MASARYK UNIVERSITY
FACULTY OF SCIENCE
DEPARTMENT OF MATHEMATICS AND STATISTICS
Mathematics I
ˇS´arka Uchytilov´a
Brno 2014
Contents
Introduction 1
1 Linear Algebra 2
1.1 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Solving systems using row operations . . . . . . . . . . . . . . . . . . . . . . 4
2 Functions of One Variable 7
2.1 Basic properties of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.3 Rational functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.4 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3 Limits, derivatives, minima and maxima 15
3.1 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.2 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.3 Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.4 L’Hospital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4 Behaviour of a function, Approximation of a function 21
4.1 Concavity and Points of Inflection . . . . . . . . . . . . . . . . . . . . . . . . 21
4.2 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.3 Behaviour of Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
5 Indefinite Integral 26
5.1 Basic Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
5.2 Basic Integration Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
5.3 Integration of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . 34
6 Definite Integral 38
6.1 Definition and Basic Properties of Definite Integral . . . . . . . . . . . . . . 38
6.2 Integration Methods for the Definite Integral . . . . . . . . . . . . . . . . . . 42
6.3 Applications of the Definite Integral . . . . . . . . . . . . . . . . . . . . . . . 43
6.4 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Literature 52
iii
Introduction
This textbook, Mathematics I, was written to meet the needs of the foreign students taking
M1010 Mathematics I and M1020 Mathematics I - seminar at Faculty of Science at Masaryk
University.
Moreover, we hope the textbook helps Czech students to get acquainted with English
Mathematical terminology as well as enables their vocabulary development, further facilitating
their studies abroad.
This textbook builds on the Czech textbook called Matematika pro chemiky, 1. d´ıl written
by prof. RNDr. Zuzana Doˇsl´a, DSc and is supported by the project MUNI/FR/0134/2014.
1
Chapter 1
Linear Algebra
1.1 Matrices
A matrix is a rectangular array of numbers arranged in rows and columns. An m×n matrix
has m rows and n columns and m × n specifies the order of a matrix.
An example of matrix which has order 2 × 3 is
A =
1 2 3
4 5 6
.
In general, matrix A which has order m × n can be written
A = aij
where i = 1, 2, 3, . . . , m; j = 1, 2, 3, . . . , n and aij is the element in the ith row and jth
column.
• Two matrices are equal if they have the same order and the elements in corresponding
positions are equal.
• A matrix consisting of only one row is called row matrix or row vector.
• A matrix consisting of only one column is called column matrix or column vector.
• A matrix in which all elements are zero is called zero matrix.
• A n × n square matrix in which elements on the main diagonal are equal to one and
are zeros elsewhere is called identity matrix or unit matrix. It is denoted by E.
If A =
a b
c d
, the value ad − bc is called the determinant of A, denoted |A| or detA.
• We can only find the determinants of square matrices.
• If |A| = 0 then we say that A is singular.
2
• If |A| = 0 then we say that A is non-singular or invertible.
If a matrix C exists such that CA=E then C is said to be the inverse matrix of A, and
we denote C by A−1
.
• The multiplicative inverse of A, denoted A−1
, satisfies A−1
A = AA−1
= E.
• If |A| = 0 then A−1
does not exist.
Matrix addition
To add two matrices A and B they must be of the same order and then we add corresponding
elements. The result of addition is another matrix C of the same order in which cij = aij +bij.
Example 1.1. Find A + B, where
A =
1 2 3
4 5 6
, B =
7 8 9
5 3 2
Solution.
A + B =
1 2 3
4 5 6
+
7 8 9
5 3 2
=
1 + 7 2 + 8 3 + 9
4 + 5 5 + 3 6 + 2
=
8 10 12
9 8 8
Multiples of matrices
If A = (aij) is a matrix and k is a scalar, then kA = (k · aij). That means that we multiply
each element in matrix A by k.
Example 1.2. Find 3 · A, where
A =
1 2 3
4 5 6
Solution.
3 · A = 3 ·
1 2 3
4 5 6
=
3 · 1 3 · 2 3 · 3
3 · 4 3 · 5 3 · 6
=
3 6 9
12 15 18
Matrix multiplication
The product of an m × n matrix A with an n × p matrix B is the m × p matrix C in which
3
the element in the ith row and jth column is the sum of the products of the elements in the
ith row of A with the corresponding elements in the jth column of B. That is
C = A · B and (cij) =
n
k=1
aikbkj = ai1b1j + ai2b2j + ai3b3j + · · · + ainbnj
The product exists only if the number of columns of A equals the number of rows of B. In
general, we cannot reverse the order in which we multiply matrices (A · B = B · A).
Example 1.3. Find A · B, where
A =
1 2
4 5
, B =
7 8 9
5 3 2
Solution.
A · B =
1 2 3
4 5 6
+
7 8 9
5 3 2
=
1 + 7 2 + 8 3 + 9
4 + 5 5 + 3 6 + 2
=
8 10 12
9 8 8
1.2 Solving systems using row operations
The system
a11x1 + a12x2 + · · · + a1nxn = b1
a21x1 + a22x2 + · · · + a2nxn = b2
. . . . . . . . . . . . . . . . . . . . . . . .
ak1x1 + ak2x2 + · · · + aknxn = bk (1.1)
where aij, bi (i = 1, . . . , k; j = 1, . . . , n) are real numbers, is called a linear system of k
equations in n variables (or k × n system).
• If (b1, b2, . . . , bk) = 0 then we say that system (1.1) is homogeneous.
• If (b1, b2, . . . , bk) = 0 then we say that system (1.1) is non-homogeneous.
A solution of this system is any ordered n-tuple (t1, t2, . . . , tn) such numbers that satisfy
the system (1.1).
In matrix form AX = B the system (1.1) is





