Calculus I
Chapter 0, About the Course
Pasi Ruotsalainen
Applied and Computational Mathematics
How to complete the course
1. exam Thursday 3 Oct. 2024 at 16:15, max points 24 p
2. exam Thursday 31 Oct. 2024 at 16:15, max points 24 p
⇒ at max 48 points from midterm exams.
Points are earned also from
▶ active participation in exercises,
▶ answering so-called Stack questions in Moodle.
These exercise and Stack points are only valid during the
fall’s lecture course, not in later final exams.
Pasi Ruotsalainen
ACM, University of Oulu
2 / 175
Points from exercises
Active participation in exercises means, that one can show that he
/ she has successfully answered the questions required for that
exercise. Points earned from 11 exercises are scaled to add to
exams points as follows:
Points from exercises
2-3,4
3,5-4,4
4,5-5,4
5,5-6,4
6,5-7,4
7,5-8,4
8,5-9,4
9,5-11
Scaled points
1
2
3
4
5
6
7
8
Assignments needed for exercise points are marked in Moodle for
each exercise. Solutions sent late or by email will not be accepted.
Pasi Ruotsalainen
ACM, University of Oulu
3 / 175
Points from Stack questions
Stack is a computer-aided system for checking math problems in
Moodle. There are 12 packages of Stack questions on this course,
each with 1 − 2 questions.
The points for correctly calculated Stack problems give points as
follows:
▶ 0.00-0.99, 0 p,
▶ 1.00-3.99, 1 p,
▶ 4.00-6.99, 2 p,
▶ 7.00-9.99, 3 p,
▶ 10.00-12.00, 4 p.
Pasi Ruotsalainen
ACM, University of Oulu
4 / 175
Thus, the maximum points available on the course are:
▶ Two midterm exams 48 p,
▶ Active participation in exercises 8 p,
▶ Stack questions 4 p,
The maximum of total score is 60 points. 26 points are needed to
pass the course. Minimum of 4 p is required in the 2nd midterm
exam. An individual midterm exam cannot be retaken and no
additional midterm exam will be held.
Pasi Ruotsalainen
ACM, University of Oulu
5 / 175
Learning outcomes of the course
Upon completion of the course, the student identifies concepts of
vector algebra, can use vector algebra for solving problems of
analytic geometry, can explain basic characteristics of elementary
functions, is able to analyse the limit and the continuity of real
valued functions of one variable, can solve problems associated
with differential and integral calculus of real valued functions of
one variable.
Contents Vector algebra and analytic geometry. Limit,
continuity, differential and integral calculus and applications of
real valued functions of one variable. Complex numbers.
Pasi Ruotsalainen
ACM, University of Oulu
6 / 175
Workload calculation
5 ECTS equals 5 · 27 h = 135 h of study. Here is example of
study time calculation:
▶ Lectures:
▶ prepairing: 14 h
▶ lectures: 28 h
▶ repetition: 14 h
▶ Exercises:
▶ prepairing: 22 h
▶ exercises: 22 h
▶ Stack assignments: 12 h
▶ Prepairing for exams: 17 h
▶ Exams: 6 h
Teaching 28+22=50 hours, self-study 85 hours (approx. 60% of
the workload).
Pasi Ruotsalainen
ACM, University of Oulu
7 / 175
Calculus I
Chapter 1, Sets and notations
Pasi Ruotsalainen
Applied and Computational Mathematics
Basic concepts of set theory
A set is a collection of well-defined objects called members or
elements of that set. Examples: the set of integers, the English
alphabet (the set of letters of the English language) or the set of
the students taking this course. Sets are often specified with curly
brace notation, for example:
M = {x ∈ R : x 2 ≥ 5}.
The left side of the colon indicates that the universal set is set of
real numbers. The condition required for a set of elements is
shown on the right. A vertical line | can also be used instead of
colon.
Pasi Ruotsalainen
ACM, University of Oulu
9 / 175
A finite number of members can be given by listing them:
A = {a, b, c, d}, B = {b, c, d, e}, C = {a, c, d}, D = {c, d, a}.
The notations a ∈ A and a ̸∈ B indicate that the object a is a
member of the set A and the object a is not a member of the set
B.
We say that C is a subset of A, if every member of C is a
member of A, written C ⊂ A.
Notation C ̸⊂ B indicates that the set C is not a subset of the
set B (because a ̸∈ B).
The sets C and D are equal, C = D.
Pasi Ruotsalainen
ACM, University of Oulu
10 / 175
Union
A ∪ B = {a, b, c, d, e} = {all the members that are in A or B (or
both)}
Intersection
A ∩ B = {b, c, d} = {all the members that are in both A and B}
Set difference
A \ C = {b} = {all the members that are in A but not in C }
Empty set
∅ = {} = the set containing no elements
Cartesian product
A × B = {(x , y ) : x ∈ A, y ∈ B} =
{(a, b), (a, c), (a, d), (a, e), (b, b), (b, c), . . . , (d, e)} = {all
possible ordered pairs, where x is a member of A and y is a
member of B}
Pasi Ruotsalainen
ACM, University of Oulu
11 / 175
Sets of numbers
The set of natural numbers:
N = {1, 2, 3, . . . .}
If the number 0 (zero) is included, notation
N0 = N ∪ {0} = {0, 1, 2, . . .} is used.
Other sets are built as extensions of N: the set of integers
Z = {. . . , −2, −1, 0, 1, 2, . . .}.
Positive integers are notated Z+ = {1, 2, . . .}, negative integers
are notated Z− = {. . . , −2, −1}.
Rational numbers are of form p/q, where p and q ̸= 0 are
integers. The set of rational numbers is
Q = {p/q : p, q ∈ Z, q ̸= 0}.
When we add, subtract or multiply two rational numbers, we get a
rational number. Also a rational number divided by another yields
a rational number, provided that the denominator is not zero.
Pasi Ruotsalainen
ACM, University of Oulu
12 / 175
It can be shown that there is no rational number whose square is
2, i.e. the equation x 2 = 2 has no solution among the rational
numbers. There are other numbers that do not have rational
representation, such as the Euler’s constant (Napier’s constant) e
and π, which is the ratio of the circumference and the diameter
of any circle. Such numbers are called irrational numbers.
Rational numbers and irrational numbers together form the set of
real numbers R.
There is no solution to the equation x 2 = −1 among real
numbers, but by extending the set of real numbers with the set of
complex numbers, this equation can also be solved.
Complex numbers C = R ∪ {imaginary numbers }.
N ⊂ N0 ⊂ Z ⊂ Q ⊂ R ⊂ C.
Pasi Ruotsalainen
ACM, University of Oulu
13 / 175
Zero-product property
xy = 0
Pasi Ruotsalainen
if and only if
x = 0 or y = 0.
ACM, University of Oulu
14 / 175
Open, closed and half-open intervals
Open interval: ]a, b[= (a, b) = {x ∈ R : a < x < b}
Closed interval: [a, b] = {x ∈ R : a ≤ x ≤ b}
Half-open intervals: ]a, b] = (a, b] = {x ∈ R : a < x ≤ b}
[a, b[= [a, b) = {x ∈ R : a ≤ x < b}
[a, ∞[= [a, ∞) = {x ∈ R : x ≥ a}
]a, ∞[= (a, ∞) = {x ∈ R : x > a}
] − ∞, a[= (−∞, a) = {x ∈ R : x < a}
] − ∞, a] = (−∞, a] = {x ∈ R : x ≤ a}
] − ∞, ∞[= (−∞, ∞) = R
R ∪ {−∞} ∪ {∞} is the so-called set of extended real numbers.
Pasi Ruotsalainen
ACM, University of Oulu
15 / 175
Summation notation
n
X
xj = x1 + x2 + · · · + xn , n ∈ N
j=1
1.
n
X
xj = xm + · · · + xn
j=m
2.
100
X
xj = x2 + x4 + x6 + · · · + x100
j=2
j/2∈N
3.
n
X
(xj + yj ) =
j=1
Pasi Ruotsalainen
n
X
j=1
xj +
n
X
yj
j=1
ACM, University of Oulu
4.
n
X
j=1
cxj = c
n
X
xj
j=1
16 / 175
Binomial formula
Let a and b ∈ R \ {0}, b ̸= −a and n ∈ N. Now
n
(a + b) =
n
X
n
j=0
where
n
j
!
=
j
!
a j b n−j ,
n!
j!(n − j)!
is the binomial coefficient and factorial n! = 1 · 2 · . . . · n and
specifically 0! = 1.
Pasi Ruotsalainen
ACM, University of Oulu
17 / 175
Absolute value
Absolute value |x | of a real number x :
(
|x | =
Pasi Ruotsalainen
x,
when x ≥ 0
−x , when x < 0.
ACM, University of Oulu
18 / 175
1. |x | ≥ 0 for every x ∈ R and |x | = 0 if and only if x = 0
2. |x | = a, a ≥ 0 if and only if x = ±a
3. |x | = | − x | and |x |2 = x 2 for every x ∈ R,
|xy | = |x ||y | for every x , y ∈ R,
|x |
x
=
for every x , y ∈ R, y ̸= 0
y
|y |
4. −|x | ≤ x ≤ |x | for every x ∈ R
5. If a ≥ 0, then |x | ≤ a if and only if −a ≤ x ≤ a,
|x | ≥ a if and only if x ≤ −a or x ≥ a.
If a > 0, then |x | < a if and only if −a < x < a,
|x | > a if and only if x < −a or x > a.
6. |x | ≤ |y | if and only if x 2 ≤ y 2 for every x , y ∈ R
Pasi Ruotsalainen
ACM, University of Oulu
19 / 175
Square root
For every x > 0 there exists a square root y ∈ R, such that
y 2 = x . Then also (−y )2 = x .
√
x > 0 is a positive
Definition Square root x of a number √
number, square of which is x . Moreover 0 = 0.
√
Therefore square root x is defined for nonnegative real
numbers x ≥ 0 with properties
√
√
x ≥0
and
( x )2 = x .
√
Notice! x 2 = |x |, x ∈ R
√
The equation x 3 = a, a ∈ R has exactly one solution x = 3 a.
√
Cube root 3 x , x ∈ R, is a real number such that
√
( 3 x )3 = x .
Pasi Ruotsalainen
ACM, University of Oulu
20 / 175
nth root
√
If n ∈ N is even and x ≥ 0, then nth root n x is a real number,
such that
√
√
n
x ≥ 0 and ( n x )n = x .
√
If n ∈ N is odd and x ∈ R, then nth root n x is a real number,
such that
√
( n x )n = x .
Pasi Ruotsalainen
ACM, University of Oulu
21 / 175
Power function (exponentiation) x r
x is variable (or base) and r is power (or exponent).
• r ∈ Z: Let x ∈ R and r ∈ N. It is defined
xr = x · . . . · x,
where variables x are multiplied r times.
If x ∈ R \ {0} and r ∈ Z− , then
xr =
1
1
,
=
x −r
x · ... · x
where variables x are multiplied −r ∈ N times in the
denominator. In addition
x 0 = 1, x ∈ R \ {0}.
