**Unformatted text preview: **Math 138
Calculus II
for Honours Mathematics
Course Notes
Barbara A. Forrest and Brian E. Forrest Version 1.4 c Barbara A. Forrest and Brian E. Forrest.
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January 1, 2020
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Author Contact Information:
Barbara Forrest ([email protected])
Brian Forrest ([email protected]) i QUICK REFERENCE PAGE 1
Right Angle Trigonometry sin θ = opposite
hypotenuse cos θ = ad jacent
hypotenuse tan θ = opposite
ad jacent csc θ = 1
sin θ sec θ = 1
cos θ cot θ = 1
tan θ Radians
Definition of Sine and Cosine The angle θ in
radians equals the
length of the directed
arc BP, taken positive
counter-clockwise and
negative clockwise.
Thus, π radians = 180◦
or 1 rad = 180
.
π For any θ, cos θ and sin θ are
defined to be the x− and y−
coordinates of the point P on the
unit circle such that the radius
OP makes an angle of θ radians
with the positive x− axis. Thus
sin θ = AP, and cos θ = OA. The Unit Circle ii QUICK REFERENCE PAGE 2
Trigonometric Identities cos2 θ + sin2 θ = 1 Pythagorean
Identity −1 ≤ cos θ ≤ 1 Range −1 ≤ sin θ ≤ 1
cos(θ ± 2π) = cos θ Periodicity sin(θ ± 2π) = sin θ
cos(−θ) = cos θ Symmetry sin(−θ) = − sin θ Sum and Difference Identities
cos(A + B) = cos A cos B − sin A sin B
cos(A − B) = cos A cos B + sin A sin B
sin(A + B) = sin A cos B + cos A sin B
sin(A − B) = sin A cos B − cos A sin B Complementary Angle Identities
cos( π2 − A) = sin A
sin( π2 − A) = cos A Double-Angle cos 2A = cos2 A − sin2 A Identities sin 2A = 2 sin A cos A Half-Angle cos2 θ = 1+cos 2θ
2 Identities sin2 θ = 1−cos 2θ
2 Other 1 + tan2 A = sec2 A iii QUICK REFERENCE PAGE 3
Table of Integrals Differentiation Rules xn+1
xn dx =
+C
n+1
R 1
dx = ln(| x |) + C
R x
e x dx = e x + C
R
sin(x) dx = − cos(x) + C
R
cos(x) dx = sin(x) + C
R
sec2 (x) dx = tan(x) + C
R
1
dx = arctan(x) + C
1 + x2
R
1
dx = arcsin(x) + C
√
1 − x2
R
−1
dx = arccos(x) + C
√
1 − x2
R
sec(x) tan(x) dx = sec(x) + C
R x
ax
+C
a dx =
ln(a)
R Function Derivative f (x) = cxa , a , 0, c ∈ R f 0 (x) = caxa−1 f (x) = sin(x) f 0 (x) = cos(x) f (x) = cos(x) f 0 (x) = − sin(x) f (x) = tan(x) f 0 (x) = sec2 (x) f (x) = sec(x) f 0 (x) = sec(x) tan(x) f (x) = arcsin(x) 1
f 0 (x) = √
1 − x2
1
f 0 (x) = − √
1 − x2
1
f 0 (x) =
1 + x2 f (x) = arccos(x)
f (x) = arctan(x)
f (x) = e x f 0 (x) = e x f (x) = a x with a > 0 f 0 (x) = a x ln(a)
1
f 0 (x) =
x f (x) = ln(x) for x > 0 Inverse Trigonometric Substitutions Integral
R √
a2 − b2 x2 dx
R √
a2 + b2 x2 dx
R √
b2 x2 − a2 dx Trig Substitution Trig Identity bx = a sin(u) sin2 (x) + cos2 (x) = 1 bx = a tan(u) sec2 (x) − 1 = tan2 (x) bx = a sec(u) sec2 (x) − 1 = tan2 (x) Additional Formulas f (x)g0 (x) dx = f (x)g(x) −
Rb
A = a |g(t) − f (t)| dt
Rb
V = a π f (x)2 dx
Rb
V = a π(g(x)2 − f (x)2 ) dx
Rb
V = a 2πx(g(x) − f (x)) dx
Rb p
S = a 1 + ( f 0 (x))2 dx
R Integration by Parts
Areas Between Curves
Volumes of Revolutions: Disk I
Volumes of Revolutions: Disk II
Volumes of Revolutions: Shell
Arc Length
Taylor Series (Maclaurin Series)
1
1−x = ex = ∞
P xn = 1 + x + x2 + x3 + · · · n=0
xn
n! =
n=0
∞
P
∞
P cos(x) =
sin(x) = 1+ x
1! + x2
2! x2n + x3
3! (−1)n (2n)! = 1 − n=0
∞
P x2n+1
(−1)n (2n+1)!
