 = y(0) = C . Thus,
y 0 e 5715k = 0.5y 0 ⇒ e 5715k = 0.5 ⇒ k =
ln 0.5
≈ −0.0001213.
5715
Finally, we use the ratio 52.5% for determining the time t when Oetzi died
y 0 e kt = 0.525y 0 ⇒ e kt = 0.525 ⇒ t =
5
ln 0.525
≈ 5212.
−0.0001213
Mixing Problem
Example 164
Mixing problems occur quite frequently in chemical industry. The tank contains
1000 gal of water in which initially 100 lb of salt is dissolved. Brine runs in a rate of
10g al /mi n, and each gallon contains 5 lb of dissolved salt. The mixture in the tank
is kept uniform by stirring. Brine runs out at 10g al /mi n. Find the amount of salt in
the tank at any time t .
SOLUTION
222
222
III. Linear Differential Equations
Let y(t ) denote the amount of salt in the tank at time t . Its time rate of change is
y ′ (t ) = Salt inflow rate − Salt outflow rate
(5.3)
5 lb times 10 gal gives an inflow of 50 lb of salt per minute. Now, the outflow is 10 gal of
10
= 0.01 = 1% of the total brine content in the tank, hence
brine per minute. This is
1000
0.01 of the salt content y(t ), that is, 0.01y(t ). Thus the model is the ODE
y ′ = 50 − 0.01y
dy
dy
= 50 − 0.01y ⇒
= dt
dt
50 − 0.01y
Z
Z
100d y
= d t ⇒ ln |y − 5000| = −0.01t + lnC
⇒
5000 − y
⇒ y − 5000 = C e −0.01t .
Initially the tank contains 100 lb of salt. Hence y(0) = 100. Substituting y = 100 and t = 0 in
the last equation gives
100 − 5000 = C e 0 ⇒ C = −4900.
Hence the amount of salt in the tank at time t is
y(t ) = 5000 − 4900e −0.01t .
III
1
Linear Differential Equations
Definition of linear equation
Definition 83
A first-order linear differential equation is one that can be put into the form
dy
+ P (x).y = Q(x)
dx
where P and Q are continuous functions on a given interval.
223
223
(5.4)
Chapter 5. Ordinary Differential Equations
2
The solution of linear equation
To
R solve the linear differential equation (5.4), multiply both sides by the integrating factor
e P (x)d x and integrate both sides.
y ′e
R
P (x)d x
+ P (x)y.e
R
P (x)d x
= Q(x).e
R
P (x)d x
³ R
´′
R
⇒ y.e P (x)d x = Q(x).e P (x)d x
Z
R
R
P (x)d x
⇒ y.e
= Q(x).e P (x)d x d x +C
y =e
R
− P (x)d x
·Z
.
Q(x).e
R
P (x)d x
¸
d x +C
Example 165
Find the solution of the problem y ′ +
1
y = 3x, y(1) = 1.
x
¸
·Z
R
R
1
P (x)d x
− P (x)d x
d x +C
Q(x).e
SOLUTION P (x) = , Q(x) = 3x. ⇒ y = e
.
x
·Z
¸
¸
·Z
R
R 1
− x1 d x
d
x
−
ln
x
ln
x
=e
3x.e x d x +C = e
.
3x.e d x +C =
=e
ln x −1
¸
·Z
3x.xd x +C =
1 3
(x +C )
x
1
Since x = 1, y = 1 ⇒ 1 = (13 +C ) ⇒ C = 0. Therefore the solution to this problem is y = x 2 .
1
3
Electric Circuit
Physical Laws. A current I in the circuit causes a voltage drop RI across the resistor (Ohm’s
dI
law) and a voltage drop LI ′ = L
across the conductor, and the sum of these two voltage
dt
drops equals the EMF (electromotive force).
According to these laws the model of the RL-circuit is
LI ′ + R I = E (t )
(5.5)
Example 166
Model the RL-circuit solves the resulting ODE for the current I (t )A (amperes), where
t is time. Assume that the circuit contains as an EMF E (t ) a battery of E = 48V
(volts), which is constant, a resistor of R = 11Ω (ohms), and an inductor of L = 0.1H
(henrys), and that the current is initially zero. Find the current I (t ).
224
224
III. Linear Differential Equations
SOLUTION According to these laws the model of the RL-circuit in standard form is
E (t )
R
·I =
L
L
Z
h
i
E (t )
e (R/L)t ·
⇒ I = e −(R/L)t
d t +C
L
In our case, R = 11, L = 0.1 and E (t ) = 48, thus
i
hZ
i
h 480
−110t
I =e
e 110t · 480d t +C = e −110t
e 110t +C
110
I′ +
The initial value I (0) = 0 gives
h 480
i
48
e 110×0 +C = 0 ⇒ C = −
110
11
³
´
48
1 − e −110t .
The particular solution is y =
11
I (0) = e −110×0
4
Hormone Level
Assume that the level of a certain hormone in the blood of a patient varies with time.
Suppose that the time rate of change is the difference between a sinusoidal input of a
24−hour period from the thyroid gland and a continuous removal rate proportional to
the level present. Set up a model for the hormone level in the blood and find its general
solution. Find the particular solution satisfying a suitable initial condition.
Example 167
Let y(t ) be the hormone level at time t . Then the removal rate is K y(t ). The input
π
rate is A + B cos ωt , where ω =
and A is the average input rate; here A Ê B to
12
make the input rate non-negative. The constants A, B, K can be determined from
measurements. Hence the model is the linear ODE
y ′ (t ) = I n − Out = A + B cos ωt − K y(t ).
The initial condition for a particular solution y p is y p (0) = y 0 with t = 0 suitably
chosen. Let A = B = 1, K = 0.05 and y 0 = 0.
225
225
Chapter 5. Ordinary Differential Equations
SOLUTION We have
y ′ (t ) + K y(t ) = A + B cos ωt
i
hZ
−K t
e K t (A + B cos ωt )d t +C
⇒ y(t ) = e
¸
´
Ae K t
B eK t ³
⇒ y(t ) = e
K cos ωt + ω sin ωt +C
+ 2
K
K + ω2
Setting t = 0 in y(t ) and choosing y 0 = 0, we have
−K t
y(0) =
·
A
BK
A
BK
+ 2
+C
=
0
⇒
C
=
−
−
K K + ω2
K K 2 + ω2
The particular solution is
y p (t ) =
IV
1
³
´ ³A
B
B K ´ −K t
A
+ 2
K
cos
ωt
+
ω
sin
ωt
−
+
e
K K + ω2
K K 2 + ω2
Bernoulli Differential Equations
Definition of Bernoulli Differential Equations
Definition 84
The first-order differential equation of the form
y ′ + P (x)y = Q(x)y α , α ̸= 0, α ̸= 1
is called a Bernoulli differential equation.
2
Solving Bernoulli Differential Equations
Substitute z =
y
= y 1−α . This substitution reduces (5.6) to a linear equation
yα
z ′ + (1 − α)P (x).z = (1 − α)Q(x).
Example 168
Solve the differential equation y ′ +
y
= x 2 y 4.
x
SOLUTION
y
The substitution z = 4 = y −3 reduces Bernoulli’s equation to a linear equation
y
1
z ′ − 3 .z = −3x 2 .
x
µ