Pasi Ruotsalainen
ACM, University of Oulu
22 / 175
• r ∈ Q: If the power r is a rational number, then it can be
written with positive denominator r = pq , p ∈ Z, q ∈ N.
If
x ∈ R+ and r ∈ Q,
then
p
xr = x q =
Pasi Ruotsalainen
√
q
x p.
ACM, University of Oulu
23 / 175
Power rules
If x > 0, y > 0, r and s ∈ R, then
x r x s = x r +s ,
xr
= x r −s ,
xs
(x r )s = x rs ,
x r y r = (xy )r ,
r
xr
x
=
.
yr
y
Pasi Ruotsalainen
ACM, University of Oulu
24 / 175
Triangle inequalities
||x | − |y || ≤ |x + y | ≤ |x | + |y |,
x , y ∈ R.
Proof From the definition of absolute value, we get x = |x | or
x = − |x |. It follows, that
− |x | ≤ x ≤ |x | and accordingly − |y | ≤ y ≤ |y | .
By adding these together, we get
− |x | − |y | ≤ x + y ≤ |x | + |y | ,
which gives us the right hand side |x + y | ≤ |x | + |y |.
To prove the left hand side, we approximate
|x | = |x + y + (−y )| ≤ |x + y | + |y |, from which it follows
|x | − |y | ≤ |x + y | and accordingly |y | − |x | ≤ |x + y |, so in all we
get ||x | − |y || ≤ |x + y |.
□
Pasi Ruotsalainen
ACM, University of Oulu
25 / 175
Calculus I
Chapter 2, Vectors
Pasi Ruotsalainen
Applied and Computational Mathematics
Rn and vectors
Definition n dimensional space Rn is a set of points, defined as
real valued n-tuples
P(x1 , x2 , . . . , xn ) = (x1 , x2 , . . . , xn ).
Vector ⃗x describes a movement from one point A to another
point B. A vector has both direction and length (magnitude). In
space Rn points can be notated A(a1 , a2 , . . . , an ) and
B(b1 , b2 , . . . , bn ) and vector is defined as n-tuple
−→
⃗x = (x1 , x2 , . . . , xn ) = (b1 − a1 , b2 − a2 , . . . , bn − an ) = AB.
(terminal point - initial point)
Pasi Ruotsalainen
ACM, University of Oulu
27 / 175
Vector calculations
Vector addition and scalar multiplication are defined in Rn , for
every ⃗u ∈ Rn , ⃗v ∈ Rn and λ ∈ R
⃗u + ⃗v = (u1 + v1 , u2 + v2 , . . . , un + vn ),
λ⃗u = (λu1 , λu2 , . . . , λun ).
Rn is closed under these calculations (for example sum of two
Rn -vectors is an Rn -vector). With these calculations defined,
Rn -vectors have the following axioms:
Pasi Ruotsalainen
ACM, University of Oulu
28 / 175
Pasi Ruotsalainen
ACM, University of Oulu
29 / 175
Axioms 2.1
1. Vector addition is commutative: ⃗u + ⃗v = ⃗v + ⃗u .
2. There exists a zero vector ⃗0, such that ⃗u + ⃗0 = ⃗u ,
⃗0 = (0, 0, . . . , 0).
⃗ ) = (⃗u + ⃗v ) + w
⃗.
3. Vector addition is associative: ⃗u + (⃗v + w
4. For every vector there exists additive inverse −⃗u , such that
⃗u + (−⃗u ) = ⃗0.
5. λ(µ⃗u ) = (λµ)⃗u .
6. (λ + µ)⃗u = λ⃗u + µ⃗u .
7. λ(⃗u + ⃗v ) = λ⃗u + λ⃗v .
8. There exists an identity element of scalar multiplication
1 ∈ R : 1 · ⃗u = ⃗u .
Pasi Ruotsalainen
ACM, University of Oulu
30 / 175
Vector space basis
Vector ⃗u is a linear combination of vectors {⃗a1 , . . . ,⃗ak }, if there
exists numbers λ1 , . . . , λk , such that
⃗u = λ1⃗a1 + λ2⃗a2 + · · · + λk⃗ak .
Definition
1. Vectors {⃗a1 , . . . ,⃗ak } are linearly independent if and only if
the only solution of the equation
λ1⃗a1 + λ2⃗a2 + · · · + λk⃗ak = ⃗0 is the trivial one:
λ1 = λ2 = · · · = λk = 0.
2. Vectors {⃗a1 , . . . ,⃗ak } are linearly dependent if and only if the
equation λ1⃗a1 + λ2⃗a2 + · · · + λk⃗ak = ⃗0 has non-zero
solutions.
Pasi Ruotsalainen
ACM, University of Oulu
31 / 175
Parallel vectors
Definition In the case of two vectors, linearly dependent vectors
are said to be parallel:
⃗u ∥ ⃗v
if and only there exists λ ∈ R : ⃗u = λ⃗v .
Basis and coordinates
Theorem 2.2 Let {⃗v1 , . . . , ⃗vn } be a set of linearly independent
vectors in Rn . Now for every vector ⃗u ∈ Rn there exists a unique
set of numbers λi ∈ Rn , i = 1, 2, . . ., n, such that
⃗u =
n
X
λi ⃗vi .
i=1
It is said, that vectors {⃗v1 , . . . , ⃗vn } are the basis for the vector
space Rn .
Pasi Ruotsalainen
ACM, University of Oulu
32 / 175
Standard basis of vector space Rn :
⃗e1 = (1, 0, . . . , 0), ⃗e2 = (0, 1, 0, . . . , 0), · · ·, ⃗en = (0, . . . , 0, 1).
Elements of the standard basis of a vector space R3 can be
notated with symbols familiar from physics courses:
⃗i = (1, 0, 0), ⃗j = (0, 1, 0), ⃗k = (0, 0, 1).
Let {⃗a1 ,⃗a2 , . . . ,⃗an } ⊂ Rn be a basis of a vector space. Now,
according the definition of basis, there exists a unique set of
numbers y1 , y2 , . . . , yn , such that ⃗x = y1⃗a1 + y2⃗a2 + · · · + yn⃗an . It
is said that these numbers are the coordinates of the vector ⃗x
with respect to the basis {⃗a1 ,⃗a2 , . . . ,⃗an }.
Pasi Ruotsalainen
ACM, University of Oulu
33 / 175
Dot product
Definition Dot product (scalar product) of two vectors of Rn ,
⃗u = (u1 , u2 , . . . , un ) and ⃗v = (v1 , v2 , . . . , vn ), is a real number.
⃗u · ⃗v =
n
X
ui vi = u1 v1 + u2 v2 + · · · + un vn .
i=1
Pasi Ruotsalainen
ACM, University of Oulu
34 / 175
Properties of dot product
Dot product is symmetric and bilinear:
⃗ ∈ Rn and for every α, β ∈ R:
for every ⃗u , ⃗v , w
⃗u · ⃗v = ⃗v · ⃗u
⃗ = α⃗u · w
⃗ + β⃗v · w
⃗
(α⃗u + β⃗v ) · w
⃗u · (α⃗v + β w
⃗ ) = α⃗u · ⃗v + β⃗u · w
⃗.
Moreover,
⃗u · ⃗u ≥ 0 and ⃗u · ⃗u = 0 if and only if ⃗u = ⃗0.
The length of a vector can be defined using dot product
√
1
∥⃗u ∥ = ⃗u · ⃗u = [u12 + u22 + . . . un2 ] 2 .
Pasi Ruotsalainen
ACM, University of Oulu
35 / 175
Distance between two points
The distance between the points A(a1 , a2 , . . . , an ) and
B(b1 , b2 , . . . , bn ) is defined
−→
|AB| = d(A, B) = ∥AB∥
=
q
(b1 − a1 )2 + (b2 − a2 )2 + · · · + (bn − an )2 .
This is so-called Euclidean or Pythagorean distance.
Pasi Ruotsalainen
ACM, University of Oulu
36 / 175
Angle and orthogonality
Definition The angle between two non zero vectors ⃗u and ⃗v ,
∠(⃗u , ⃗v ), is defined by
cos(∠(⃗u , ⃗v )) =
⃗u · ⃗v
.
∥⃗u ∥∥⃗v ∥
The angle gets values 0 ≤ ∠(⃗u , ⃗v ) ≤ π.
The vectors ⃗u and ⃗v are said to be orthogonal, if and only if the
dot product is zero
⃗u · ⃗v = 0.
Notation for orthogonality:
⃗u ⊥ ⃗v .
Pasi Ruotsalainen
ACM, University of Oulu
37 / 175
3•
A
2•
−→
−→
Kuva: Angle between vectors BA and BC , (A = (2, 1, 3), B = (0, 1, 1)
and C = (−1, 0, 1)) (α = 120◦ ).
Pasi Ruotsalainen
ACM, University of Oulu
38 / 175
Cross product
Definition Cross product (vector product) is an operation on two
vectors in three-dimensional space R3
⃗u = (u1 , u2 , u3 ), ⃗v = (v1 , v2 , v3 ),
and it can be expressed as the formal determinant
⃗u × ⃗v =
⃗i ⃗j ⃗k
u1 u2 u3 .
v1 v2 v3
Thus, cross product of two vectors is also a vector. Coordinates of
this vector can be computed using cofactor expansion along the
first row
⃗u × ⃗v = (u2 v3 − u3 v2 , u3 v1 − u1 v3 , u1 v2 − u2 v1 ).
Pasi Ruotsalainen
ACM, University of Oulu
39 / 175
Properties of cross product
Cross product is anticommutative and bilinear :
⃗ ∈ R3 and for every α, β ∈ R:
for every ⃗u , ⃗v , w
⃗u × ⃗v = −⃗v × ⃗u
⃗ = α⃗u × w
⃗ + β⃗v × w
⃗
(α⃗u + β⃗v ) × w
⃗u × (α⃗v + β w
⃗ ) = α⃗u × ⃗v + β⃗u × w
⃗.
Cross product is a zero vector, if and only if the vectors are
linearly dependent:
⃗v ∥⃗u
Pasi Ruotsalainen
if and only if ⃗u × ⃗v = ⃗0.
ACM, University of Oulu
40 / 175
Cross product ⃗u × ⃗v of two non zero vectors ⃗u and ⃗v is always
orthogonal to both vectors
⃗u ⊥ ⃗u × ⃗v , ⃗v ⊥ ⃗u × ⃗v .
If the vectors ⃗u and ⃗v are linearly independent, then also the
vectors {⃗u , ⃗v , ⃗u × ⃗v } are linearly independent.
These vectors form a basis for three-dimensional space R3 .
Pasi Ruotsalainen
ACM, University of Oulu
41 / 175
Area of parallelogram
Consider a parallelogram with the vectors ⃗u and ⃗v for sides. The
length of the cross product ⃗u × ⃗v is equal to the area of the
parallelogram
A = ∥⃗u × ⃗v ∥.