n=0 + ··· x2
2! = x− R=1 +
x3
3! x4
4! + −
x5
5! x6
6! − + ···
x7
7! + ··· iv R f 0 (x)g(x) dx Differential Equations R=∞ Solve y0 = f (x)g(y)
R 1
R
?
g(y) = 0, g(y)
dy = f (x)dx R=∞ FOLDE y0 = f (x)y + g(x) R=∞ Solve y= Separable R g(x)I(x)dx
,
I(x) I(x) = e− R f (x) dx QUICK REFERENCE PAGE 4
LIST of THEOREMS:
Chapter 1: Integration
Integrability Theorem
for Continuous Functions
Properties of Integrals Theorem
Integrals over Subintervals Theorem
Average Value Theorem
(Mean Value Theorem for Integrals)
Fundamental Theorem of Calculus (Part 1)
Extended Version of the
Fundamental Theorem of Calculus
Power Rule for Antiderivatives
Fundamental Theorem of Calculus (Part 2)
Change of Variables Theorem
Chapter 2: Techniques of Integration
Integration by Parts Theorem
Integration of Partial Fractions
p-Test for Type I Improper Integrals
Properties of Type I Improper Integrals
The Monotone Convergence Theorem
for Functions
Comparison Test
for Type I Improper Integrals
Absolute Convergence Theorem
for Improper Integrals
p-Test for Type II Improper Integrals
Chapter 3: Applications of Integation
Area Between Curves
Volumes of Revolution: Disk Methods
Volumes of Revolution: Shell Method
Arc Length Chapter 5: Numerical Series
Geometric Series Test
Divergence Test
Arithmetic for Series Theorems
The Monotone Convergence Theorem
for Sequences
Comparison Test for Series
Limit Comparison Test
Integral Test for Convergence
p-Series Test
Alternating Series Test (AST) and
the Error in the AST
Absolute Convergence Theorem
Rearrangement Theorem
Ratio Test
Polynomial versus Factorial Growth Theorem
Root Test
Chapter 6: Power Series
Fundamental Convergence Theorem
for Power Series
Test for the Radius of Convergence
Equivalence of Radius of Convergence
Abel’s Theorem: Continuity of Power Series
Addition of Power Series
Multiplication of Power Series by (x − a)m
Power Series of Composite Functions
Term-by-Term Differentiation of Power Series
Uniqueness of Power Series Representations
Term-by-Term Integration of Power Series
Taylor’s Theorem
Taylor’s Approximation Theorem I
Convergence Theorem for Tayor Series
Binomial Theorem
Generalized Binomial Theorem Chapter 4: Differential Equations
Theorem for Solving
First-order Linear Differential Equations
Existence and Uniqueness Theorem
for FOLDE v Table of Contents
Page
1 Integration
1.1 Areas Under Curves . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Estimating Areas . . . . . . . . . . . . . . . . . . . . . .