¶
Z
R 3
R 3
dx
− x dx
2
x
⇒z =e
.
e
.(−3x )d x +C .
1
⇒ z = x 3 (−3 ln |x| +C ) ⇒ y = p
x 3 −3 ln |x| +C
226
226
(5.6)
V. Homogeneous Equations of the form f
V
1
³y´
x
Homogeneous Equations of the form f
³y´
x
Definition of homogeneous equations of the form f
³y´
x
Definition 85
The first-order differential equation of the form
³y´
dy
=f
dx
x
(5.7)
is said to be of homogeneous type.
y
⇒ y = u.x ⇒ y ′ = u ′ .x + u. transforms (5.7) into a separable
x
equation in the variables u and x
The change of variables u =
dx
du
=
f (u) − u
x
Example 169
Find the solution of the initial-value problem y ′ =
SOLUTION Substitute u =
rewritten as
y
y
π
+ sin , y(1) = .
x
x
2
y
⇒ y = ux ⇒ y ′ = u ′ .x + u. Then the given equation can be
x
du
du
dx
.x = sin u ⇒
=
dx
sin u
x
Z
Z
¯
¯
dx
u¯
du
¯
=
⇒ ln ¯tan ¯ = ln |x| + lnC
⇒
sin u
x
2
u
⇒ tan = C x.
2
u ′ .x + u = u + sin u ⇒
Substituting u =
y
, we have
x
tan
Using the initial condition y(1) =
y
= C x.
2x
π
π
, that means x = 1, y = , we have
2
2
tan
π
= C .1 ⇒ C = 1.
4
Therefore the solution of the initial-value problem is
tan
2
y
= x.
2x
Another form of homogeneous equations
227
227
Chapter 5. Ordinary Differential Equations
Proposition 5.1
The differential equation of the form
y ′ = f (x, y),
(5.8)
where f (x, y) is homogeneous of degree 0, that means
f (t x, t y) = t 0 . f (x, y) = f (x, y),
can be reduced to a homogeneous equation.
Example 170
Solve the differential equation (x 2 + 2x y)d x + x yd y = 0
SOLUTION The given equation can be rewritten as
y′ = −
x 2 + 2x y
= f (x, y).
xy
where f (x, y) is homogeneous of degree 0 since
f (t x, t y) = −
t 2 x 2 + 2t 2 x y
x 2 + 2x y
=
−
= f (x, y).
t 2x y
xy
y
⇒ y = u.x ⇒ y ′ = u ′ .x + u.
x
So the given equation can be rewritten as
Let u =
1
du
1 + 2u + u 2
− 2 ⇒ x.
=−
u
dx
u
Z
Z
ud u
ud u
dx
dx
⇒
=−
⇒
=−
2
2
(u + 1)
x
(u + 1)
x
Z
Z
u +1−1
dx
⇒
du = −
2
(u + 1)
x
1
⇒ ln |u + 1| +
= − ln |x| +C
u +1
u′x + u = −
Substituting u =
y
, we have
x
¯y +x¯
x
x
¯
¯
ln |x| + ln ¯
= C ⇒ ln |x + y| +
= C.
¯+
x
x+y
x+y
Proposition 5.2
The differential equation of the form
a1 x + b1 y + c1
y =f
a2 x + b2 y + c2
′
µ
228
228
¶
(5.9)
V. Homogeneous Equations of the form f
³y´
x
can be reduced to a homogeneous or a separable equation.
¯
¯
¯ a1 b1 ¯
¯ ̸= 0
Case 1: ¯¯
a2 b2 ¯
¯
¯
½
¯ a1 b1 ¯
a1 x + b1 y + c1 = 0
¯
¯
If ¯
̸= 0 then the system of linear equations
has a unique
¯
a2 b2
a2 x + b2 y + c2 = 0
solution (x 0 , y 0 ).
Let X = x − x 0 , Y = y − y 0 ⇒ y ′ = Y ′ . Use this transformation to obtain the differential
equation
Ã
!
µ
¶
Y
a
+
b
a
X
+
b
Y
1
1
1
1
X
Y′=f
⇒Y′= f
Y
a2 X + b2 Y
a2 + b2 X
which is¯ a homogeneous
equation.
¯
¯ a1 b1 ¯
¯=0
Case 2: ¯¯
a2 b2 ¯
¯
¯
¯ a1 b1 ¯
¯
¯ = 0 ⇒ a1 = b1 = K .
If ¯
a2 b2 ¯
a2 b2
Let u = a 2 x + b 2 y ⇒ u ′ = a 2 + b 2 y ′ . The given equation can be rewritten as
µ
¶
a1 x + b1 y + c1
⇒ b2 y ′ = b2 f
a2 x + b2 y + c2
µ
¶
K u + c1
′
⇒ u − a2 = b2 . f
u + c2
µ
¶
K u + c1
du
= a2 + b2 . f
⇒
dx
u + c2
is a separable equation.
Example 171
Solve the differential equation (2x + y + 1)d x + (x + 2y − 1)d y = 0
SOLUTION The given equation can be rewritten as
y′ = −
¯
¯ 2 1
where ¯¯
1 2
2x + y + 1
x + 2y − 1
¯
¯
¯ = 3 ̸= 0. Solving the system of equations
¯
½
2x + y + 1 = 0
x + 2y − 1 = 0
we find the unique solution (x 0 , y 0 ) = (−1, 1).
Let X = x + 1, Y = y − 1 ⇒ Y ′ = y ′ . The given equation can be rewritten as
Y′=
Let u =
2 + YX
2(X − 1) + (Y + 1) + 1 2X + Y
=
=
·
(X − 1) + 2(Y + 1) − 1 X + 2Y
1 + 2 YX
Y
⇒ Y = u.X ⇒ Y ′ = u ′ .X + u.
X
⇒ u ′ .X + u = −
2+u
(1 + 2u)d u
dX
⇒ 2
= −2
1 + 2u
u +u +1
X
229
229
Chapter 5. Ordinary Differential Equations
⇒ X 2 .(u 2 + u + 1) = C .
Substituting u =
Y
we have Y 2 + X Y + X 2 = C ⇒ (y − 1)2 + (x + 1)(y − 1) + (x + 1)2 = C .
X
Example 172
Solve the differential equation (x + y + 2)d x + (2x + 2y − 1)d y = 0
SOLUTION The given equation can be rewritten as
dy
x + y +2
=−
dx
2x + 2y − 1
¯
¯ 1 1
where ¯¯
2 2
¯
¯
¯ = 0. Let u = x + y ⇒ u ′ = 1 + y ′ . Then
¯
u′ − 1 = −
⇒
u −3
u +2
⇒ u′ =
2u − 1
2u − 1
du
u −3
(2u − 1)d u
=
⇒
=x
d x 2u − 1
u −3
Z
Z
2u − 1
⇒
d u = xd x +C
u −3
⇒ 2u + 5 ln |u − 3| = x +C
Substituting u = x + y gives
x + 2y + 5 ln |x + y − 3| = C
as the implicit solution to the differential equation.
VI
The second order differential equations with
constant coefficients
1
Definition
Definition 86
Homogeneous Second-order Differential Equation with constant coefficients has a
form
Ay ′′ + B y ′ +C y = 0,
(A, B,C ∈ R, A ̸= 0).
(5.10)
Definition 87
Nonhomogeneous Second-order Differential Equation with constant coefficients
has a form
Ay ′′ + B y ′ +C y = f (x),
230
230
(A, B,C ∈ R, A ̸= 0)
(5.11)
VII. Initial and boundary value problems
VII
Initial and boundary value problems
Definition 88
An initial value problem consists of a differential equation such as (5.11) or
(5.10) together with a pair of initial conditions y(x 0 ) = α, y ′ (x 0 ) = β, where
α, β = const are given numbers prescribing values for y and y ′ at the initial
point x 0 .
A boundary value problem consists of a differential equation such as (5.11)
or (5.10) together with a pair of boundary conditions y(x 0 ) = α, y(x 1 ) = β, or
y ′ (x 0 ) = α, y ′ (x 1 ) = β, or y(x 0 ) = α, y ′ (x 1 ) = β, where α, β = const are given
numbers prescribing values for y or y ′ at different points x 0 , x 1 .
Corollary 7.0
Solution of Second-order Differential Equations with constant coefficients
general solution of nonhomogeneous=homogeneous solution+particular solution of
nonhomogeneous.
VIII Homogeneous Differential Equation with constant coefficients
1
Linear Independence and the Wronskian
Theorem VIII.1