Pasi Ruotsalainen
ACM, University of Oulu
42 / 175
Calculus I
Chapter 3, Lines and planes
Pasi Ruotsalainen
Applied and Computational Mathematics
Lines and planes in R3
In this chapter, we use vectors to study lines and planes. By
representing a point P(x , y , z) with a corresponding position
vector ⃗p = x⃗i + y⃗j + z ⃗k we can use vector algebra to define the
points a line (or a plane) consist of.
Pasi Ruotsalainen
ACM, University of Oulu
44 / 175
Parametric forms of a line
Definition Let A(⃗a) ∈ R3 be a point and ⃗u ̸= ⃗0 a vector. We are
defining a line l that passes through the point A in the direction
of vector ⃗u . Vector parametric equation of a line l is
⃗p = ⃗a + t⃗u , t ∈ R.
Scalar parametric equation of a line l is
(x , y , z) = (x0 , y0 , z0 ) + t(u1 , u2 , u3 ), t ∈ R,
where A(x0 , y0 , z0 ) is the position vector of any fixed point on the
line l and ⃗u = u1⃗i + u2 ⃗j + ⃗u3⃗k is the direction vector of the line l.
Pasi Ruotsalainen
ACM, University of Oulu
45 / 175
Distance of a point from a line
Let d be the distance of a point P1 (⃗p1 ) from a line
l : ⃗p = ⃗a + t⃗u , t ∈ R. We determine d by evaluation the area A
of the parallelogram span by vectors ⃗p1 − ⃗a and ⃗u in two separate
ways (i.e. the parallelogram whose adjacent sides are the
mentioned vectors). In the previous chapter we derived the area
of a such parallelogram using cross product
A = ∥⃗u × (⃗p1 − ⃗a)∥.
On the other hand, if we consider the base of the parallelogram to
be direction vector ⃗u , then the height is the distance d and area is
A = base · height = ∥⃗u ∥ · d.
Combining these two equations gives us the distance of a point
P1 from a line
∥⃗u × (⃗p1 − ⃗a)∥
.
d=
∥⃗u ∥
Pasi Ruotsalainen
ACM, University of Oulu
46 / 175
Parametric and normal form of a plane
Definition Let ⃗u and ⃗v be two nonparallel vectors in R3 and let
A(⃗a) be a point. We define a plane, such that the point A is on
the plane and vectors ⃗u and ⃗v are parallel to the plane.
Parametric equation of the plane T is
⃗p = ⃗a + s⃗u + t⃗v , s, t ∈ R.
It is a property of the cross product that vector
⃗n = ⃗u × ⃗v
is orthogonal to the vectors of the plane. It is called normal vector
of the plane.
Pasi Ruotsalainen
ACM, University of Oulu
47 / 175
Let P(⃗p ) be an arbitrary point of the plane T . Now vector ⃗p − ⃗a
is a direction vector of the plane. So it is orthogonal to the
normal vector giving us the normal form of the plane :
(⃗p − ⃗a) · ⃗n = 0.
If ⃗n = a⃗i + b⃗j + c ⃗k, ⃗p = x⃗i + y⃗j + z ⃗k and −⃗n · ⃗a = d, then the
equation of the plane can be written
ax + by + cz + d = 0.
Pasi Ruotsalainen
ACM, University of Oulu
48 / 175
Pasi Ruotsalainen
ACM, University of Oulu
49 / 175
Distance from a point to a plane
Let A(⃗a) be a point of a plane and let ⃗n be a normal vector of
that plane. Distance from a point P1 (⃗p1 ) to a plane is
h = ∥⃗p1 − ⃗a∥| cos α|,
where α is the angle between vectors ⃗p1 − ⃗a and ⃗n. We can solve
the value of the cosine from the dot product of these vectors
cos α =
⃗n · (⃗p1 − ⃗a)
,
∥⃗n∥∥⃗p1 − ⃗a∥
giving us the result
h=
Pasi Ruotsalainen
|⃗n · (⃗p1 − ⃗a)|
.
∥⃗n∥
ACM, University of Oulu
50 / 175
Pasi Ruotsalainen
ACM, University of Oulu
51 / 175
Calculus I
Chapter 4, Functions
Pasi Ruotsalainen
Applied and Computational Mathematics
Function
Definition Let X and Y be nonempty sets. Function f from set
X to set Y ,
f : X → Y,
is a rule that assigns to each number x ∈ X a single number
y ∈ Y . This is usually written y = f (x ). Domain X is usually
notated Df .
The range R(f ) of f is the subset of Y consisting of all values
f (x ) of function
Rf = {f (x ) | x ∈ Df } ⊂ Y .
Pasi Ruotsalainen
ACM, University of Oulu
53 / 175
A function f is fully defined, when we know
1. relation between variable x and values y of the function
y = f (x ) and
2. domain Df .
In this course, we consider only so-called real-valued functions, for
which Df ⊂ R and Rf ⊂ R.
Unless given explicitly, the domain of a function are those values
of x for which the right hand side of the expression y = f (x ) (or
f (x ) = . . .) is a real number.
Pasi Ruotsalainen
ACM, University of Oulu
54 / 175
Adding, subtracting, multiplying and
dividing functions. Composite function.
Two functions f and g are equal, the same function, if and only if
Df = Dg and f (x ) = g(x ) for every x ∈ Df .
Example Are f and g the same function, if
√
√
f (x ) = x 2 and g(x ) = ( x )2 ?
Pasi Ruotsalainen
ACM, University of Oulu
55 / 175
For functions f and g the sum f + g, the difference f − g, the
product fg and the quotient gf and multiplying with a constant
c ∈ R are defined:
(f + g)(x ) = f (x ) + g(x ), Df +g = Df ∩ Dg
(f − g)(x ) = f (x ) − g(x ), Df −g = Df ∩ Dg
(fg)(x ) = f (x )g(x ), Dfg = Df ∩ Dg
f (x )
f
, D f = Df ∩ Dg \ {x ∈ Dg | g(x ) = 0}
( )(x ) =
g
g
g(x )
(cf )(x ) = cf (x ), Dcf = Df
Pasi Ruotsalainen
ACM, University of Oulu
56 / 175
Composite function f ◦ g (read as ”f of g”or ”f circle g”) is the
notation for function x → f (g(x ))
(f ◦ g)(x ) = f (g(x ))
and
Df ◦g = {x ∈ R | x ∈ Dg and g(x ) ∈ Df }.
Here function g is so-called inside function and f outside function.
Note! It is true, that (f ◦ g) ◦ h = f ◦ (g ◦ h), but usually
f ◦ g ̸= g ◦ f .
Pasi Ruotsalainen
ACM, University of Oulu
57 / 175
Monotonic function (increasing/decreasing
function)
Let A ⊂ R. Function f is said to be increasing/strictly increasing
in the set A, if for every x1 , x2 ∈ A
(i)
from x1 < x2 it follows f (x1 ) ≤ f (x2 )
(ii)
from x1 < x2 it follows f (x1 ) < f (x2 ).
In the same manner, we define a function to be
decreasing/strictly decreasing in A:
(iii) if x1 < x2 , then f (x1 ) ≥ f (x2 ) for every x1 , x2 ∈ A,
(iv)
if x1 < x2 , then f (x1 ) > f (x2 ) for every x1 , x2 ∈ A.
Pasi Ruotsalainen
ACM, University of Oulu
58 / 175
Even and odd functions, periodic functions
If f (−x ) = f (x ), for every x ∈ Df , then f is even function.
If f (−x ) = −f (x ), for every x ∈ Df , then f is odd function.
Function f : R → R is periodic, if there exists a number a ̸= 0
such, that
f (x + a) = f (x ) for every x ∈ R.
Number a is called the period. If all the periods are of the form
na (n ∈ Z, n ̸= 0, a > 0), then a is fundamental period.
Pasi Ruotsalainen
ACM, University of Oulu
59 / 175
Classes of functions
A. Polynomial functions and rational functions
Polynomial P is of the form
P(x ) = an x n + an−1 x n−1 + · · · + a1 x + a0 ,
where a0 , a1 , . . . , an ∈ R are constants, an ̸= 0 (n is the degree of
the polynomial P).
Let P1 and P2 be polynomials. The function defined by equation
R(x ) =
a n x n + · · · + a1 x + a0
P1 (x )
, an , bm ̸= 0,
=
P2 (x )
bm x m + · · · + b1 x + b0
is said to be a rational function.
Pasi Ruotsalainen
ACM, University of Oulu
60 / 175
B. Power functions and root functions
√
q
f (x ) = x r , f (x ) = x p
C. Trigonometric functions and arcus functions
Sine, cosine, tangent and cotangent functions (along with secant
and cosecant functions) are trigonometric functions. Arcus
functions (also called cyclometric functions) are inverse
trigonometric functions.
D. Exponential functions and logarithmic functions
f (x ) = ax , f (x ) = loga x .
Pasi Ruotsalainen
ACM, University of Oulu
61 / 175
Elementary functions are functions that are defined as taking
sums, differences, products, quotients and compositions of
functions from the classes A–D above.
Function f is algebraic, if it satisfies a polynomial equation
[f (x )]n + Rn−1 (x )[f (x )]n−1 + · · · + R0 (x ) = 0,
where functions Ri , i = 0, 1 . . . , n − 1 are rational functions. A
function which is not algebraic is called a transcendental function.
Pasi Ruotsalainen
ACM, University of Oulu
62 / 175
Trigonometric functions
Measurements of angle
Degrees ↔ radians: π rad = 180◦
1 rad =
180◦
π
and
1◦ =
π
rad.
180
Trigonometric functions can be calculated using the unit circle.
Sine function is odd and and fundamental period is 2π.
Dsin = R, Rsin = [−1, 1]
Cosine function is even and and fundamental period is 2π.
Dcos = R, Rcos = [−1, 1]
sin x
is odd and and fundamental
Tangent function tan x =
cos x
period is π.
Dtan = R \ { π2 + kπ, k ∈ Z}, Rtan = R.
Pasi Ruotsalainen
ACM, University of Oulu
63 / 175
Pasi Ruotsalainen
ACM, University of Oulu
64 / 175
Formulas
For every x , y ∈ R
sin2 x + cos2 x = 1,
cos(x − y ) = cos x cos y + sin x sin y ,
cos(x + y ) = cos x cos y − sin x sin y ,
π
− x = sin x ,
cos
2
π
sin
− x = cos x ,
2
sin(x + y ) = sin x cos y + cos x sin y ,
sin(x − y ) = sin x cos y − cos x sin y ,
sin(2x ) = 2 sin x cos x ,
cos(2x ) = cos2 x − sin2 x = 2 cos2 x − 1 = 1 − 2 sin2 x .
Pasi Ruotsalainen
ACM, University of Oulu
65 / 175
Power functions
Power function f is of the form
f (x ) = x r
where variable x ∈ Df and power r ∈ R. In this case of general
power Df =]0, ∞[, which gives Rf =]0, ∞[. Depending on the
value of r , Df can be greater (see ”Exponentiation” in the first
set of slides).