1.1.2 Approximating Areas Under Curves . . . . . . . . . . . .
1.1.3 The Relationship Between Displacement and Velocity . .
1.2 Riemann Sums and the Definite Integral . . . . . . . . . . . . . .
1.3 Properties of the Definite Integral . . . . . . . . . . . . . . . . . .
1.3.1 Additional Properties of the Integral . . . . . . . . . . . .
1.3.2 Geometric Interpretation of the Integral . . . . . . . . . .
1.4 The Average Value of a Function . . . . . . . . . . . . . . . . . .
1.4.1 An Alternate Approach to the Average Value of a Function
1.5 The Fundamental Theorem of Calculus (Part 1) . . . . . . . . . .
1.6 The Fundamental Theorem of Calculus (Part 2) . . . . . . . . . .
1.6.1 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . .
1.6.2 Evaluating Definite Integrals . . . . . . . . . . . . . . . .
1.7 Change of Variables . . . . . . . . . . . . . . . . . . . . . . . . .
1.7.1 Change of Variables for the Indefinite Integral . . . . . . .
1.7.2 Change of Variables for the Definite Integral . . . . . . . . 1
1
1
2
8
13
18
19
22
27
28
30
40
41
43
47
48
52 2 Techniques of Integration
2.1 Inverse Trigonometric Substitutions . . . . . . . . . . .
2.2 Integration by Parts . . . . . . . . . . . . . . . . . . .
2.3 Partial Fractions . . . . . . . . . . . . . . . . . . . . .
2.4 Introduction to Improper Integrals . . . . . . . . . . . .
2.4.1 Properties of Type I Improper Integrals . . . .
2.4.2 Comparison Test for Type I Improper Integrals
2.4.3 The Gamma Function . . . . . . . . . . . . . .
2.4.4 Type II Improper Integrals . . . . . . . . . . . . .
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. 56
56
62
71
80
86
88
94
96 3 Applications of Integration
3.1 Areas Between Curves . . . . . . . .
3.2 Volumes of Revolution: Disk Method .
3.3 Volumes of Revolution: Shell Method .
3.4 Arc Length . . . . . . . . . . . . . . . .
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. 4 Differential Equations
122
4.1 Introduction to Differential Equations . . . . . . . . . . . . . . . . 122
4.2 Separable Differential Equations . . . . . . . . . . . . . . . . . . 124
4.3 First-Order Linear Differential Equations . . . . . . . . . . . . . . 131 vi 4.4 Initial Value Problems . . . . . . . . . . . . . . . . . . . . .