Let y 1 , y 2 be any 2 solutions of the homogeneous linear differential equation (5.10),
then
C 1 y 1 +C 2 y 2
is also a solution of equation (5.10) where C 1 ,C 2 are arbitrary constants.
Example 173
Consider the differential equation y ′′ + y = 0. We note that y 1 (x) = sin x and
y 2 (x) = cos x are solutions. Thus, y(x) = C 1 sin x +C 2 cos x is also a solution for any
constants C 1 and C 2 .
Definition 89
1. The 2 functions y 1 , y 2 are linearly dependent on a É x É b if there exist constants C 1 ,C 2 , not all zero, such that
C 1 y 1 +C 2 y 2 = 0
for all x such that a É x É b.
231
231
Chapter 5. Ordinary Differential Equations
2. The 2 functions y 1 , y 2 are linearly independent on a É x É b if
C 1 y 1 +C 2 y 2 = 0
for all x such that a É x É b implies C 1 = C 2 = 0.
Example 174
The functions x and −3x are linearly dependent on 0 É x É 1 since there are constants C 1 and C 2 , not both zero, such that
C 1 x +C 2 (−3x) = 0
for all x on 0 É x É 1. Just take C 1 = 3 and C 2 = 1.
Example 175
Show that the functions x and x 2 are linearly independent on −1 É x É 1.
SOLUTION
We show that
C 1 x +C 2 x 2 = 0
for all x on −1 É x É 1 implies both C 1 = 0 and C 2 = 0.
Since the equation holds for all x, it must hold when x = 1. This gives
C 1 .1 +C 2 .12 = 0.
It must also hold when x = −1 and this gives
C 1 .(−1) +C 2 .(−1)2 = 0.
Solve the system of equations
½
C 1 .1 +C 2 .12 = 0.
⇔
C 1 .(−1) +C 2 .(−1)2 = 0.
½
C 1 = 0.
C 2 = 0.
Definition 90
Let y 1 , y 2 be 2 real functions each of which has the first derivative on the interval
a É x É b. The determinant
¯
¯
¯ y 1 (x) y 2 (x) ¯
¯
W (x) = ¯¯ ′
y 1 (x) y 2′ (x) ¯
is called the Wronskian of these 2 functions.
Note that W (x) is a real function defined on a É x É b.
232
232
VIII. Homogeneous Differential Equation with constant coefficients
Theorem VIII.2
Let y 1 , y 2 be 2 real functions each of which has the first derivative on the interval
a É x É b. Suppose that the functions y 1 , y 2 are each a solution of the second order
linear homogeneous differential equation. Then
1. W (x) ̸= 0 for all x ∈ (a, b) if and only if the functions y 1 , y 2 are linearly independent on (a, b).
2. W (x) = 0 for all x ∈ (a, b) if and only if the functions y 1 , y 2 are linearly dependent on (a, b).
Example 176
Suppose that the functions y 1 = e x , y 2 = e 2x are each a solution of the second order
linear homogeneous differential equation. Show that the functions e x and e 2x are
linearly independent.
SOLUTION
¯
¯ y 1 (x) y 2 (x)
W (x) = ¯¯ ′
y 1 (x) y 2′ (x)
¯ ¯ x
¯ ¯ e
¯=¯
¯ ¯ ex
¯
e 2x ¯¯
=
2e 2x ¯
= 2e x .e 2x − e x .e 2x = e 3x ̸= 0, ∀x ∈ R.
Therefore, the functions e x and e 2x are linearly independent.
Theorem VIII.3 (General solution of homogeneous equation)
If y 1 , y 2 are 2 linearly independent solutions of equation (5.10), then every homogeneous solution y h of (5.10) can be expressed as a linear combination
y h = C 1 y 1 +C 2 y 2
(5.12)
of these 2 linearly independent solutions by proper choice of the constants C 1 ,C 2 .
2
Characteristic equation
We suppose that y = e kx , where k is a parameter to be determined. Then it follows that
y ′ = ke kx and y ′′ = k 2 e kx . By substituting these expressions for y, y ′ and y ′′ in equation
(5.10), we obtain
(Ak 2 + Bk +C )e kx = 0 ⇒ Ak 2 + Bk +C = 0(since e kx ̸= 0)
Definition 91
Equation
Ak 2 + Bk +C = 0
is called the characteristic equation for differential equation (5.10)
Case 1: Distinct Real Roots
233
233
(5.13)
Chapter 5. Ordinary Differential Equations
If the characteristic equation has real and different roots k 1 ̸= k 2 , then
½
Ak 12 + Bk 1 +C = 0
Ak 22 + Bk 2 +C = 0
and therefore, y 1 = e k1 x , y 2 = e k2 x are 2 solutions of (5.10), because
½
Ay 1′′ + B y 1′ +C y 1 = e k1 x (Ak 12 + Bk 1 +C ) = 0
Ay 2′′ + B y 2′ +C y 2 = e k2 x (Ak 22 + Bk 2 +C ) = 0
¯
¯ y 1 (x) y 2 (x)
W (x) = ¯¯ ′
y 1 (x) y 2′ (x)
¯ ¯ k1 x
¯ ¯ e
¯=¯
¯ ¯ k e k1 x
1
¯
e k2 x ¯¯
=
k 2 e k2 x ¯
= k 2 e k1 x .e k2 x − k 1 e k1 x .e k2 x = (k 2 − k 1 )e (k1 +k2 )x ̸= 0, ∀x ∈ R.
So y 1 = e k1 x , y 2 = e k2 x are linearly independent. Thus, the homogeneous solution of (5.10)
is
y h = C 1 y 1 +C 2 y 2 = C 1 e k1 x +C 2 e k2 x .
(5.14)
Example 177
Find homogeneous solution of the differential equation
y ′′ − y ′ − 6y = 0
SOLUTION
The characteristic equation is
k2 − k − 6 = 0
which has the roots k 1 = −2 and k 2 = 3.
Thus, y 1 = e k1 x = e −2x and y 2 = e k2 x = e 3x are linearly independent solutions and thus the
homogeneous solution is
y h = C 1 y 1 +C 2 y 2 = C 1 e −2x +C 2 e 3x .
3
Reduction of order
Theorem VIII.4
Let y 1 be a solution of the second-order homogeneous linear differential equation
(5.10). Then the transformation
y 2 = y 1 (x).v(x)
reduces equation (5.10) to the first order homogeneous linear differential equation
in the dependent variable
dv
w=
·
dx
At that time, y 1 , y 2 are linearly independent solutions of equation (5.10)
Case 2: Repeated Real Roots
234
234
VIII. Homogeneous Differential Equation with constant coefficients
If the characteristic equation has double real root k = k 0 = −
Ak 02 + Bk 0 +C
B
, then B 2 = 4AC and
2A
¶
µ
B 2
=0
= 0 ⇔ A k0 +
2A
and therefore, y 1 = e k0 x is a solution of (5.10), because
Ay 1′′ + B y 1′ +C y 1 = e k0 x (Ak 02 + Bk 0 +C ) = 0
We have to find another solution y 2 that is linearly independent of y 1 by reducing the order
of the equation. Let y 2 = y 1 .v = e k0 x v ⇒ y 2′ = v ′ e k0 x + k 0 e k0 x v,
y 2′′ = e k0 x v ′′ + 2k 0 e k0 x v ′ + k 02 e k0 x v.
³
´
Substituting these derivatives into Ay ′′ +B y ′ +C y = 0 gives A e k0 x v ′′ + 2k 0 e k0 x v ′ + k 02 e k0 x v +
´
³