Pasi Ruotsalainen
ACM, University of Oulu
66 / 175
Exponential functions
Exponential function f is of the form
f (x ) = ax ,
where exponent x ∈ Df = R and base a ∈ R+ . Clearly
Rf =]0, ∞[.
The most common base in science is Euler’s number
e ≈ 2.71828 . . . which leads to (natural) exponential function
f (x ) = ex .
Notations: ax = expa x , ex = exp x .
Pasi Ruotsalainen
ACM, University of Oulu
67 / 175
Hyperbolic functions
1. hyperbolic sine: sinh x = 12 ( ex − e−x )
2. hyperbolic cosine: cosh x = 12 ( ex + e−x )
sinh x
3. hyperbolic tangent: tanh x = cosh
x
x
4. hyperbolic cotangent: coth x = cosh
sinh x .
These functions have same similar identities as trigonometric
functions, for example
cosh2 x − sinh2 x = 1,
sinh(2x ) = 2 sinh x cosh x ,
cosh(2x ) = cosh2 x +sinh2 x = 1+2 sinh2 x = 2 cosh2 x −1, x ∈ R.
Pasi Ruotsalainen
ACM, University of Oulu
68 / 175
Inverse function
x
y=
y=3
x-1
Example Let a) y = 3x − 1, b) y = 3x 2 − 1.
Solve x as a function of y from the equations. Is x a function of
y?
y=3x2-1
3
+1)/
y=(x
Pasi Ruotsalainen
ACM, University of Oulu
69 / 175
What kind of functions have inverse
functions?
A function f : X → Y is called
• injection, if for every x1 , x2 ∈ X , from the condition
f (x1 ) = f (x2 ) it follows, that x1 = x2 (one-to-one function)
• surjection, if Rf = {f (x ) | x ∈ Df } = Y
• bijection, if f is both, injection and surjection.
Note! Equivalently with the first definition: f is an injection, if
from the condition x1 ̸= x2 , it follows, that f (x1 ) ̸= f (x2 ).
Pasi Ruotsalainen
ACM, University of Oulu
70 / 175
Let f : X → Y be an injection. For every number y ∈ Rf there
exists exactly one number x ∈ X , for which f (x ) = y , so
f : X → Rf is a bijection. This means that there exists a function
g : Rf → X , which maps number y to the number x .
We use notation g = f −1 and call f −1 the inverse function of f ,
with domain and range
Df −1 = Rf ,
Rf −1 = Df
(a)
(f −1 ◦ f )(x ) = f −1 (f (x )) = x ,
x ∈ Df
(b)
(f ◦ f −1 )(y ) = f (f −1 (y )) = y ,
y ∈ Rf .
In order to prove that function g is the inverse function of f , we
need to show, that (a) and (b) are true when we substitute f −1
with g.
Pasi Ruotsalainen
ACM, University of Oulu
71 / 175
Finding the inverse function
1. Define the domain Df and the range Rf of the function f .
2. Solve the variable x as a function of y from the original
equation y = f (x ). If unique x can be found, then the
inverse function f −1 exists and x = f −1 (y ).
3. Give the definition of f −1 obtained in step 2 as a function of
x.
4. Check that Df −1 = Rf and Rf −1 = Df .
Pasi Ruotsalainen
ACM, University of Oulu
72 / 175
Properties of inverse function
The existence of an inverse function cannot always be proved the
way described above (by solving x as s function of y ). In those
cases, it is useful to know:
A strictly increasing or decreasing function is an injection, so it
has an inverse function.
It is not necessary for a function to be strictly monotonic in order
to have an inverse function.
Theorem 4.1 Inverse function f −1 of a strictly increasing (strictly
decreasing) function f is strictly increasing (strictly decreasing).
Theorem 4.2 (f ◦ g)−1 = g −1 ◦ f −1 , D(f ◦g)−1 = Rf ◦g .
Pasi Ruotsalainen
ACM, University of Oulu
73 / 175
Logarithmic function
If 0 < a < 1, then f (x ) = ax is strictly decreasing, and if a > 1
then f (x ) = ax is strictly increasing function. Thus exponential
function f (x ) = ax , a ̸= 1, has an inverse function. This inverse
function is called logarithmic function and is notated loga .
y = ax if and only if x = loga y , x ∈ R, y > 0, a > 0, a ̸= 1
Dloga = Rexpa =]0, ∞[, Rloga = Dexpa = R.
From theorem 4.1 it follows, that function loga is strictly
increasing, when a > 1 and strictly decreasing when 0 < a < 1.
Pasi Ruotsalainen
ACM, University of Oulu
74 / 175
Theorem 4.3 Let a, x , y > 0, a ̸= 1. Following formulas hold for
logarithmic function
1. loga xy = loga x + loga y
x
2. loga = loga x − loga y
y
3. loga x r = r loga x , r ∈ R.
Note! loga (x + y ) ̸= loga x + loga y
Notations: log e x = ln x natural logarithm,
log10 x = lg x decimal logarithm.
Pasi Ruotsalainen
ACM, University of Oulu
75 / 175
Useful formulas
1. ax = ex ln a , a > 0, x ∈ R
loga x
2. logb x = log
, a, b, x > 0, a, b ̸= 1
ab
√
1
n
3. loga x = n loga x , n ∈ N, x > 0, a > 0, a ̸= 1
Pasi Ruotsalainen
ACM, University of Oulu
76 / 175
Inverse hyperbolic functions (area functions)
(sinh−1 )(x ) = arsinh x = ln(x +
p
x 2 + 1),
Darsinh = Rarsinh = R.
(tanh−1 )(x ) = artanh x =
1 1+x
ln
,
2 1−x
Dartanh =] − 1, 1[
1 x +1
ln
,
2 x −1
Darcoth =] − ∞, −1[∪]1, ∞[
(coth−1 )(x ) = arcoth x =
Pasi Ruotsalainen
ACM, University of Oulu
77 / 175
Inverse trigonometric functions (arcus
functions)
Function f (x ) = sin x , x ∈ R, does not have inverse function.
Why?
Pasi Ruotsalainen
ACM, University of Oulu
78 / 175
Consider function f (x ) = sin x , x ∈ [− π2 , π2 ]. Function f is strictly
increasing, so the inverse function exists.
x ∈ [− π2 , π2 ], y ∈ [−1, 1] : y = sin x if and only if x = arcsin y
f −1 (x ) = arcsin x , Darcsin = Rf = [−1, 1], Rarcsin = Df = [− π2 , π2 ]
Pasi Ruotsalainen
ACM, University of Oulu
79 / 175
Let f (x ) = cos x , x ∈ [0, π].
x ∈ [0, π], y ∈ [−1, 1] : y = cos x if and only if x = arccos y
f −1 (x ) = arccos x , Darccos = Rf = [−1, 1], Rarccos = Df = [0, π]
Inverse of function f (x ) = tan x , x ∈] − π2 , π2 [:
x ∈] − π2 , π2 [, y ∈ R : y = tan x if and only if x = arctan y
f −1 (x ) = arctan x , Darctan = R, Rarctan =] − π2 , π2 [.
Pasi Ruotsalainen
ACM, University of Oulu
80 / 175
Pasi Ruotsalainen
ACM, University of Oulu
81 / 175
Calculus I
Chapter 5, Limits and continuity
Pasi Ruotsalainen
Applied and Computational Mathematics
Limits and continuity
Example Consider the function f (x ) = 3x − 1. Find the values x ,
x ̸= 1, for which it holds
|f (x ) − 2| < ε,
a) ε = 1
b) ε = 0.1
c) ε = 0.01.
y=
3
x-
1
when
Pasi Ruotsalainen
ACM, University of Oulu
83 / 175
Definition Let f be defined on the interval ]x0 − r , x0 + r [, where
0 < r ≤ ∞, except possibly at x0 itself. If for every given ε > 0,
there exists δ > 0 such that,
(1)
|f (x ) − b| < ε always, when 0 < |x − x0 | < δ,
then the number b is called the limit of f at the point x0 and is
notated
lim f (x ) = b,
x →x0
Pasi Ruotsalainen
or
f (x ) → b, when x → x0 .
ACM, University of Oulu
84 / 175
Pasi Ruotsalainen
ACM, University of Oulu
85 / 175
If we substitute, in the equation (1), the condition
0 < |x − x0 | < δ with one-sided condition 0 < x − x0 < δ
(respectively −δ < x − x0 < 0), then b is the right limit
(respectively left limit) of f at the point x0 . This is denoted
lim f (x ) = b
x →x0+
(respectively lim f (x ) = b).
x →x0−
In this case, the interval on which f is defined could be
]x0 , x0 + r [ or [x0 , x0 + r [ (respectively ]x0 − r , x0 [ or ]x0 − r , x0 ]).
Superscript notation + (or –) could be left out if it is obvious
that we are talking about one-sided limit.
Pasi Ruotsalainen
ACM, University of Oulu
86 / 175
Limit is unique
Theorem 5.1 If lim f (x ) = a and lim f (x ) = b, then a = b.
x →x0
x →x0
Moreover
lim f (x ) = b if and only if lim f (x ) = b and lim+ f (x ) = b.
x →x0
Pasi Ruotsalainen
x →x0−
ACM, University of Oulu
x →x0
87 / 175
Properties of limits
Theorem 5.2 lim (ax + b) = ax0 + b (a, b ∈ R)
x →x0
Theorem 5.3 If the limits on the right-hand side are defined,
then
1. lim (f + g)(x ) = lim f (x ) + lim g(x )
x →x0
x →x0
x →x0
2. lim (fg)(x ) = lim f (x ) lim g(x )
x →x0
x →x0
x →x0
3. lim (cf )(x ) = c lim f (x ), c ∈ R
x →x0
x →x0
4. lim
f (x )
x →x0 g(x )
lim f (x )
=
x →x0
lim g(x )
.
x →x0
5. If f (x ) ≥ 0 and lim f (x ) = b ≥ 0, then lim
x →x0
x →x0
p
f (x ) =
√
b.
6. If lim f (x ) = b, then lim |f (x )| = |b|.
x →x0
Pasi Ruotsalainen
x →x0
ACM, University of Oulu
88 / 175
Result If P(x ) and Q(x ) are polynomials, Q(x0 ) ̸= 0. Then
(a)
lim P(x ) = P(x0 ),
x →x0
(b)
P(x )
P(x0 )
=
.
x →x0 Q(x )
Q(x0 )
lim
Note! If P(x0 ) = Q(x0 ) = 0, we divide both numerator and
P(x )
denominator of fraction Q(x
) with factor (x − x0 ). Then we can
possibly use result (b).