4.5 Graphical and Numerical Solutions to Differential Equations
4.5.1 Direction Fields . . . . . . . . . . . . . . . . . . . .
4.5.2 Euler’s Method . . . . . . . . . . . . . . . . . . . . .
4.6 Exponential Growth and Decay . . . . . . . . . . . . . . . .
4.7 Newton’s Law of Cooling . . . . . . . . . . . . . . . . . . .
4.8 Logistic Growth . . . . . . . . . . . . . . . . . . . . . . . . .
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140
140
142
145
149
152 5 Numerical Series
5.1 Introduction to Series . . . . . . . . . . . . . . . . . . .
5.2 Geometric Series . . . . . . . . . . . . . . . . . . . . .
5.3 Divergence Test . . . . . . . . . . . . . . . . . . . . . .
5.4 Arithmetic of Series . . . . . . . . . . . . . . . . . . . .
5.5 Positive Series . . . . . . . . . . . . . . . . . . . . . . .
5.5.1 Comparison Test . . . . . . . . . . . . . . . . . .
5.5.2 Limit Comparison Test . . . . . . . . . . . . . .
5.6 Integral Test for Convergence of Series . . . . . . . . .
5.6.1 Integral Test and Estimation of Sums and Errors
5.7 Alternating Series . . . . . . . . . . . . . . . . . . . . .
5.8 Absolute versus Conditional Convergence . . . . . . . .
5.9 Ratio Test . . . . . . . . . . . . . . . . . . . . . . . . . .
5.10 Root Test . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. 162
162
167
169
173
177
180
185
189
199
202
213
218
227 6 Power Series
6.1 Introduction to Power Series . . . . . . . . . . . . . .
6.1.1 Finding the Radius of Convergence . . . . . .
6.2 Functions Represented by Power Series . . . . . . . .
6.2.1 Building Power Series Representations . . . .
6.3 Differentiation of Power Series . . . . . . . . . . . . .
6.4 Integration of Power Series . . . . . . . . . . . . . . .
6.5 Review of Taylor Polynomials . . . . . . . . . . . . . .
6.6 Taylor’s Theorem and Errors in Approximations . . . .
6.7 Introduction to Taylor Series . . . . . . . . . . . . . . .
6.8 Convergence of Taylor Series . . . . . . . . . . . . . .
6.9 Binomial Series . . . . . . . . . . . . . . . . . . . . .
6.10 Additional Examples and Applications of Taylor Series .
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253
257
269
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285
290 vii .
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. Chapter 1
Integration
Many operations in mathematics have an inverse operation: addition and subtraction;
multiplication and division; raising a number to the nth power and finding its nth
root; taking a derivative and finding its antiderivative. In each case, one operation
“undoes” the other. In this chapter, we begin the study of the integral and integration.
Soon you will understand that integration is the inverse operation of differentiation. 1.1 Areas Under Curves The two most important ideas in calculus - differentiation and integration - are both
motivated from geometry. The problem of finding the tangent line led to the definition
of the derivative. The problem of finding area will lead us to the definition of the
definite integral. 1.1.1 Estimating Areas Our objective is to find the area under the curve of some function.
What do we mean by the area under a curve?
The question about how to calculate areas is actually thousands of years old and it is
one with a very rich history. To motivate this topic, let’s first consider what we know
about finding the area of some familiar shapes. We can easily determine the area of
a rectangle or a right-angled triangle, but how could we explain to someone why the
area of a circle with radius r is πr2 ?
The problem of calculating the area of a circle was studied by the ancient Greeks. In
particular, both Archimedes and Eudoxus of Cnidus used the Method of Exhaustion
to calculate areas. This method used various regular inscribed polygons of known
area to approximate the area of an enclosed region. Chapter 1: Integration 2 In the case of a circle, as the number of sides of the inscribed polygon increased, the
error in using the area of the polygon to approximate the area of the circle decreased.
As a result, the Greeks had effectively used the concept of a limit as a key technique
in their calculation of the area. 1.1.2 Approximating Areas Under Curves Let’s use the ideas from the Method of Exhaustion and try to find the area underneath
a parabola by using rectangles as a basis for the approximation.
y f (x) = x2 2
1.5 Suppose that we have the
function f (x) = x2 . Consider the
region R bounded by the graph of
f , by the x-axis, and by the lines
x = 0 and x = 1. (1, 1) 1
0.5
R
−1 −0.5 0 0.5 x 1 How could we determine the area of this irregular region?
y
For our first estimate, we can
approximate the area of R by
constructing a rectangle R1 of
length 1 (from x = 0 to x = 1)
and height 1 (y = f (1) = 12 = 1).
This rectangle (in this case a
square) has area
length × height = 1 × 1 = 1. f (x) = x2 2
1.5
1 R1 (1, 1)
height 0.5
R
−1 −0.5 0 x 0.5
1
length y f (x) = x2 2
1.5 The diagram shows that the area
of rectangle R1 is larger than the
area of region R. Moreover, the
error is actually quite large. 1
0.5 R1 (1, 1) error
R −1 −0.5 0 0.5 1 x We can find a better estimate if we split the interval [0, 1] into 2 equal subintervals,
[0, 12 ] and [ 21 , 1].