B v ′ e k0 x + k 0 e k0 x v +C (e k0 x v) = 0
⇔ Ae k0 x v ′′ + (2Ak 0 + B )e k0 x v ′ + e k0 x v(Ak 02 + Bk 0 +C ) = 0
⇔ Ae k0 x v ′′ + (2Ak 0 + B )e k0 x v ′ = 0 ⇔ Ae k0 x v ′′ = 0 ⇔ v ′′ = 0
Now let w =
dv
= v ′ ⇒ w ′ = v ′′ so that
dx
w′ = 0 ⇔ w′ = 0 ⇔
dw
=0⇔ w =C
dx
⇒ v ′ = C ⇒ v = C x +C 1
Choose C = 1,C 1 = 0 we have
v = x ⇒ y 2 = v y 1 = xe k0 x
¯
¯ y 1 (x) y 2 (x)
W (x) = ¯¯ ′
y 1 (x) y 2′ (x)
¯ ¯ k0 x
¯ ¯ e
¯=¯
¯ ¯ k e k0 x
0
¯
¯
xe k0 x
¯
k0 x
k0 x ¯ =
e + xk 0 e
= (1 + k 0 x)e 2k0 x − k 0 xe 2k0 x = e 2k0 x ̸= 0, ∀x ∈ R.
So y 1 = e k0 x , y 2 = xe k0 x are linearly independent. Thus, the homogeneous solution of (5.10)
is
y h = C 1 y 1 +C 2 y 2 = C 1 e k0 x +C 2 xe k0 x
Example 178
Find homogeneous solution of the differential equation
y ′′ − 6y ′ + 9y = 0
SOLUTION
The characteristic equation is
k 2 − 6k + 9 = 0
235
235
(5.15)
Chapter 5. Ordinary Differential Equations
which has the double root k 0 = 3.
Thus, y 1 = e k0 x = e 3x and y 2 = xe k0 x = xe 3x are linearly independent solutions and thus the
homogeneous solution is
y h = C 1 y 1 +C 2 y 2 = C 1 e 3x +C 2 xe 3x .
Case 3: Complex Roots
If the characteristic equation has complex conjugate roots k 1 = a + bi , k 2 = a − bi , then
e (a+i b)x and e (a−i b)x are linearly independent solutions.
We are interested in real linearly independent solutions. So we use
Remark 20 (Euler’s formula)
e i ϕ = cos ϕ + i sin ϕ
(5.16)
e (a+i b)x = e ax .e i bx = e ax (cos bx + i sin bx)
e (a−i b)x = e ax .e −i bx = e ax (cos bx − i sin bx)
Then
e (a+i b)x + e (a−i b)x
y1 =
= e ax cos bx,
2
and
y2 =
e (a+i b)x − e (a−i b)x
= e ax sin bx
2i
are also 2 solutions of (5.10).
¯
¯ y 1 (x) y 2 (x)
W (x) = ¯¯ ′
y 1 (x) y 2′ (x)
¯
¯
¯=
¯
¯
¯
e ax cos bx
e ax sin bx
= ¯¯
ae ax cos bx − be ax sin bx ae ax sin bx + be ax cos bx
¯
¯
¯=
¯
= e 2ax (a sin bx + b cos bx) cos bx − e 2ax (a cos bx − b sin bx) sin bx =
= e 2ax b(cos2 bx + sin2 bx) = be 2ax ̸= 0, ∀x ∈ R.
So y 1 = e ax cos bx and y 2 = e ax sin bx are 2 linearly independent, real-valued solutions and
the homogeneous solution is
y h = C 1 y 1 +C 2 y 2 = C 1 e ax cos bx +C 2 e ax sin bx
y h = e ax (C 1 cos bx +C 2 sin bx)
Example 179
Find homogeneous solution of the differential equation y ′′ + 9y = 0.
SOLUTION The characteristic equation is
k2 + 9 = 0
236
236
(5.17)
IX. Non-homogeneous Equation with constant coefficients
which has the complex root
k 1 = 3i , k 2 = −3i ⇒ a = 0, b = 3
Thus, y 1 = e ax cos bx = cos 3x and y 2 = e ax sin bx = sin 3x are 2 linearly independent solutions and thus the homogeneous solution is
y h = C 1 y 1 +C 2 y 2 = C 1 cos 3x +C 2 sin 3x.
Summary
Evaluate ∆ = B 2 − 4AC
1. If ∆ > 0 then the characteristic equation has real and different roots k 1 ̸= k 2 .
The homogeneous solution is y h = C 1 e k1 x +C 2 e k2 x
2. If ∆ = 0 then the characteristic equation has double real root k = k 0 . The
homogeneous solution is y h = C 1 e k0 x +C 2 x.e k0 x
3. If ∆ < 0 then the characteristic equation has complex conjugate roots k 1 =
a+bi , k 2 = a−bi . The homogeneous solution is y h = e ax (C 1 cos bx +C 2 sin bx).
IX Non-homogeneous Equation with constant coefficients
1
Method of Undetermined Coefficients
Case 1: f (x) = e αx .P n (x)
The particular solution of non-homogeneous equation has a form y p = x s .e αx .Q n (x), where
Q n (x) is polynomial of the same degree n as P n (x).
1. If α is not the root of the characteristic equation then s = 0 and particular solution
has a form y p = e αx .Q n (x)
2. If α is one of 2 different roots of characteristic equation then s = 1 and the particular
solution has a form y p = x.e αx .Q n (x)
3. If α is the double root of the characteristic equation then s = 2 and the particular
solution has a form y p = x 2 .e αx .Q n (x)
h
i
Case 2: f (x) = e αx . P n (x). cos βx +Q m (x). sin βx
The particular solution of non-homogeneous equation has a form y p = x s .e αx .[Hk (x). cos βx+
Tk (x). sin βx], where Hk (x), Tk (x) are polynomials of degree k = max{m, n}.
1. If α + i β is not the root of the characteristic equation then s = 0 and the particular
solution has a form y p = e αx .[Hk (x). cos βx + Tk (x). sin βx]
2. If α + i β is one of complex conjugate roots of the characteristic equation then s = 1
and the particular solution has a form y p = x.e αx .[Hk (x). cos βx + Tk (x). sin βx]
237
237
Chapter 5. Ordinary Differential Equations
2
Principle of Superposition
Theorem IX.1 (Principle of Superposition)
Suppose that f (x) is the sum of 2 terms, f (x) = f 1 (x) + f 2 (x)
1. If y p 1 is a particular solution of the equation
Ay ′′ + B y ′ +C y = f 1 (x)
2. and y p 2 is a particular solution of the equation
Ay ′′ + B y ′ +C y = f 2 (x).
then the particular solution of the equation Ay ′′ + B y ′ + C y = f (x) is y p =
y p1 + y p2 .
Example 180
Solve the differential equation y ′′ − 2y ′ − 3y = e 4x and find the solution that satisfies
the conditions y(ln 2) = 1, y(2 ln 2) = 1.
SOLUTION
Step 1. Solve the homogeneous equation
y ′′ − 2y ′ − 3y = 0
The characteristic equation k 2 − 2k − 3 = 0 has 2 real different roots k 1 = −1, k 2 = 3
Step 2. The homogeneous solution is
y h = C 1 e −x +C 2 e 3x
Step 3. Find a particular solution of non-homogeneous equation
y ′′ − 2y ′ − 3y = e 4x .
The particular solution has a form y p = x s .e 4x .A. Since α = 4 is not the root of characteristic
equation then s = 0 and y p = A.e 4x .
−3
−2
1
′′
y p − 2y p′ − 3y p
yp
=
Ae 4x
y p′
=
4Ae 4x
y p′′
=
16Ae 4x
4x
= 5Ae =
e 4x
⇒A=
1
5
1
Step 4. The general solution is y g en = y h + y p = C 1 e −x +C 2 e 3x + e 4x .
5
Since the general solution satisfies conditions y(ln 2) = 1, y(2 ln 2) = 1 then