Theorem 5.4 (Squeeze Theorem) If f (x ) ≤ h(x ) ≤ g(x ),
x ∈ (x0 − r , x0 + r ), and lim f (x ) = b = lim g(x ), then
x →x0
x →x0
lim h(x ) = b.
x →x0
Pasi Ruotsalainen
ACM, University of Oulu
89 / 175
Theorem 5.5 lim sinx x = 1
x →0
Pasi Ruotsalainen
ACM, University of Oulu
90 / 175
Properties of infinity
In the set of extended real numbers R = R ∪ {−∞} ∪ {∞} the
usual arithmetic operations of R can be extended as follows
(when x ∈ R)
−∞ < x < ∞,
x − ∞ = −∞,
x · (±∞) = ∓∞, x < 0,
∞ + ∞ = ∞,
∞ · ∞ = ∞,
−∞ · ∞ = −∞,
Pasi Ruotsalainen
x + ∞ = ∞,
x · (±∞) = ±∞, x > 0,
x
= 0,
±∞
−∞ − ∞ = −∞,
∞ · (−∞) = −∞,
−∞ · (−∞) = ∞.
ACM, University of Oulu
91 / 175
Note! Indeterminate forms
0
∞ ∞
, 0 · ∞, ∞ − ∞,
,
, ∞0 , 00 , 0∞ , 1∞ .
0
∞ 0
Pasi Ruotsalainen
ACM, University of Oulu
92 / 175
Limits at infinity
If for every ε > 0 there is an N so that
whenever x > N, it follows |f (x ) − b| < ε,
and we notate
lim f (x ) = b, or f (x ) → b, when x → ∞.
x →∞
If for every ε > 0 there is an N so that
whenever x < N, it follows |f (x ) − b| < ε,
and we notate
lim f (x ) = b, or f (x ) → b, when x → −∞.
x →−∞
Theorem 5.6
lim
a
x →±∞ x
Pasi Ruotsalainen
=0
for every
ACM, University of Oulu
a ∈ R.
93 / 175
Infinite limits
If for every K ∈ R there is an δ > 0 so that
whenever 0 < |x − x0 | < δ, it follows f (x ) > K
(respectively f (x ) < K ), then
lim f (x ) = ∞
x →x0
(respectively lim f (x ) = −∞).
x →x0
Corresponding one-sided infinite limits are defined by substituting
condition 0 < |x − x0 | < δ with condition 0 < x − x0 < δ or with
condition −δ < x − x0 < 0.
Pasi Ruotsalainen
ACM, University of Oulu
94 / 175
If for every K ∈ R there is an N, so that
whenever x > N, it follows f (x ) > K
and we notate
lim f (x ) = ∞.
x →∞
Respectively, we define
lim f (x ) = −∞, lim f (x ) = ∞, lim f (x ) = −∞.
x →∞
x →−∞
x →−∞
Rules given in Theorem 5.3 also apply to limits at infinity and
infinite limits, if calculations needed are defined in R.
Pasi Ruotsalainen
ACM, University of Oulu
95 / 175
Theorem 5.7 Let x0 ∈ R or x0 = ∞ or x0 = −∞. If
lim f (x ) = ∞ (or −∞), then
x →x0
a
lim
= 0, a ∈ R.
x →x0 f (x )
Moreover, if f (x ) → 0+ , when x → x0 , it holds
lim
1
x →x0 f (x )
= ∞,
and respectively, if f (x ) → 0− , when x → x0 , it holds
lim
1
x →x0 f (x )
Pasi Ruotsalainen
= −∞.
ACM, University of Oulu
96 / 175
Continuity of function
Example Study a piecewise function f
f (x ) =
Pasi Ruotsalainen

3

x − x ,
x ̸= ±1

1,
x = 1 or x = −1.
1 − x2
ACM, University of Oulu
97 / 175
Definition Function f is continuous at point x0 ∈ Df if
lim f (x ) = f (x0 ).
x →x0
Function f is continuous on an open interval ]a, b[ if f is
continuous at every point x ∈]a, b[.
Function f is right continuous (left continuous) at point x0 ∈ Df
if
lim f (x ) = f (x0 )
x →x0+
(respectively lim f (x ) = f (x0 )).
x →x0−
Function f is continuous on a closed interval [a, b] if f is
continuous on the interval ]a, b[ and right continuous at the point
a and left continuous at the point b.
Function f is continuous if f is continuous at every point x0 ∈ Df .
Pasi Ruotsalainen
ACM, University of Oulu
98 / 175
Continuous functions
From the properties of limits (Theorem 5.3) we get:
Theorem 5.8 If functions f and g are continuous
at a point x0 ,
√
f
then f + g, fg, g (if g(x0 ) ̸= 0), |f | and f (f (x0 ) > 0) are
continuous at the point x0 .
Results
a) f1 + · · · + fn and f1 f2 · · · fn are continuous, if each fi is
continuous
b) polynomial P(x ) = a0 + a1 x + . . . + an x n is continuous
c) rational function R(x ) is continuous in its domain
d) trigonometric functions sin x , cos x , tan x , cot x are continuous.
Pasi Ruotsalainen
ACM, University of Oulu
99 / 175
Theorem 5.9 If f is continuous at a point x0 and g is continuous
at the point f (x0 ), then the composite function g(f (x )) is
continuous at the point x0 .
Theorem 5.10 If lim f (x ) = b ∈ R, where x0 ∈ R or x0 = ±∞
x →x0
and if g is continuous at the point b, then
lim g(f (x )) = g( lim f (x )) = g(b).
x →x0
Pasi Ruotsalainen
x →x0
ACM, University of Oulu
100 / 175
Theorem 5.11 Exponential function f (x ) = ax , a > 0, a ̸= 1, is
continuous.
Theorem 5.12 Strictly increasing (decreasing) continuous
function has a strictly increasing (respectively decreasing)
continuous inverse function.
It follows that
Theorem 5.13 Arcus functions are continuous. Logarithmic
function loga x (a > 0, a ̸= 1) is continuous.
Theorem 5.14 Power function f (x ) = x r , x > 0, r ∈ R, is
continuous.
Pasi Ruotsalainen
ACM, University of Oulu
101 / 175
Calculus I
Chapter 6, Derivatives and complex numbers
Pasi Ruotsalainen
Applied and Computational Mathematics
Derivative
Function f , defined over an interval ]x0 − r , x0 + r [, r > 0, is
differentiable at x0 , if the limit
(∗)
f ′ (x0 ) = lim
x →x0
f (x ) − f (x0 )
x − x0
exists. Value f ′ (x0 ) is called derivative of the function f at x0 .
Function f is differentiable on an open interval ]a, b[, if f ′ (x )
exists for every x ∈]a, b[.
Pasi Ruotsalainen
ACM, University of Oulu
103 / 175
df
Notations f ′ (x0 ) = (Df )(x0 ) = dx
x =x0
The equation (∗) can also be written in the form
f (x0 + h) − f (x0 )
.
h→0
h
f ′ (x0 ) = lim
Geometric interpretation: When x → x0 , then the slope of the line
f (x ) − f (x0 )
S,
→ f ′ (x0 ), which is the slope of the tangent line
x − x0
T touching the curve at the point (x0 , f (x0 )). Line
y = f (x0 ) + f ′ (x0 )(x − x0 )
is the tangent line of the curve y = f (x ) passing through the
point (x0 , f (x0 )).
Pasi Ruotsalainen
ACM, University of Oulu
104 / 175
h=5
h=10
Pasi Ruotsalainen
ACM, University of Oulu
105 / 175
If f (t) represents how the value of f depends on time t, then the
derivative f ′ (t) is interpreted as the rate of change of the
function f .
Example Let T (t) be the temperature of an object at an instant
t. Then T ′ (t) represents the rate of change of the temperature of
the object.
Pasi Ruotsalainen
ACM, University of Oulu
106 / 175
The tangent line of the curve y = f (x ) passing through the point
(x◦ , f (x◦ ))
y − f (x◦ ) = f ′ (x◦ )(x − x◦ ).
Example The tangent line of the curve f (x ) = −2x 2 passing
through a) the point (1, −2), b) the point (−1, 0).
b)
a)
•
•
Pasi Ruotsalainen
ACM, University of Oulu
107 / 175
Derivation rules and derivatives of
elementary functions
Theorem 6.1
If the function f is differentiable at the point x0 , then f is
continuous at this point
Note! This is not true in the other direction. For example, the
function f (x ) = |x | is continuous everywhere, but not
differentiable at the point x = 0.
Pasi Ruotsalainen
ACM, University of Oulu
108 / 175
Theorem 6.2 If f and g are differentiable at the point x , then
(D1)
(f + g)′ (x ) = f ′ (x ) + g ′ (x )
(D2)
(f − g)′ (x ) = f ′ (x ) − g ′ (x )
(D3)
(fg)′ (x ) = f ′ (x )g(x ) + f (x )g ′ (x )
(D4)
(cf )′ (x ) = cf ′ (x )
(D5)
f ′ (x )g(x ) − f (x )g ′ (x )
f
, g(x ) ̸= 0.
( )′ (x ) =
g
(g(x ))2
Pasi Ruotsalainen
ACM, University of Oulu
109 / 175
Derivatives of elementary functions
(1)
Dc ≡ 0
(2)
Dx ≡ 1
(3)
Dx r = rx r −1 , r ∈ R
(4)
D sin x = cos x
(5)
D cos x = − sin x
1
D tan x =
= 1 + tan2 x
cos2 x
1
D cot x = − 2 = −(1 + cot2 x )
sin x
1
D ln x =
x
D ex = ex
(6)
(7)
(8)
(9)
Pasi Ruotsalainen
ACM, University of Oulu
110 / 175
1
1 − x2
1
(11) D arccos x = − √
1 − x2
1
(12) D arctan x =
1 + x2
(13) D sinh x = cosh x
(10) D arcsin x = √
(14) D cosh x = sinh x
1
(15) D tanh x =
= 1 − tanh2 x
cosh2 x
1
(16) D coth x = −
sinh2 x
(17) Dax = ax ln a
Pasi Ruotsalainen
ACM, University of Oulu
111 / 175
Derivative of composite function
Theorem 6.3 (Chain rule) Let f be differentiable at the point
x0 and let g be differentiable at the point y0 = f (x0 ). Then
function (g ◦ f )(x ) = g(f (x )) is differentiable at the point x0 and
(g ◦ f )′ (x0 ) = g ′ (f (x0 ))f ′ (x0 ).
Pasi Ruotsalainen
ACM, University of Oulu
112 / 175
Proof Let y = f (x ) and y0 = f (x0 ) such that y ̸= y0 . Then
(g ◦ f )(x ) − (g ◦ f )(x0 )
g(f (x )) − g(f (x0 )) f (x ) − f (x0 )
=
·
x − x0
f (x ) − f (x0 )
x − x0
g(y ) − g(y0 ) f (x ) − f (x0 )
·
=
y − y0
x − x0
→ g ′ (f (x0 ))f ′ (x0 ),
when x → x0 , for then by the continuity of f f (x ) → f (x0 ) and
y → y0 .