Calculus 2 (B. Forrest)2 Section 1.1: Areas Under Curves 3 y
Using these intervals, two
rectangles are constructed. The
first rectangle R1 has its length
from x = 0 to x = 21 with height
equal to f ( 12 ) = 212 = 41 . f (x) = x2 2
1.5 R2 1
0.5
0.25 The second rectangle R2 has its
length from x = 21 to x = 1 with
height f (1) = 12 = 1. −1 −0.5 0 R1 (1, 1) ( 12 , 14 ) 0.5 1 x The area of rectangle R1 is equal to
R1 = length × height = 1
1 1
× 2 =
2 2
8 while the area of rectangle R2 is equal to
R2 = length × height = 1
1
×1=
2
2 Our second estimate for the area of the original region R is obtained by adding the
areas of these two rectangles to get
R1 + R2 =
Observe from the diagram that
our new estimate using two
rectangles for the area under
f (x) = x2 on the interval [0, 1] is
much better than our first
estimate since the error is
smaller. The region containing
the dashed lines indicates the
improvement in our estimate (this
is the amount by which we have
reduced the error from our first
estimate). 1 1 5
+ = = 0.625
8 2 8
y f (x) = x2 2
1.5
1 R2
error (1, 1) 0.5
R1
0.25 error ( 12 , 41 )
−1 −0.5 0 0.5 1 x To improve our estimate even further, divide the interval [0, 1] into five equal
subintervals of the form
i−1 i
[
, ]
5 5
where i ranges from 1 to 5.
This produces the subintervals
1 1 2 2 3 3 4 4 5
[0, ], [ , ], [ , ], [ , ], [ , ]
5 5 5 5 5 5 5 5 5
each having equal lengths of 15 .
Calculus 2 (B. Forrest)2 Chapter 1: Integration 4 Next we construct five new rectangles where the ith rectangle forms its length from
i−1
to 5i and has height equal to the value of the function at the right-hand endpoint
5
of the interval. That is, the height of a rectangle is f (x) = x2 where x = 5i or
i
i
f ( ) = ( )2
5
5
i2
= 2
5
The area of the ith rectangle is given by
length × height = 1 i2
i2
× 2 = 3
5 5
5 y f (x) = x2
error
area under curve 2
1.5 R5 1 (1,1) R4
0.5 R3
R2 R1
0 1
5 2
5 3
5 4
5 5
5 =1 x 1
5 Our new estimate is the sum of the areas of these rectangles which is
2 1
3 1
4 1
5 1
1 1
R1 + R2 + R3 + R4 + R5 = [( )2 ( )] + [( )2 ( )] + [( )2 ( )] + [( )2 ( )] + [( )2 ( )]
5 5
5 5
5 5
5 5
5 5
12 22 32 42 52
= 3+ 3+ 3+ 3+ 3
5
5
5
5
5
= 1 2
(1 + 22 + 32 + 42 + 52 )
53 = 5
1 X 2
i
53 i=1 Note: It can be shown that for any n
n
X
i=1 Calculus 2 i2 = (n)(n + 1)(2n + 1)
6
(B. Forrest)2 Section 1.1: Areas Under Curves 5 This means that the sum of the areas of the rectangles is
5
1 (5)(5 + 1)(2(5) + 1)
1 X 2
i = 3×
3
5 i=1
5
6 = 1 (5)(6)(11)
×
53
6 = 11
25 = 0.44
So far the estimates for the area under the curve of f (x) = x2 on the interval [0, 1]
are:
Number of Subintervals
(Rectangles) Length of Subinterval
(Width of Rectangle) Estimate for Area
under Curve 1 1 1 2 1
2 0.625 5 1
5 0.44 Observe from the diagram that the estimate for the area is getting better while the
error in the estimate is getting smaller.