1

 C 1 e − ln 2 +C 2 e 3 ln 2 + e 4 ln 2 = 1
5
1

 C 1 e −2 ln 2 +C 2 e 6 ln 2 + e 8 ln 2 = 1
5
238
238
IX. Non-homogeneous Equation with constant coefficients
1
16
C 1 + 8C 2 +
=1
652
491
2
5
⇒
⇒ C1 =
,C 2 = −
.
1
256

75
600
 C 1 + 64C 2 +
=1
4
5
652 −x 491 3x 1 4x
So the solution is y =
e −
e + e .
75
600
5



Example 181
Solve the differential equation y ′′ − 2y ′ + 2y = x 2 .
Step 1. Solve the homogeneous equation
y ′′ − 2y ′ + 2y = 0
The characteristic equation k 2 −2k +2 = 0 has 2 complex conjugate roots k 1 = 1+i , k 2 = 1−i
Step 2. The homogeneous solution is
y h = e x (C 1 cos x +C 2 sin x)
Step 3. Find a particular solution of non-homogeneous equation
y ′′ − 2y ′ + 2y = x 2
The particular solution has a form y p = x s .e 0x .(Ax 2 + B x + C ). Since α = 0 is not the root of
characteristic equation, so s = 0 and y p = Ax 2 + B x +C .
2
−2
1
′′
y p − 2y p′ + 2y p
yp
= Ax 2 + B x +C
′
yp
= 2Ax + B
′′
yp
= 2A
2
= 2Ax + (2B − 4A)x + 2C − 2B + 2A
= x2
1
1
⇒ A = , B = 1,C =
2
2
Step 4. The general solution is
1
y g en = e x (C 1 cos x +C 2 sin x) + (x + 1)2 .
2
Example 182
Find the solution of the differential equation y ′′ + y ′ − 2y = cos x − 3 sin x that satisfies the initial conditions y(0) = 1, y ′ (0) = 2.
SOLUTION
Step 1. Solve the homogeneous equation
y ′′ + y ′ − 2y = 0
The characteristic equation k 2 + k − 2 = 0 has 2 real different roots k 1 = −2, k 2 = 1
Step 2. The homogeneous solution is
y h = C 1 e −2x +C 2 e x
239
239
Chapter 5. Ordinary Differential Equations
Step 3. Find a particular solution of non-homogeneous equation y ′′ +y ′ −2y = cos x −3 sin x
The particular solution has a form y p = x s .e 0x .(A cos x + B sin x). Since 0 + i is not the root
of the characteristic equation, then s = 0 and y p = A cos x + B sin x
yp
= A cos x + B sin x
y p′
= −A sin x + B cos x
y p′′
= −A cos x − B sin x
= (B − 3A) cos x +(−3B − A) sin x =
= cos x − 3 sin x
−2
1
1
′′
y p + y p′ − 2y p
½
⇒
B − 3A = 1
⇒ A = 0, B = 1.
3B + A = 3
Step 4. The general solution is
y g en = C 1 e −2x +C 2 e x + sin x.
Since the general solution satisfies the conditions y(0) = 1, y ′ (0) = 2 then
½
C 1 e 0 +C 2 e 0 + sin 0 = 1
−2C 1 e 0 +C 2 e 0 + cos 0 = 2
½
⇒
C 1 +C 2 = 1
⇒ C 1 = 0,C 2 = 1.
−2C 1 +C 2 = 1
Therefore, the solution is y = e x + sin x.
Example 183
Solve the differential equation y ′′ + y = xe x + 2e −x .
SOLUTION
Step 1. Solve the homogeneous equation
y ′′ + y = 0
The characteristic equation k 2 + 1 = 0 has 2 conjugate roots k 1 = −i , k 2 = i
Step 2. The homogeneous solution is
y h = C 1 cos x +C 2 sin x
f (x) = f 1 (x) + f 2 (x) = xe x + 2e −x
Step 3a. Find a particular solution y p 1 of equation
y ′′ + y = xe x
The particular has a form y p 1 = x s .e x .(Ax + B ). Since α = 1 is not the root of the characteristic equation, then s = 0 and y p 1 = e x (Ax + B )
1
0
1
′′
y p1 + y p1
y p1
y p′ 1
y p′′1
= 2Axe x + (2A + 2B )e x
240
240
= e x (Ax + B )
= Ae x + (Ax + B )e x
= 2Ae x + (Ax + B )e x
= xe x
IX. Non-homogeneous Equation with constant coefficients
½
⇒
1
1
2A = 1
⇒ A = ,B = − ·
2A + 2B = 0
2
2
Step 3b. Find a particular solution y p 2 of equation
y ′′ + y = 2e −x
The particular has a form y p 2 = x s .e −x .C . Since α = −1 is not the root of the characteristic
equation, then s = 0 and y p 2 = C e −x
1
0
1
′′
y p2 + y p2
y p2
y p′ 2
y p′′2
= 2C e −x
= C e −x
= −C e −x
= C e −x
= 2e −x
⇒ C = 1.
Step 4. The general solution is
1
y g en = y h + y p 1 + y p 2 = C 1 cos x +C 2 sin x + (x − 1)e x + e −x .
2
3
Mechanical Vibration
Example 184
Suppose the spring has unstretched length L. The mass m is attached to the spring
and the spring stretches to its equilibrium position, stretching the spring an amount
ℓ. The stretched length is L + ℓ. Find the differential equation for the motion of the
mass on the spring.
Theorem IX.2 (Hooke’s law)
It is experimentally observed that the magnitude of the force needed to produce a
certain elongation of a spring is directly proportional to the amount of the elongation,
provided the elongation is not too great. That is,
|F | = kx
(5.18)
where |F | is the magnitude of the force F, x is the amount of elongation, and k is
a constant of proportionality, called the spring constant, which depends upon the
characteristics of the spring.
241
241
Chapter 5. Ordinary Differential Equations
We assume the origin of a coordinate system at the equilibrium position.
We also assume the positive direction as down.
When a mass is hung upon a spring that has a spring constant k and produces
elongation x, the force F of the mass upon the spring has magnitude kx. At the same
time, the spring exerts a force upon the mass called the restoring force. This restoring
force is equal in magnitude, but opposite in sign to F and hence is −kx.
The value x is positive, zero, or negative, depending upon whether the mass is below,
at, or above equilibrium, respectively.
Forces acting upon the mass
1. F 1 , the force of gravity: F 1 = mg is positive since it acts downward.
2. F 2 , the restoring force of the spring. Since x + ℓ is the total elongation of the spring,
then F 2 = −k(x + ℓ). At the equilibrium point, the force of gravity is equal to the
restoring force, so that mg = k(0 + ℓ) ⇒ mg = kℓ ⇒ F 2 = −kx − mg .
3. F 3 , the resisting force of the medium (damping force) F 3 = −b.x ′ (t ), where b > 0 is
the damping constant.
4. F 4 , any external forces that act upon the mass F 4 = F (t )
We can apply Newton’s second law
F = ma = F 1 + F 2 + F 3 + F 4
⇔ mx ′′ (t ) = mg − kx − mg − bx ′ (t ) + F (t )
⇔ mx ′′ (t ) + bx ′ (t ) + kx = F (t )
This is a nonhomogeneous second-order linear differential equation with constant coefficients for the motion of the mass on a spring.
Definition 92
1. If b = 0, the motion is called undamped; otherwise, it is called damped.
2. If F (t ) = 0 for all t , the motion is called free; otherwise, it is called forced.
Example 185
A spring with a mass of 2 kg has natural length 0.5m. A force of 25.6N is required to
maintain it stretched to a length of 0.7m. If the spring is stretched to a length of 0.7m
and then released with initial velocity 0, find the position of the mass at any time t .
SOLUTION. From Hooke’s Law, the force required to stretch the spring is
F = k × (0.7 − 0.5) = 25.6 ⇒ k = 128.
Using this value of the spring constant k, together with m = 2, we have
2x ′′ (t ) + 128x = 0.
242
242
IX. Non-homogeneous Equation with constant coefficients
The solution of this equation is
x(t ) = C 1 cos 8t +C 2 sin 8t .
We are given the initial condition that x(0) = 0.2 and x ′ (0) = 0. Therefore,
x ′ (0) = 8C 2 ⇒ C 1 = 0.2,
x(0) = C 1 ;
C 2 = 0.
So the solution is
x(t ) = 0.2 cos 8t .
Example 186
Suppose that the spring of Example 3 is immersed in a fluid with damping constant
b = 40. Find the position of the mass at any time t if it starts from the equilibrium
position and is given a push to start it with an initial velocity of 0.6m/s.
SOLUTION. The mass is m = 2 and the spring constant is k = 128, so the differential
equation becomes
2x ′′ (t ) + 40x ′ (t ) + 128x = 0.
The characteristic equation is
2k 2 + 40k + 128 = 0 ⇔ k = −4 ∨ k = −16.
The solution is
x(t ) = C 1 e −4t +C 2 e −16t
We are given that x(0) = 0 and x ′ (0) = 0.6. Therefore,
½
½
x(0) = C 1 +C 2 = 0
C 1 = 0.05
⇔
′
x (0) = −4C 1 − 16C 2 = 0.6
C 2 = −0.05
So
³
´
x(t ) = 0.05 e −4t − e −16t .
Example 187
A 1−kg is attached to a spring with stiffness k = 1.25N /m. At time t = 0, an external
force F (t ) = 3 cos t (N ) is applied to the system. The damping constant for the system
is 1 N-sec/m. If the spring is stretched to a length of 2m from the natural length and
then released with initial velocity 3, find the position of the mass at any time t .
SOLUTION. Substituting the given parameters into the differential equation, we obtain
x ′′ (t ) + x ′ (t ) + 1.25x = 3 cos t .
and the initial conditions are x(0) = 2, x ′ (0) = 3.
The characteristic equation is
k 2 + k + 1.25 = 0 ⇔ k = −0.5 ± i .
Thus the homogeneous solution is
x h (t ) = e −0.5t (C 1 cos t +C 2 sin t ).
243
243
Chapter 5. Ordinary Differential Equations
The particular solution has the form x p (t ) = A cos t +B sin t ⇒ x p′ (t ) = −A sin t +B cos t and
x p′′ (t ) = −A cos t − B sin t . Thus, we obtain
(0.25A + B ) cos t + (−A + 0.25B ) sin t = 3 cos t ⇒ A =
12
48
,B = ·
17
17
The general solution is
x(t ) = x h (t ) + x p (t ) = e −0.5t (C 1 cos t +C 2 sin t ) +
12
48
cos t +
sin t .
17
17
The constants C 1 and C 2 are determined by the initial conditions