If y = y0 , x ̸= x0 , then
lim
x →x0
(g ◦ f )(x ) − (g ◦ f )(x0 )
g(y ) − g(y0 )
= lim
x
→x
0
x − x0
x − x0
0
= lim
= 0 = g ′ (f (x0 )) · 0.
x →x0 x − x0
Pasi Ruotsalainen
ACM, University of Oulu
113 / 175
Derivative of the inverse function
Theorem 6.4 Let f be differentiable at the point x0 and
f ′ (x0 ) ̸= 0. If there exists f −1 which is continuous at the point
y0 = f (x0 ), then
1
.
(f −1 )′ (y0 ) = ′
f (x0 )
Proof Denote y = f (x ) if and only if x = f −1 (y ), and let
y ̸= y0 (= f (x0 )). Then x ̸= x0 and
f −1 (y ) − f −1 (y0 )
x − x0
1
=
= f (x )−f (x ) .
0
y − y0
f (x ) − f (x0 )
x −x0
f −1 is continuous at the point y0 = f (x0 ), so when y → y0 , then
f −1 (y ) → f −1 (y0 ), which means that x → x0 . We get
1
1
1
f −1 (y ) − f −1 (y0 )
lim
= lim f (x )−f (x ) =
= ′
.
f (x )−f (x0 )
0
y →y0
x →x0
y − y0
f
(x
0)
lim
x −x0
Pasi Ruotsalainen
ACM, University of Oulu
x →x0
x −x0
114 / 175
Higher derivatives
If f ′ (x ) is differentiable, its derivative is called second derivative
of f and denoted f ′′ (x ):
f ′ (x + h) − f ′ (x )
.
h→0
h
f ′′ (x ) = lim
It can also be denoted
f ′′ (x ) = D 2 f (x ) =
d 2 f (x )
d 2y
=
= y ′′ .
dx 2
dx 2
The nth derivative of function f is f (n) , n ∈ N
f (n) (x ) = D n f (x ) =
Pasi Ruotsalainen
d ny
d n f (x )
=
= y (n) .
dx n
dx n
ACM, University of Oulu
115 / 175
One-sided derivative
The limit
lim+
h→0
f (x + h) − f (x )
= f+′ (x )
h
( lim
h→0−
f (x + h) − f (x )
= f−′ (x ))
h
is called the right (respectively, left) derivative of f at the point x .
One-sided derivative can exist at a point where the function is not
differentiable. If function f is differentiable at the point x0 , then
f+′ (x0 ) = f−′ (x0 ) = f ′ (x0 ).
Pasi Ruotsalainen
ACM, University of Oulu
116 / 175
Important theorems on differentiation
Theorem 6.5 (Rolle’s Theorem) If f is continuous on the
interval [a, b] and differentiable on the open interval ]a, b[ and if
f (a) = f (b), then f ′ (t) = 0 for some t ∈]a, b[.
•
g'(t)=0
g(a)=g(b)
a
Pasi Ruotsalainen
b
ACM, University of Oulu
117 / 175
Proof
1) If f (x ) ≡ C (= f (a) = f (b)), then f ′ (x ) = 0 for every x ∈]a, b[.
2) If f (x ) > f (a) for some x ∈]a, b[, then there is t ∈]a, b[, such
that f (x ) ≤ f (t) for every x ∈ [a, b] (continuous function on a
closed interval [a, b] attains its minimum and maximum value).
Then
≤0
≤0
z
}|
{
z
}|
{
f (x ) − f (t)
f (x ) − f (t)
= f ′ (t) = lim+
≤ 0,
0 ≤ lim
x| {z
− t}
x| {z
− t}
x →t
x →t −
<0
>0
so we have f ′ (t) = 0.
3) If f (x ) < f (a) for some x ∈]a, b[, then proof is similar with the
earlier one.
Pasi Ruotsalainen
ACM, University of Oulu
118 / 175
Mean value theorem
Theorem 6.6 (Mean value theorem) If f is continuous on an
interval [a, b] and differentiable on an interval ]a, b[, then there is
t ∈]a, b[, with
f (b) − f (a) = f ′ (t)(b − a).
f (b) − f (a)
x . This
b−a
function satisfies the Rolle’s theorem, which means that there is
at least one t ∈]a, b[ such that
Proof Lets define a function g(x ) = f (x ) −
g ′ (t) = 0, if and only if f ′ (t) −
Pasi Ruotsalainen
ACM, University of Oulu
f (b) − f (a)
= 0.
b−a
119 / 175
Pasi Ruotsalainen
ACM, University of Oulu
120 / 175
Behavior of a function
Let f be differentiable on an interval I ⊂ R. Then on the interval
I function f is
▶ increasing, if f ′ (x ) ≥ 0 for every x ∈ I,
▶ strictly increasing, if f ′ (x ) > 0 for every x ∈ I,
▶ decreasing, if f ′ (x ) ≤ 0 for every x ∈ I,
▶ strictly decreasing, if f ′ (x ) < 0 for every x ∈ I.
Proof (for the first part): Let x1 , x2 ∈ I and x1 < x2 . Based on
the mean value theorem there is t ∈]x1 , x2 [, such that
f (x2 ) − f (x1 ) = f ′ (t)(x2 − x1 ).
Because f ′ (x ) ≥ 0, then f ′ (t) ≥ 0 and furthermore x2 − x1 > 0,
so f (x2 ) − f (x1 ) ≥ 0, which means that f (x2 ) ≥ f (x1 ). So, we get
that f is increasing on the interval I.
Pasi Ruotsalainen
ACM, University of Oulu
121 / 175
In a point x0 of its domain, function f has
▶ local maximum , if is there is r > 0, so that f (x ) ≤ f (x0 ) for
every x ∈]x0 − r , x0 + r [,
▶ local minimum , if is there is r > 0, so that f (x ) ≥ f (x0 ) for
every x ∈]x0 − r , x0 + r [,
▶ local extremum, if at the point x0 there is either local
minimum or maximum .
Local extremum is strict, if equality is valid only if x = x0 .
If f ′ (x0 ) = 0, it is said that x0 is critical point of the function f .
Then, if f ′ (x0 ) = 0 and
▶ f ′′ (x0 ) > 0, then x0 is strict local minimum point,
▶ f ′′ (x0 ) < 0, then x0 is strict local maximum point,
▶ f ′′ (x0 ) = 0, then other results must be used to examine the
quality of the critical point.
Pasi Ruotsalainen
ACM, University of Oulu
122 / 175
L’Hospital’s rule
Theorem 6.7 (L’Hospital’s rule ) Let f and g be differentiable
in the neighborhood of the point x0 ∈ R, with g ′ (x ) ̸= 0. If
(i) lim f (x ) = 0 and lim g(x ) = 0 or
x →x0
x →x0
(ii) lim f (x ) = ∞ and lim g(x ) = ∞, then
x →x0
x →x0
f ′ (x )
f (x )
= lim ′
x →x0 g (x )
x →x0 g(x )
lim
if the right hand side limit exists (or is infinite). This result is also
true when x0 = ∞ or x0 = −∞.
Pasi Ruotsalainen
ACM, University of Oulu
123 / 175
L’Hospital’s rule is useful when in those situations, where limit
lim f (x )
f (x )
=
is of indeterminate form
lim
g(x )
lim g(x )
0
0
or
∞
.
∞
Note Other indeterminate forms 0 · ∞ or ∞ − ∞ may often be
manipulated into forms 00 or ∞
∞ using substitution.
Note Forms ∞0 , 00 or 1∞ can be changed into limit of form
0 · ∞ using equation
[f (x )]g(x ) = eg(x ) ln f (x ) .
Pasi Ruotsalainen
ACM, University of Oulu
124 / 175
Pasi Ruotsalainen
ACM, University of Oulu
125 / 175
Implicit differentiation
Example Let a function y = f (x ) be defined with an equation
e2y − x = 0, x > 0. Evaluate f ′ (x ).
Solution e2y − x = 0 2y = ln x y = 21 ln x = f (x )
We get
1
1 1
, x > 0.
f ′ (x ) = · =
2 x
2x
Often, differentiation is practically easiest to do implicitly as
follows. Because y is a function of x , y = y (x ), it holds
e2y (x ) − x = 0.
Differentiate each side of the equation. Using the derivation rule
of composite function (chain rule) we get
e2y (x ) · 2y ′ (x ) − 1 = 0 ⇒ y ′ (x ) =
Pasi Ruotsalainen
ACM, University of Oulu
1
2 e2y (x )
=
1
.
2x
126 / 175
Parametric equation
Example Sketch on the xy -plane the curve defined by equations
(
x =t +1
y = (t + 1)(t − 1), t ∈ [−2, 2]
Identify the curve.
Take
(
(1)
x = u(t)
y = v (t), t ∈ I ⊂ R,
where u and v are continuous functions on an interval I. Value
t ∈ I makes up a point (x , y ) = (u(t), v (t)). The collection of
points {(x (t), y (t)) | t ∈ I} is called curve, t ∈ I is parameter and
equations (1) are the parametric equations of a curve.
Pasi Ruotsalainen
ACM, University of Oulu
127 / 175
When does (1) define y as a function of x , i.e. y = f (x )?
Let u : I → R an injection. Then there is u −1 : Ru → I, and
t = u −1 (x ). Furthermore,
=f
y = v (u
−1
z
}|
{
(x )) = (v ◦ u −1 )(x ) = f (x ),
where f = v ◦ u −1 is a continuous function: Ru → Rv .
In the previous example
u(t) = t + 1, u −1 (x ) = x − 1, Ru = [−1, 3].
Note! If v : I → R is an injection, then there is v −1 : Rv → I,
and t = v −1 (y ). Furthermore
x = u(t) = u(v −1 (y )) = (u ◦ v −1 )(y ) = g(y ).
Pasi Ruotsalainen
ACM, University of Oulu
128 / 175
When y = f (x ) can be(represented in parametric form (1)?
x =t
ALWAYS! For example
y = f (t), t ∈ Df .
Example Identify the curves
(
a)
x = cos t
y = sin t, 0 ≤ t < 2π.
(
c)
Pasi Ruotsalainen
(
b)
x = cosh t
y = sinh t, t ∈ R.
x =1−t
y = t + 2, t ∈ R.
ACM, University of Oulu
129 / 175
Derivative of a function defined by
parametric equations (1)
Let u : I → R be an injection, such that parametric equations (1)
define a function
y = (v ◦ u −1 )(x ) = f (x ), x ∈ Ru .
Suppose u and v to be differentiable. Then according the chain
rule of the derivative of a composite function we get
dy
= f ′ (x ) = v ′ (u −1 (x ))(u −1 )′ (x ).
dx
Because (u −1 )′ (x ) = u′1(t) , when u ′ (t) ̸= 0 (derivative of a inverse
function), we get
dy
v ′ (t)
= f ′ (x ) = ′ , t ∈ I.
dx
u (t)
Pasi Ruotsalainen
ACM, University of Oulu
130 / 175
If (slightly inconsistent) notations are used
(
(2)
x = x (t)
y = y (t),
it can be written formally (provided that x ′ (t) ̸= 0)
dy
y ′ (t)
dy
dt
= dx
= ′ .
dx
x (t)
dt
Pasi Ruotsalainen
ACM, University of Oulu
131 / 175
Pasi Ruotsalainen
ACM, University of Oulu
132 / 175
Polar coordinates
Let r ≥ 0 be the distance of a point P = (x , y ) from the origin
O = (0, 0), and φ ∈ [0, 2π[ (convention!) the angle between the
vector OP and the positive x -axis.