y
f (x) = x2
2 1.5 Let’s repeat this process again by
using 10 equal subintervals. 1 (1,1)
Ri 0.5 0 Calculus 2 i−1
10 i
10 x (B. Forrest)2 Chapter 1: Integration 6 In this case, the sum of the areas of the 10 rectangles will be
10
X Ri = R1 + R2 + R3 + R4 + R5 + R6 + R7 + R8 + R9 + R10 i=1 = 10
1 X 2
i
103 i=1 = 1 (10)(10 + 1)(2(10) + 1)
103
6 = 77
200 = 0.385
If we were to use 1000 subintervals, the estimate for the area would be
1000
X Ri = R1 + R2 + R3 + . . . + R1000 i=1 = 1000
1 X 2
i
10003 i=1 = 1 (1000)(1000 + 1)(2(1000) + 1)
10003
6 = 0.3338335
You should begin to notice that as we increase the number of rectangles (number
of subintervals), the total area of these rectangles seems to be getting closer and
closer to the actual area of the original region R. In particular, if we were to produce
an accurate diagram that represents 1000 rectangles, we would see no noticeable
difference between the estimated area and the true area. For this reason we would
expect that our latest estimate of 0.3338335 is actually very close to the true value of
the area of region R.
We could continue to divide the interval [0, 1] into even more subintervals. In fact,
we can repeat this process with n subintervals for any n ∈ N. In this generic case, the
estimated area Rn would be
n
1 X 2
Rn = 3
i
n i=1
=
= 1 (n)(n + 1)(2(n) + 1)
n3
6
1
(2n3
n3 + 3n2 + n)
6 2 + n3 +
=
6
Calculus 2 1
n2 (B. Forrest)2 Section 1.1: Areas Under Curves 7 Note that if we let the number of subintervals n approach ∞, then
2 + n3 +
n→∞
6 lim Rn = lim n→∞ = 2
6 = 1
3 1
n2 By calculating the area under the graph of f (x) = x2 using an increasing number of
rectangles, we have constructed a sequence of estimates where each estimate is larger
than the actual area. Though it appears that the limiting value 31 is a plausible guess
for the actual value of the area, at this point the best that we can say is that the area
should be less than or equal to 13 .
Number of Subintervals
(Rectangles) Length of Subinterval
(Width of Rectangle) Estimate for Area
under Curve 1 1 1 2 1
2 0.625 5 1
5 0.44 10 1
10 0.385 1000 1
1000 0.3338335 approaches ∞ approaches 0 approaches 1
3 y
f (x) = x2
Alternately, we can use a
similar process that would
produce an estimate for the
area that will be less than the
actual value. To do so we
again divide the interval
[0, 1] into n subintervals of
length n1 with the i-th interval
[ i−1
, i ]. This interval again
n n
forms the length of a
rectangle Li , but this time we
will use the left-hand
endpoint of the interval so
that the value f ( i−1
) is the
n
height of the rectangle. Calculus 2 2 1.5 1 (1,1) Li 0.5
First
rectangle has
height 0 0 i−1
n i
n x (B. Forrest)2 Chapter 1: Integration 8 In this case, notice that since f (0) = 0 the first rectangle is really just a horizontal
line with area 0. Then the estimated area Ln for this generic case would be
Ln = n
1 X
(i − 1)2
n3 i=1 = 1 (n − 1)(n + 1 − 1)(2(n − 1) + 1)
n3
6 = 1 (n − 1)(n)(2n − 1)
n3
6 = 1 2n3 − 3n2 + n
n3
6 2 − n3 +
=
6
Finally, observe that 1
n2 2 − 3n +
n→∞
6 lim Ln = lim n→∞ 1
n2 1
= .
3 In summary, we have now shown that if R is the area of the region under the graph
of f (x) = x2 , above the x-axis, and between the lines x = 0 and x = 1, then for each
n ∈ N,
Ln ≤ R ≤ Rn .
y
It would be reasonable to
f (x) = x2
conclude that the area under the
2
graph of f (...

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