12
22

 C1 =
=2
17
17
⇔
48
14


′
 x (0) = −0.5C 1 +C 2 +
 C2 =
=3
17
17



x(0) = C 1 +
Thus we finally obtain the solution of the given initial value problem, namely,
x(t ) = e
−0.5t
µ
¶
22
14
12
48
cos t +
sin t +
cos t +
sin t .
17
17
17
17
X Homogeneous linear systems with constant coefficients
1
Elimination method
We now consider the homogeneous linear system
½
x ′ (t ) = ax + b y
y ′ (t ) = cx + d y
(A)
(B )
The first equation (A) from (5.19) gives y in terms of x and x ′
x ′ − ax
y=
,
b
(b ̸= 0)
Differentiating this will give us y ′ in terms of x ′ and x ′′
y′ =
x ′′ − ax ′
b
and we can now substitute these into the second equation (B) from (5.19) to give
x ′′ − ax ′
x ′ − ax
= cx + d ·
b
b
Rearranging this gives a second order equation for x
x ′′ − (a + d )x ′ + (ad − bc)x = 0
244
244
(5.19)
X. Homogeneous linear systems with constant coefficients
We shoud be able to solve this to find x(t ), using the techniques for second order linear
homogeneous equations. Once we know x(t ) we can then use the formula
x ′ − ax
b
y=
to work out y(t ).
Example 188
Find the homogeneous solution of the system
½
x ′ (t ) = 7x + 3y
y ′ (t ) = 6x + 4y
SOLUTION Rearranging the first equation we can find y in terms of x
x ′ − 7x
3
y=
and so, differentiating this,
x ′′ − 7x ′
3
Substituting these into the second equation gives
y′ =
x ′′ − 7x ′
x ′ − 7x
= 6x + 4 ·
⇔ x ′′ − 11x ′ + 10x = 0
3
3
The characteristic equation is
½
2
k − 11k + 10 = 0 ⇔
k1 = 1
k 2 = 10
Since x 1 = e k1 t = e t and x 2 = e k2 t = e 10t are linearly independent solutions and thus the
homogeneous solution is
x h = C 1 x 1 +C 2 x 2 = C 1 e t +C 2 e 10t ⇒ x h′ = C 1 e t + 10C 2 e 10t .
⇒ yh =
x h′ − 7x h
3
= −2C 1 e t +C 2 e 10t .
Example 189
Find the homogeneous solution of the system
½
x ′ (t ) = 4x − 3y
y ′ (t ) = 3x + 4y
SOLUTION Rearranging the first equation we can find y in terms of x
y=
x ′ − 4x
−3
y′ =
x ′′ − 4x ′
−3
and so, differentiating this,
245
245
Chapter 5. Ordinary Differential Equations
Substituting these into the second equation gives
x ′ − 4x
x ′′ − 4x ′
= 3x + 4 ·
⇔ x ′′ − 8x ′ + 25x = 0
−3
−3
The characteristic equation is
k 2 − 8k + 25 = 0 ⇔
½
k 1 = 4 + 3i
k 2 = 4 − 3i
We are interested in real linearly independent solutions. So we use Euler’s formula
e k1 t = e (4+3i )t = e 4t .e 3i t = e 4t (cos 3t + i sin 3t )
e k2 t = e (4−3i )t = e 4t .e −3i t = e 4t (cos 3t − i sin 3t )
Then
x1 =
e (4+3i )t + e (4−3i )t
= e 4t cos 3t ,
2
x2 =
e (4+3i )t − e (4−3i )t
= e 4t sin 3t
2i
and
are also 2 solutions of the equation x ′′ − 8x ′ + 25x = 0.
Since x 1 = e 4t cos 3t and x 2 = e 4t sin 3t are linearly independent solutions and thus the
homogeneous solution is
x h = C 1 x 1 +C 2 x 2 = C 1 e 4t cos 3t +C 2 e 4t sin 3t
⇒ x h′ = C 1 .4e 4t cos 3t +C 1 e 4t .3.(− sin 3t )+
+C 2 .4e 4t . sin 3t +C 2 e 4t .3 cos 3t
⇒ yh =
x h′ − 4x h
−3
= C 1 e 4t sin 3t −C 2 e 4t cos 3t
Therefore,
½
x h (t ) = e 4t (C 1 cos 3t +C 2 sin 3t )
y h (t ) = e 4t (C 1 sin 3t −C 2 cos 3t )
Example 190
Find the homogeneous solution of the system
½
x ′ (t ) = 5x − y
y ′ (t ) = x + 3y
SOLUTION Rearranging the first equation we can find y in terms of x
y = 5x − x ′
and so, differentiating this,
y ′ = 5x ′ − x ′′
246
246
XI. Non-homogeneous linear systems with constant coefficients
Substituting these into the second equation gives
5x ′ − x ′′ = x + 3(5x − x ′ ) ⇔ x ′′ − 8x ′ + 16x = 0
The characteristic equation is
k 2 − 8k + 16 = 0 ⇔ k 1 = k 2 = k 0 = 4.
So x 1 = e 4t , x 2 = t e 4t are linearly independent. Thus, the homogeneous solution is
x h = C 1 x 1 +C 2 x 2 = C 1 e 4t +C 2 t e 4t .
x h′ = C 1 .4.e 4t +C 2 e 4t +C 2 .t .4e 4t .
⇒ y h = 5x h − x h′ = C 1 e 4t +C 2 t .e 4t −C 2 e 4t
Therefore,
½
x h (t ) = C 1 e 4t +C 2 t e 4t
y h (t ) = C 1 e 4t +C 2 (t − 1).e 4t
XI
Non-homogeneous linear systems with constant coefficients
1
Elimination method
Consider the non-homogeneous linear system
½
x ′ (t ) = ax + b y + f 1 (t )
y ′ (t ) = c x + d y + f 2 (t )
(C )
(D)
(5.20)
The first equation (C) from (5.20) gives y in terms of x and x ′
y=
x ′ − ax − f 1 (t )
,
b
(b ̸= 0)
Differentiating this will give us y ′ in terms of x ′ and x ′′
y′ =
x ′′ − ax ′ − f 1′ (t )
b
and we can now substitute these into the second equation (D) from (5.20) to give
x ′′ − ax ′ − f 1′ (t )
x ′ − ax − f 1 (t )
= cx + d ·
+ f 2 (t ).
b
b
Rearranging this gives a second order equation for x
x ′′ − (a + d )x ′ + (ad − bc)x = f 1′ (t ) + b. f 2 (t ) − d . f 1 (t ).
We shoud be able to solve this to find x(t ), using the techniques for second order linear
nonhomogeneous equations. Once we know x(t ) we can then use the formula
y=
x ′ − ax − f 1 (t )
b
247
247
Chapter 5. Ordinary Differential Equations
to work out y(t ).
Example 191
Find the general solution of the system
½
x ′ (t ) = 4x − 3y + e −t
y ′ (t ) =
2x − y
(1)
(2)
SOLUTION 1 The first equation (1) gives y in terms of x and x ′
y=
x ′ − 4x − e −t
−3
Differentiating this will give us y ′ in terms of x ′ and x ′′
y′ = −
x ′′ − 4x ′ + e −t
3
and we can now substitute these into the equation (2) to give
x ′′ − 4x ′ + e −t
x ′ − 4x − e −t
= 2x −
·
3
−3
Rearranging this gives a second order equation for x
−
x ′′ − 3x ′ + 2x = 0.
The characteristic equation is
k 2 − 3k + 2 = 0
which has the roots k 1 = 1 and k 2 = 2.
Thus, x 1 = e k1 t = e t and x 2 = e k2 t = e 2t are linearly independent solutions and thus the
homogeneous solution is
x = C 1 x 1 +C 2 x 2 = C 1 e t +C 2 e 2t .
⇒ x ′ (t ) = C 1 e t + 2C 2 e 2t
x ′ − 4x − e −t
C 1 e t + 2C 2 e 2t − 4(C 1 e t +C 2 e 2t ) − e −t
=−
−3
3
2
1
⇒ y = C 1 e t + · C 2 e 2t + e −t .
3
3
Therefore, the solution of the nonhomogeneous linear system is

 x = C 1 e t +C 2 e 2t .
2
1
 y = C 1 e t + · C 2 e 2t + e −t .
3
3
⇒y=
SOLUTION 2 The second equation (2) gives x in terms of y and y ′
x=
y′ + y
2
248
248
XI. Non-homogeneous linear systems with constant coefficients
Differentiating this will give us x ′ in terms of y ′ and y ′′
x′ =
y ′′ + y ′
2
and we can now substitute these into the equation (1) to give
y ′′ + y ′
y′ + y
= 4·
− 3y + e −t
2
2
Rearranging this gives a second order equation for y
y ′′ − 3y ′ + 2y = 2e −t .
The characteristic equation of homogeneous equation is
k 2 − 3k + 2 = 0
which has the roots k 1 = 1 and k 2 = 2.
Thus, y 1 = e k1 t = e t and y 2 = e k2 t = e 2t are linearly independent solutions and thus the
homogeneous solution is
y h = C 1 y 1 +C 2 y 2 = C 1 e t +C 2 e 2t .
Find a particular solution of non-homogeneous equation
y ′′ − 3y ′ + 2y = 2e −t .
The particular solution has a form y p = t s .e −t .A. Since α = −1 is not the root of characteristic equation then s = 0 and y p = A.e −t .
2
−3
1
′′
y p − 3y p′ + 2y p
yp
=
Ae −t
y p′
=
−Ae −t
y p′′
=
Ae −t
= 6Ae −t = 2e −t
⇒A=
1
3
1
The general solution is y = y h + y p = C 1 e t +C 2 e 2t + e −t .
3
1
⇒ y ′ (t ) = C 1 e t + 2C 2 e 2t − e −t
3
1 −t
1 −t
t
2t
t
2t
C
e
+
2C
e
−
e
+C
e
+C
e
+
e
1
2
1
2
y +y
3
3
⇒x=
=
2
2
3
⇒ x = C 1 e t + · C 2 e 2t .
2
Therefore, the solution of the nonhomogeneous linear system is