Polar coordinates (r , φ) can be converted to the
(x , y )-coordinates using equations
(
x = r cos φ
y = r sin φ.
Pasi Ruotsalainen
ACM, University of Oulu
133 / 175
(x , y )-coordinates ((x , y ) ̸= (0, 0)) can be converted to unique
polar coordinates (r , φ) using equations:

p

x2 + y2
r
=



x
cos φ = √ 2
2
x +y



sin φ = √ y
.
x 2 +y 2
Pasi Ruotsalainen
ACM, University of Oulu
134 / 175
Pasi Ruotsalainen
ACM, University of Oulu
135 / 175
Complex numbers
The set of complex numbers C is defined as a set of ordered pairs
of real numbers, where addition and multiplication are defined as
follows:
If z1 = (a, b) and z2 = (c, d), a, b, c, d ∈ R then
1. z1 + z2 = (a + c, b + d)
2. z1 · z2 = (ac − bd, ad + bc).
Thus z1 = z2 if and only if a = c and b = d.
We associate the set of real numbers R and the set
{(a, 0) | a ∈ R}, (a, 0) = a, a ∈ R, so that
a(c, d) = (a, 0)(c, d) = (ac, ad).
Denote
i = (0, 1) = 0 + 1 · i
where i is the imaginary unit.
Pasi Ruotsalainen
ACM, University of Oulu
136 / 175
Then a = (a, 0) = a + 0 · i, (0, b) = b(0, 1) = bi = 0 + bi
and furthermore (a, b) = (a, 0) + (0, b) = a + bi.
Using the product
i 2 = (0, 1) · (0, 1) = (0 · 0 − 1 · 1, 0 · 1 + 1 · 0) = (−1, 0) = −1,
we get the addition and multiplication of complex numbers a + bi
and c + di
a + bi + c + di = a + c + (b + d)i
(a + bi)(c + di) = ac + adi + bci − bd = ac − bd + (ad + bc)i.
Therefore, complex numbers can also be defined as
C = {a + bi | a, b ∈ R, i 2 = −1},
where similar assumptions as those of R apply.
Pasi Ruotsalainen
ACM, University of Oulu
137 / 175
Complex number z = a + ib
▶ real part: Re z = a
▶ imaginary part: Im z = b
▶ conjugate: z = a − ib
▶ inverse (reciprocal): z −1 =
a
ib
1
= 2
− 2
,
2
z
a +b
a + b2
z = (a, b) ̸= (0, 0)
▶ absolute value (modulus or magnitude):
√
√
|z| = z z = a2 + b 2
Pasi Ruotsalainen
ACM, University of Oulu
138 / 175
Properties of conjugate
z + w = z + w,
zw =zw
z =z
Properties of absolute value
|z| ≥ 0 and |z| = 0 if and only if z = 0 = (0, 0)
|z| = |z|, |Re z| ≤ |z|, |Im z| ≤ |z|
|zw | = |z||w |, |z + w | ≤ |z| + |w |
Pasi Ruotsalainen
ACM, University of Oulu
139 / 175
Polar form of complex numbers
Complex number z = a + ib and a point of real plane (a, b) can
be associated (see previous slides for polar coordinates).
Polar coordinate representation of a complex number z = a + ib
with absolute value r ≥ 0 and argument (or angle) φ ∈ [0, 2π[
(convention):
z = r (cos φ + i sin φ).
Pasi Ruotsalainen
ACM, University of Oulu
140 / 175
Euler’s and De Moivre’s formulas
On the course Calculus 2 it will be shown that there are power
series expansions for exponential, sine and cosine functions,
x ∈ R,
ex =
cos x =
sin x =
∞
X
xk
k=0
∞
X
k!
=1+x +
x2 x3 x4 x5 x6 x7
+
+
+
+
+
+ ··· ,
2!
3!
4!
5!
6!
7!
(−1)k x 2k
x2 x4 x6 x8
=1−
+
−
+
− ··· ,
(2k)!
2!
4!
6!
8!
k=0
∞
X
(−1)k x 2k+1
k=0
Pasi Ruotsalainen
(2k + 1)!
=x−
x3 x5 x7
+
−
+ ··· .
3!
5!
7!
ACM, University of Oulu
141 / 175
Substitute x = iφ in the series of ex :
iφ
e
=
∞
X
(iφ)k
k=0
k!
2
3
4
5
6
7
(iφ)
(iφ)
(iφ)
(iφ)
(iφ)
= 1 + iφ + (iφ)
2! + 3! + 4! + 5! + 6! + 7! + · · ·
2
3
4
5
6
7
8
φ
iφ
φ
iφ
φ
= 1 + iφ − φ2! − iφ
3! + 4! + 5! − 6! − 7! + 8! + · · ·
2
4
6
3
5
7
= 1 − φ2! + φ4! − φ6! + · · · + i φ − φ3! + φ5! − φ7! + · · ·
=
∞
X
(−1)k φ2k
(2k)!
k=0
Pasi Ruotsalainen
+i
∞
X
(−1)k φ2k+1
(2k+1)!
.
k=0
ACM, University of Oulu
142 / 175
This gives us formally so called Euler’s formula
eiφ = cos φ + i sin φ,
which is set as the definition. From this and the polar coordinate
representation of complex numbers we get the exponential form
of a complex number
z = r eiφ .
Furthermore, the mathematical induction and trigonometric
formulas can be used to prove De Moivre’s formula
(cos(φ) + i sin(φ))n = cos(nφ) + i sin(nφ), n ∈ N.
Pasi Ruotsalainen
ACM, University of Oulu
143 / 175
Solving the equation z n = w
Use the exponential form for numbers w = ρ eiα and z = r eiφ .
The equation can now be written
r n einφ = ρ eiα ,
from which we further get two equations to solve the absolute
value r and the argument φ of the roots zk , k = 0, . . . , n − 1
(
rn = ρ
nφ = α + k2π, k ∈ Z
Pasi Ruotsalainen
or

√
r = n ρ
φ = α + k 2π , k = 0, . . . , n − 1.
n
ACM, University of Oulu
n
144 / 175
Theorem 6.8 (Fundamental Thm. of algebra) every equation
an z n + an−1 z n−1 + · · · + a1 z + a0 = 0, ak ∈ C, k = 0, 1, . . . , n,
n ∈ N, has (at least) one root in C.
Corollary
1. Every polynomial with complex coefficents (ak ∈ C)
Pn (z) = an z n + an−1 z n−1 + · · · + a1 z + a0
can be factored into a product of first order polynomials, i.e.
Pn (z) = an (z − z1 ) · . . . · (z − zn ), zk ∈ C.
2. Every polynomial with real coefficents Pn (x )(ak ∈ R) can be
factored into a product of second order polynomials, i.e.
Pn (x ) = an (x −x1 )·. . .·(x −xr )(x 2 +c1 x +d1 )·. . .·(x 2 +cs x +ds ),
where r + 2s = n.
Pasi Ruotsalainen
ACM, University of Oulu
145 / 175
Calculus I
Chapter 7, Integrals
Pasi Ruotsalainen
Applied and Computational Mathematics
Integral calculus
Theorem 7.1 If the function f is continuous on an interval [a, b]
with f ′ (x ) = 0 for every x ∈]a, b[, then the function f is
constant, f (x ) ≡ C for every x ∈ [a, b].
Proof Let x0 , x ∈ [a, b] such that x0 < x . According the Mean
value theorem there is t ∈]x0 , x [ such that
f (x ) − f (x0 ) = f ′ (t)(x − x0 ).
Because f ′ (t) = 0, then f (x ) − f (x0 ) = 0 for every x0 , x ∈ [a, b]
which means that f (x ) = f (x0 ) for every x0 , x ∈ [a, b]. So f (x ) is
constant for every x ∈ [a, b].
Theorem 7.2 If f and g are continuous on an interval [a, b] and
f ′ = g ′ on the interval ]a, b[, then
f (x ) = g(x ) + C , x ∈ [a, b].
Pasi Ruotsalainen
ACM, University of Oulu
147 / 175
Integral function
Integral function F of a function f is a function with property
F′ = f .
Integral function is also notated
Z
f (x ) dx .
Integral function is unique up to a constant (Theorem 7.2) and as
a differentiable function it is continuous on its domain (Theorem
6.1).
According the rules of differentiation (linearity) we have
Z
Z
[f (x ) + g(x )] dx =
Z
Z
cf (x ) dx = c
Pasi Ruotsalainen
Z
f (x ) dx +
g(x ) dx ,
f (x ) dx , c ∈ R.
ACM, University of Oulu
148 / 175
Integrals of elementary functions
Z
(1)
x r dx =
1
x r +1 + C , r ̸= −1
r +1
dx
= ln |x | + C
Z x
ex dx = ex + C
Z
(2)
(3)
Z
(4)
sin x dx = − cos x + C
Z
(5)
Pasi Ruotsalainen
cos x dx = sin x + C
ACM, University of Oulu
149 / 175
dx
= (1 + tan2 x ) dx = tan x + C
cos2 x
Z
Z
dx
= (1 + cot2 x ) dx
2
sin x
Z
dx
√
= arcsin x + C = − cot x + C
1 − x2
Z
dx
= arctan x + C
1 + x2
Z
(6)
(7)
(8)
(9)
Pasi Ruotsalainen
Z
ACM, University of Oulu
150 / 175
From differentiation rule of composite function (chain rule)
Dg(f (x )) = g ′ (f (x ))f ′ (x ) we get
Z
g ′ (f (x ))f ′ (x ) dx = g(f (x )) + C
Z
f ′ (x )
dx = ln |f (x )| + C
f (x )
Z
ef (x ) f ′ (x ) dx = ef (x ) + C
(10)
(11)
(12)
Z
(13)
sinh x dx = cosh x + C
Z
(14)
cosh x dx = sinh x + C
Z
(15)
Pasi Ruotsalainen
Z
tan x dx =
sin x
dx = − ln | cos x | + C
cos x
ACM, University of Oulu
151 / 175
Definite integral
Let f be a function defined on an interval [a, b] and bounded by a
constant M > 0 such that
|f (x )| ≤ M always, when x ∈ [a, b].
Furthermore, let {xk }nk=0 be a partition of the interval [a, b], so
that a = x0 < x1 < x2 < . . . < xn = b. From each subinterval
[xk−1 , xk ] we can choose a evaluation point tk . The sum
In =
n
X
f (tk )(xk − xk−1 )
k=1
is called the Riemann sum of the function f .
Pasi Ruotsalainen
ACM, University of Oulu
152 / 175
Kuva: Lower and upper Riemann sums of the function 2t − t 2 on the
interval t ∈ [0, 2].
Pasi Ruotsalainen
ACM, University of Oulu
153 / 175
We add points xk so that the length of the longest subinterval
max (xk − xk−1 ) → 0. If there exists I ∈ R such that In → I,
1≤k≤n
independent of choice of the partition points xk and the
evaluation points tk , it is said that f is integrable on the interval
[a, b] and I is the definite integral of f on [a, b]. We notate
Zb
I=
f (x ) dx .
a
If f (x ) ≥ 0, then
Rb
f (x ) dx is the area bounded by the graph
a
y = f (x ) and the x -axis on the interval [a, b].
Pasi Ruotsalainen
ACM, University of Oulu
154 / 175
Example Riemann sum of the constant function f (x ) ≡ c is
In =
n
X
f (tk )(xk − xk−1 ) =
k=1
n
X
c(xk − xk−1 )
k=1
= c(x1 − x0 + x2 − x1 + · · · + xn − xn−1 ) = c(b − a)
for every choise of the partition points {xk } and the evaluation
points {tk }. This gives us
Zb
c dx = c(b − a).
a
Pasi Ruotsalainen
ACM, University of Oulu
155 / 175
Example Dirichlet function
(
f (x ) =
1, x rational
0, x irrational
is not integrable on any closed interval [a, b]: divide [a, b] to n
subintervals. If every tk is chosen to be rational we get
In = b − a. If every tk is chosen to be irrational we get In = 0, so
lim In is not independent of the choice of theR evaluation points tk .
(Dirichlet function is Lebesgue-integrable: L f (x ) dx = 0).
Pasi Ruotsalainen
ACM, University of Oulu
156 / 175
Theorem 7.3 If a bounded function f has a finite number
discontinuities on the interval [a, b], then f is integrable on the
interval [a, b].
Corollary Because continuous function is bounded on a closed
interval, function is integrable on every closed interval on which it
is continuous. Thus every elementary function is integrable on
every closed interval of its domain.
Pasi Ruotsalainen
ACM, University of Oulu
157 / 175
We define
Za
Za
f (x ) dx = 0 and
a
f (x ) dx = −
Zb
f (x ) dx , when a < b.
a
b
Corollary If
(
f (x ) =
then
g(x ), x ∈ [a, b]\{c1 , c2 , . . . , cm }
h(x ), x ∈ {c1 , . . . , cm },
Zb
Zb
f (x ) dx =
a
g(x ) dx ,
a
if g is integrable on the interval [a, b].
Pasi Ruotsalainen
ACM, University of Oulu
158 / 175
From the definition of integral it follows
Theorem 7.4 If f and g are integrable on the interval [a, b], then
f + g and αf , α ∈ R, are integrable and
Zb
Zb
Zb
[f (x ) + g(x )] dx =
a
f (x ) dx +
a
Zb
g(x ) dx
a
Zb
(αf )(x ) dx = α
a
f (x ) dx .
a
Furthermore,
Zb
Zc
f (x ) dx =
a
Pasi Ruotsalainen
Zb
f (x ) dx +
a
f (x ) dx , a ≤ c ≤ b.
c
ACM, University of Oulu
159 / 175
Connection between integral and integral
function
Theorem 7.5 If F is the integral function of a function f
integrable on the interval [a, b], then
Zb
f (x ) dx = F (b) − F (a) =
F (x ).
a
a
Pasi Ruotsalainen
.b
ACM, University of Oulu
160 / 175
Corollary If F is the integral function of a function f integrable
on the interval [a, b], then
Zx
F (x ) =
f (t) dt + C , a ≤ x ≤ b (C = F (a)).
a
On the other hand, we have
Theorem 7.6 If f is continuous on the interval [a, b], then the
function F defined with the equation
Zx
F (x ) =
f (t) dt + C ,
a ≤ x ≤ b,
a
is the integral function of the function f on the interval [a, b] i.e.
F ′ (x ) = f (x ), a < x < b.
Pasi Ruotsalainen
ACM, University of Oulu
161 / 175
Partial integration
From the product rule of differentiation,
d
[u(x )v (x )] = u ′ (x )v (x ) + u(x )v ′ (x ),
dx
it follows
Z
d
[u(x )v (x )] dx =
dx
Z
Z
′
u (x )v (x ) dx +
u(x )v ′ (x ) dx ,
which gives the formula of partial integration (integration by
parts):
Z
u(x )v ′ (x ) dx = u(x )v (x ) −
Z
u ′ (x )v (x ) dx .
This formula is valid if these integrals exist (particularly if u ′ and
v ′ are continuous).
Pasi Ruotsalainen
ACM, University of Oulu
162 / 175
Corresponding formula for definite integral is
Zb
′
u(x )v (x ) dx =
.b
u(x )v (x ) −
a
a
Zb
u ′ (x )v (x ) dx .
a
Note! Formula
Z
Z
u(x ) dx =
u(x ) · 1 dx = u(x ) · x −
Z
xu ′ (x ) dx
is useful, for example, when we have to integrate u(x ) = ln x or
u(x ) = arctan x .
Pasi Ruotsalainen
ACM, University of Oulu
163 / 175
Integration by substitution
Example Evaluate the integral
Z
√
sin( x )
√
dx .
x
The aim is to convert the integral to a form that makes the
integral function easier to see.
Pasi Ruotsalainen
ACM, University of Oulu
164 / 175
If function f is continuous on some interval and v ′ is continuous
with v ′ (t) ̸= 0, we can make the substitutions
x = v (t),
dx
= v ′ (t) which gives dx = v ′ (t) dt,
dt
and furthermore
Z
Z
f (x ) dx =
′
Z
f (v (t))v (t) dt =
g(t) dt = G(t) + C
= G(v −1 (x )) + C ,
where G, the integral function of g, is easy to find.
Correspondingly, for definite integral we get
v −1 (b)
Zb
f (x ) dx =
a
Pasi Ruotsalainen
.
G(t) = G(v −1 (b)) − G(v −1 (a)).
v −1 (a)
ACM, University of Oulu
165 / 175
Finding area of plane region
Zb
A=
f (x ) dx , when f (x ) ≥ 0 for every x ∈ [a, b]
a
−A =
Zb
f (x ) dx , when f (x ) ≤ 0 for every x ∈ [a, b]
a
f(x)=e-xcos(4x)
Pasi Ruotsalainen
f(x)=|e-xcos(4x)|
ACM, University of Oulu
166 / 175
Zb
A=
[f (x ) − g(x )] dx , when f (x ) ≥ g(x ) for every x ∈ [a, b]
a
cos(x)
Pasi Ruotsalainen
ACM, University of Oulu
167 / 175
Integrating rational function
On this section we study integrating a rational (fractional)
function
R(x ) =
a n x n + · · · + a1 x + a0
P(x )
, an , bm ̸= 0.
=
m
bm x + · · · + b1 x + b0
Q(x )
If n ≥ m, i.e. if the numerator is of the same or higher degree as
the denominator, we first do the division (for example with
polynomial long division).
In the following, it is assumed that the division has already been
done, i.e. m > n.
Pasi Ruotsalainen
ACM, University of Oulu
168 / 175
Consider the polynomial in the numerator Q(x ).
A. Q(x ) = (x − x1 )m
Partial Fraction Decomposition (PFD) of a rational function R(x )
is
A1
A2
Am
R(x ) =
+
+ ··· +
,
2
x − x1 (x − x1 )
(x − x1 )m
coefficients of which A1 , . . . , Am can be defined (by multiplying to
have a common denominator and comparing the coefficients of
the numerator). Partial fractions are easy to integrate.
Pasi Ruotsalainen
ACM, University of Oulu
169 / 175
B. Q(x ) = bm (x − x1 )k1 (x − x2 )k2 . . . (x − xr )kr ,
where k1 + · · · + kr = m.
Now the partial fraction decomposition is of the form
R(x ) =
A1
Ak1
B1
Bk2
+ ··· +
+
+ ··· +
k
1
x − x1
(x − x1 )
x − x2
(x − x2 )k2
Ckr
C1
+ ··· +
.
+ ··· +
x − xr
(x − xr )kr
Pasi Ruotsalainen
ACM, University of Oulu
170 / 175
As a consequence of the fundamental theorem of algebra
(Theorem 6.8) every polynomial with real-valued coefficients
Q(x ) = bm x m + bm−1 x m−1 + · · · + b1 x + b0
can be written, by using its roots, in the form
C. Q(x )=bm (x −x1 )k1 ···(x −xr )kr (x 2 +c1 x +d1 )h1 ···(x 2 +cs x +ds )hs ,
where none of the factors can be divided with each other and
none the factors of the form x 2 + cx + d can be factored into
polynomials of first degree (c 2 − 4d < 0).
Factors of the form (x 2 + cx + d)k give the partial fraction
decomposition the terms
D2 x + E2
D k x + Ek
D1 x + E1
+ 2
+ ··· + 2
.
x 2 + cx + d
(x + cx + d)2
(x + cx + d)k
Pasi Ruotsalainen
ACM, University of Oulu
171 / 175
Conclusion
In the integral function of a rational function there can be terms
with rational, logarithmic and arctan functions.
Finding the partial fraction decomposition
1. The degree of the numerator has to be less than the degree of
the denominator. Do the division if needed.
2. Factor the denominator into polynomials of first and second
degree.
3. Every factor of the denominator gives the partial fraction
decomposition one or more terms (listed on the next slide).
4. Solve the coefficients A, B, C , . . ..
Pasi Ruotsalainen
ACM, University of Oulu
172 / 175
Factors of the
denominator
x −a
(x − a)2
(x − a)3
..
.
x 2 + ax + b
(indivisible!)
(x 2 + ax + b)2
..
.
Pasi Ruotsalainen
Terms of PFD
(A, B, C , . . . unknown constants)
A
x −a
A
B
+
x − a (x − a)2
A
B
C
+
+
2
x − a (x − a)
(x − a)3
..
.
Dx + E
2
x + ax + b
Dx + E
x 2 + ax + b
+
Fx + G
(x 2 + ax + b)2
..
. ACM, University of Oulu
173 / 175
Calculating volume by integration
Let A(xk ) be the area of the cross-section of the plane x = xk
and a solid.
We divide the interval [a, b] of x-axis into n subintervals by the
partition points {xk } and choose evaluation points tk ∈ [xk−1 , xk ].
The volume of the solid can be estimated with a sum
n
X
A(tk )(xk − xk−1 ),
k=1
which in the case of integrable function A approaches the integral
Rb
A(x ) dx when the partition gets finer. Thus
a
Zb
V =
A(x ) dx .
a
Pasi Ruotsalainen
ACM, University of Oulu
174 / 175
Volume of solid of revolution
If a curve y = f (x ), a ≤ x ≤ b, where f is continuous, rotates
about x -axis, then the volume of the resulting solid is
Zb
V =π
(f (x ))2 dx .
a
Funktion f(x)=2*sqrt(x2-1) pyörähdyskappale, x=1..5.
Pasi Ruotsalainen
ACM, University of Oulu
175 / 175