3

 x = C 1 e t + · C 2 e 2t .
2
1

 y = C 1 e t +C 2 e 2t + e −t .
3
′
249
249
Chapter 5. Ordinary Differential Equations
2
Interconnected fluid tanks
Example 192
Two large tanks, each holding 24 liters of a brine solution, are interconnected by
pipes. Fresh water flows into tank A at a rate of 6L/mi n, and fluid is drained out of
tank B at the same rate; also 8L/mi n of fluid are pumped from tank A to tank B, and
2L/mi n from tank B to tank A. The liquids inside each tank are kept well stirred so
that each mixture is homogeneous. If, initially, the brine solution in tank A contains
x 0 (kg ) of salt and that in tank B initially contains y 0 (kg ) of salt, determine the mass
of salt in each tank at time t > 0. Assume that the lengths and volumes of the pipes
are sufficiently small that we can ignore the diffusive and advective dynamics taking
place therein.
SOLUTION Note that the volume of liquid in each tank remains constant at 24L because of
the balance between the inflow and outflow volume rates.
1. Let x(t ) be the mass of salt in tank A and y(t ) be the mass of salt in tank B.
x
2. The salt concentration in tank A is
(kg /L), the concentration of salt in tank B
24
y
is
(kg /L), so the upper interconnecting pipe carries salt out of tank A at a rate
24
8x
of
(kg /mi n); similarly, the lower interconnecting pipe brings salt into tank A at
24
2y
the rate
(kg /mi n). The fresh water inlet, of course, transfers no salt (it simply
24
maintains the volume tank A at 24L.
3. From our premise,
of salt in tank A is
dx
= input rate − output rate, so the rate of change of the mass
dt
d x 2y 8x
y
x
=
−
=
−
dt
24 24 12 3
The rate of change of salt in tank B is determined by the same interconnecting pipes
6y
and by the drain pipe, carrying away
(kg /mi n) :
24
d y 8x 2y 6y x y
=
−
−
= −
dt
24 24 24 3 3
4. The interconnected tanks are thus governed by a system of differential equations.
250
250
XII. Exercises
XII
1
Exercises
Essay Questions
The first order differential equations
75 A tank initially contains 80 grams of salt dissolved in 100 liters of water. Pure
water flows into the tank at the rate of 12 liters per minute, and the well stirred mixture
flows out of the tank at the same rate. How much salt does the tank contain after 2
minutes?
SOLUTION
dQ
is equal to the rate at which salt is
dt
flowing in minus the rate at which it is flowing out.
1. Step 1. The rate of change of salt in the tank,
dQ
= rate in − rate out
dt
2. Step 2. The rate at which salt enters the tank is the concentration 0 g r /l times the
flow rate 12 l /mi n or 0 g r /mi n.
3. Step 3. To find the rate at which salt leaves the tank, we need to multiply the concentration of salt in the tank by the rate of outflow, 12 l /mi n. The volume of water
in the tank remains constant at 100 liters, and since the mixture is well-stirred, the
Q
g r /l . Therefore, the rate
concentration throughout the tank is the same, namely,
100
12Q
at which salt leaves the tank is
g r /mi n.
100
4. Step 4. Solve the differential equation using separable method
Z
Z
dQ
12Q
dQ
12
dQ
12
= 0−
⇒
=−
dt ⇒
= −
dt
dt
100
Q
100
Q
100
⇒ ln |Q| = −
12
t + lnC ⇒ Q = C e −12t /100
100
5. Step 5. The initial condition is Q(0) = 80, so 80 = C e 0 = C . Thus Q = 80e −12t /100
⇒ Q(2) = 80e −12×2/100 = 62.9302.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
76 A tank with a capacity of 130 liters initially contains 80 grams of salt dissolved
in 100 liters of water. Pure water flows into the tank at the rate of 12 liters per minute,
and the well stirred mixture flows out of the tank at the rate of 2 liters per minute. How
much salt does the tank contain when the tank is full?
SOLUTION
251
251
Chapter 5. Ordinary Differential Equations
dQ
is equal to the rate at which salt is
dt
flowing in minus the rate at which it is flowing out.
1. Step 1. The rate of change of salt in the tank,
dQ
= rate in − rate out
dt
2. Step 2. The rate at which salt enters the tank is the concentration 0 g r /l times the
flow rate 12 l /mi n or 0 g r /mi n.
3. Step 3. To find the rate at which salt leaves the tank, we need to multiply the concentration of salt in the tank by the rate of outflow, 2 l /mi n. The volume of water in the
tank after t minutes is 100 + (12 − 2)t = 100 + 10t liters, and since the mixture is wellQ
stirred, the concentration throughout the tank is the same, namely,
g r /l .
100 + 10t
2Q
g r /mi n.
Therefore, the rate at which salt leaves the tank is
100 + 10t
4. Step 4. Solve the differential equation using separable method
Z
Z
2Q
dQ
2
2
dQ
dQ
= 0−
⇒
=−
dt ⇒
= −
dt
dt
100 + 10t
Q
100 + 10t
Q
100 + 10t
⇒ ln |Q| = −
2
2
ln |100 + 10t | + lnC ⇒ Q = C (100 + 10t )− 10
10
2
2
5. Step 5. The initial condition is Q(0) = 80, so 80 = C × 100− 10 ⇒ C = 80 × 100 10 . Thus
³
´
2
2
Q = 80 × 100 10 (100 + 10t )− 10 .
130 − 100
= 3 minutes.
12 − 2
³
´
2
2
⇒ Q(3) = 80 × 100 10 (100 + 10 × 3)− 10 = 75.9104.
The tank is full after
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
77 Recall that a simple electric circuit with a resistor of R ohms, an inductor of L
henries and a battery with a voltage of E volts will produce a current I (t ) amperes (t is
measured in seconds) satisfying the following differential equation LI ′ (t ) + R I = E . Let
study one such circuit with R = 2 ohms, L = 3 henries and E = 12 volts. Given initial
condition I (0) = 3 amperes. Evaluate current’s value at t = 3 seconds.
SOLUTION
1. Step 1. Let consider the differential equation LI ′ (t ) + R I = E in our case, we have
2
12
3I ′ (t ) + 2I = 12 ⇔ I ′ (t ) + I =
3
3
2. Step 2. Solving this linear differential equation with p(t ) =
obtain
Z
I (t ) = e
−
2
12
and q(t ) =
, we
3
3
Z
p(t )d t hZ
252
252
q(t )e
p(t )d t
i
d t +C .
XII. Exercises
Therefore,
Z
2
2
Z
h
i
−
dt
dt
12
3
I (t ) = e
e 3 d t +C
3
Z
h
i
2
12 2 t
e 3 d t +C
I (t ) = e − 3 t
3
µ
¶
− 23 t 12 32 t
I (t ) = e
e +C
2
Z
µ
¶
12
12
− 32 t 12 23 t
3. Step 3. The initial condition is I (0) = +C = 3, so C = 3− . Thus I (t ) = e
e +C
2
2
2
µ
¶
− 32 ×3 12 32 ×3
e
⇒ I (3) = e
+C = 5.594.
2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
The second order differential equations
78
d
d2
Solve the differential equation −108y(x) − 3
y(x) +
y(x) = 0.
dx
d x2
SOLUTION
1. Step 1. Solve the homogeneous equation
d
d2
−108y(x) − 3
y(x) +
y(x) = 0.
dx
d x2
The characteristic equation k 2 − 3k − 108 = 0 has 2 real different roots
k 1 = 12, k 2 = −9
2. Step 2. The homogeneous solution is
y h = C 1 e 12x +C 2 e −9x = C 1 e −9x +C 2 e 12x .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
79
Solve the differential equation 4y(x) − 4
d
d2
y(x) +
y(x) = 0.
dx
d x2
SOLUTION
1. Step 1. Solve the homogeneous equation
4y(x) − 4
d
d2
y(x) +
y(x) = 0.
dx
d x2
The characteristic equation k 2 − 4k + 4 = 0 has double real root
k1 = k2 = 2
253
253
Chapter 5. Ordinary Differential Equations
2. Step 2. The homogeneous solution is
y h = C 1 e 2x +C 2 xe 2x = (C 1 +C 2 x) e 2x .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
80
d
d2
Solve the differential equation 269y(x) − 26
y(x) +
y(x) = 0.
dx
d x2
SOLUTION
1. Step 1. Solve the homogeneous equation
269y(x) − 26
d
d2
y(x) +
y(x) = 0.
dx
d x2
The characteristic equation k 2 − 26k + 269 = 0 has 2 complex conjugate roots
k 1 = 10i + 13; k 2 = 13 − 10i
2. Step 2. The homogeneous solution is
h
i
y h = C 1 sin (10x) +C 2 cos (10x) e 13x .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
Applications of the second order differential equations
81 Recall that a simple electric circuit with a resistor of R ohms, an inductor of L
henries, a capacitor of C farads and a battery with a voltage E (t ) = 100 sin(ωt ) (volts)
will produce a current I (t ) amperes (t is measured in seconds) satisfying the following
1
differential equation LI ′′ (t )+R I ′ + I = E ′ (t ) = 100ω cos(ωt ). Let study one such circuit
C
with R = 8 ohms, L = 5 henries, C = 0.2 farads and ω = 11 rad/seconds. Given initial
condition I (0) = 0, I ′ (0) = 0, evaluate current’s value at t = 6.2 seconds.
SOLUTION
1. Step 1. Solve the homogeneous equation
LI ′′ (t ) + R I ′ +
The characteristic equation Lk 2 + Rk +
conjugate roots
1
I = 0.
C
1
= 0 ⇒ 5k 2 + 8k + 5 = 0 has 2 complex
C
4 3
4 3
k1 = − + i ; k2 = − − i
5 5
5 5
2. Step 2. The homogeneous solution is
·
µ ¶
µ ¶¸
3
3
− 45 t
Ih = e
C 1 cos t +C 2 sin t
5
5
254
254
XII. Exercises
3. Step 3. Find a particular solution of nonhomogeneous equation
LI ′′ (t ) + R I ′ +
1
I = 100ω cos(ωt ).
C
h
i
The particular solution has a form I p = t s .e 0t A cos(ωt ) + B sin(ωt ) . Since 0 + ωi is
not the root of characteristic equation then s = 0 and I p = A cos(ωt ) + B sin(ωt ).
5×
8×
5×
5I "p + 8I p′ + 5I p
½
⇒
Ip
I p′
I "p
=
=
A cos(ωt ) + B sin(ωt )
= −Aω sin(ωt ) + B ω cos(ωt )
= −Aω2 cos(ωt ) − B ω2 sin(ωt )
100ω cos(ωt )

ω(5 − 5ω2 ) × 100


 A=
2
(5 − 5ω )A + 8ωB = 100ω
(5 − 5ω2 )2 + (8ω)2
⇔
2
8ω2 × 100

−8ωA + (5 − 5ω )B = 0

 B=
(5 − 5ω2 )2 + (8ω)2
4. Step 4. The general solution is
µ ¶¸
·
µ ¶
3
3
− 54 t
I g en = I h + I p = e
C 1 cos t +C 2 sin t + A cos(ωt ) + B sin(ωt ).
5
5
·µ
¶
µ ¶ µ
¶
µ ¶¸
4
3
3
4
3
3
′
− 45 t
We have I g en = e
− C 1 + C 2 cos t + − C 2 − C 1 sin t −Aω sin(ωt )+
5
5
5
5
5
5
B ω cos(ωt ) From the initial condition I (0) = 0, I ′ (0) = 0, we obtain


C 1 = −A
5
4A
 C2 = − B ω −
3
3
5. Step 5. Evaluate I g en (6.2) = e
6.2) + B sin(ω × 6.2) ≈ −1.3042.
− 45 ×6.2
¶
µ
¶¸
·
µ
3
3
C 1 cos × 6.2 +C 2 sin × 6.2 + A cos(ω ×
5
5
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
82 Recall that a spring with mass m, spring constant k relaxing after being stretched
will be in motion, where its position x(t ) satisfies the following differential equation
mx ′′ (t ) + kx = 0. Let study one spring with m = 2 and k = 32. Given initial condition
x(0) = 6, x ′ (0) = 4, evaluate numerically its position at t = 2.
SOLUTION
1. Step 1. Solve the homogeneous equation
mx ′′ (t ) + kx = 0.
The characteristic equation mr 2 + k = 0 has 2 complex conjugate roots
s
s
r
r
k
32
k
32
r1 =
i=
i ; r2 = −
i =−
i
m
2
m
2
255
255
Chapter 5. Ordinary Differential Equations
2. Step 2. The homogeneous solution is
Ãr
x = C 1 cos
r
3. Step 3. x ′ = −C 1
we have











32
sin
2
Ãr
!
Ãr
!
32
32
t +C 2 sin
t
2
2
!
Ãr
!
r
32
32
32
t + C2
cos
t . From the initial condition,
2
2
2
Ãr
!
Ãr
!
32
32
x(0) = C 1 cos
× 0 +C 2 sin
× 0 = C1 = 6
2
2
Ãr
!
Ãr
!
r
r
r
32
32
32
32
32
′
x (0) = −C 1
sin
× 0 +C 2
cos
× 0 = C2
=4
2
2
2
2
2
⇔
Ãr
4. Step 4. Evaluate x(2) = C 1 cos


C1 = 6
r
2
 C2 = 4 ×
32
!
Ãr
!
32
32
× 2 +C 2 sin
× 2 ≈ 0.1164.
2
2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .□
2
Multiple-choice Questions
The first order differential equations
Question 1 (L.O.2): Which function is a solution of the following differential equation
7
y ′ + y = 3x, where C is any constant?
x
1
3
1
A y = x 2 +C x −7
B y = x 2 +C x −7
C y = x 2 +C x 7
3
8
3
3 2
3 2
7
−7
D y = x +C x
E y=
x +C x
8
10
SOLUTION
This is the first order linear differential equation with p(x) =
Z
y =e
Z
y =e
−
−
7
and q(x) = 3x. Therefore,
x
Z
p(x)d x hZ
7
d x hZ
x
3xe
Z
q(x)e
p(x)d x
d x +C
i
7
i
h x9
i 1
dx
x
d x +C = x −7 3 +C = x 2 +C x −7
9
3
¤ The correct answer is A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
256
256
XII. Exercises
The second order differential equations
Question 2 (L.O.2): Let y(x) be a function satisfying y ′′ − 4y ′ − 32y = 0, y(0) = 6, y ′ (0) =
24. Evaluate y(1.88).
A 13609714.8215
B 13609714.2745
C 13609713.4938
D 13609713.4611
E 13609714.0102
SOLUTION
1. Step 1. Solve the homogeneous equation
y ′′ − 4y ′ − 32y = 0.
The characteristic equation k 2 − 4k − 32 = 0 has 2 real different roots
k 1 = 8, k 2 = −4
2. Step 2. The homogeneous solution is
y h = C 1 e 8x +C 2 e −4x
3. Step 3. Thus,
y h′ = 8C 1 e 8x + (−4)C 2 e −4x .
Since y(0) = 6 ⇒ C 1 +C 2 = 6 and y ′ (0) = 8C 1 +(−4)C 2 = 24 ⇒ C 1 = 4,C 2 = 2. Therefore,
y(1.88) ≈ 13609714.0102
¤ The correct answer is E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 3 (L.O.2): Let y(x) be a function satisfying y ′′ − 5y ′ = 2x 2 + 5, y(0) = 0, y ′ (0) =
0. Evaluate y(2).
A 4543.1199
B 4542.7416
C 4542.6055
D 4542.9902
E 4541.7701
SOLUTION
1. Step 1. Solve the homogeneous equation
y ′′ − 5y ′ = 0.
The characteristic equation k 2 − 5k = 0 has 2 real different roots
k 1 = 5, k 2 = 0
2. Step 2. The homogeneous solution is
y h = C 1 e 5x +C 2 e 0x = C 1 e 5x +C 2
257
257
Chapter 5. Ordinary Differential Equations
3. Step 3. Find a particular solution of nonhomogeneous equation
y ′′ − 5y ′ = 2x 2 + 5.
The particular solution has a form y p = x s .e 0x .(Ax 2 + B x + C ). Since α = 0 is a root of
characteristic equation then s = 1 and y p = x(Ax 2 + B x +C ) = Ax 3 + B x 2 +C x.
0×
−5×
1×
′′
y p − 5y p′
yp
=
Ax 3 + B x 2 +C x
y p′
=
3Ax 2 + 2B x +C
′′
yp
=
6Ax + 2B
2
= −15Ax + (−10B + 6A)x + 2B − 5C =
2x 2 + 5
⇒ A=−
2
129
2
; B = − ;C = −
15
25
125
4. Step 4. The general solution is
¶
µ
¶
µ
¶
µ
2
2
129
3
2
y g en = y h + y p = C 1 e +C 2 + −
x + −
x + −
x
15
25
125
µ
¶
µ
¶
µ ¶
2 2
4
129
′
5x
x+ −
.
⇒ y g en = 5C 1 e + − x + −
5
25
125
5x
Since y(0) = 0 ⇒ C 1 +C 2 = 0 và y ′ (0) = 0 ⇒ C 1 =
y(2) = −
129
129
,C 2 = −C 1 = −
. Therefore,
625
625
6857 129e 10
+
≈ 4542.6055
1875
625
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
Question 4 (L.O.2): Let y(x) = axe 4x be a function satisfying y ′′ − 10y ′ + 24y =
−6.3894e 4x . Find the value of a.
A 3.5403
B 3.6631
C 3.1947
D 3.2759
E 3.0536
SOLUTION
Since y(x) = axe 4x is the solution of y ′′ − 10y ′ + 24y = −6.3894e 4x . Then,
24×
−10×
1×
y ′′ − 10y ′ + 24y
yp
=
axe 4x
′
4x
yp
=
ae (1 + 4x)
′′
=
ae 4x (8 + 16x)
yp
= −2ae 4x = −6.3894e 4x
⇒ a = 3.1947.
¤ The correct answer is C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . □
258
258