a11 a12 · · · a1n
a21 a22 · · · a2n
...
...
...
ak1 ak2 · · · akn










x1
x2
...
xk





=





b1
b2
...
bk





,
where A is a coefficient matrix.
4
The matrix 




a11 a12 · · · a1n b1
a21 a22 · · · a2n b2
...
...
...
...
ak1 ak2 · · · akn bk,





is the system’s augmented matrix form. To solve such a system, we need to reduce the
augmented matrix to the echelon form using elementary row operations.
This method is called the Gaussian-elimination algorithm and is based on the fact that
we do not change a solution if we
a) interchange rows,
b) multiply any row by a non-zero number,
c) add a multiple of any row to another row.
Number of solutions
1. Linear system of equations has a solution if and only if a rank of coefficient matrix is
equal to rank of augmented matrix.
2. Linear system of k equations in n unknowns has only one solution if a rank of coefficient
matrix is equal to rank of augmented matrix and it is equal to number of unknowns
n, i.e. r = n.
3. Linear system of k equations in n unknowns has infinitely many solutions if a rank of
coefficient matrix is equal to rank of augmented matrix and it is smaller than number
of unknowns n, i.e. r < n.
Example 1.4. Solve the system: x − 2y + z = 1
−x + 3y + 2z = 0
2x − y + 5z = 5
Solution.
In augmented form, the system is


1 −2 1 1
−1 3 2 0
2 −1 5 5

 .
The first row stays the same. We add the first row to the second one and subtract double
first row from the third one 

1 −2 1 1
0 1 3 1
0 3 3 3

 .
Now, the first and the second rows stay the same and we subtract triple second row from
the third row 

1 −2 1 1
0 1 3 1
0 0 −6 0

 .
5
The last row corresponds to equation −6z = 0 and therefore z = 0.
Substituting z = 0 into the second equation y + 3z = 1 we get y = 1.
Using the first row and z = 0 and y = 1 we have x = 3.
Thus we have a unique solution (x, y, z) = (3, 1, 0).
6
Chapter 2
Functions of One Variable
2.1 Basic properties of functions
Let D ⊂ R and H ⊂ R be given sets. A function f is a rule that assigns to each element x
in set D exactly one element, called f(x), in set H.
• The set D is called the domain of the function and is denoted D(f). It gives us all
permissible values that x may attain. Symbol x is called an independent variable.
• The set H is called the range of the function and is denoted H(f). The range of f is
the set of all possible values of f(x) and symbol f(x) is called a dependent variable.
The graph of function f : D(f) → R is the set of ordered pairs {(x, f(x)) ∈ R2
: x ∈ D(f)}.
Properties of functions
• Function f is said to be bounded if there exists a constant K ∈ R, K < 0 for which
|f(x)| ≤ K for every x ∈ D(f). For example, y = sin x is a bounded function.
• We say that the function f is an even function if it satisfies f(−x) = f(x) for every
x ∈ D(f). For example, y = cos x is an even function.
• We say that the function f is an odd function if it satisfies f(−x) = −f(x) for every
x ∈ D(f). For example, function y = sin x is an odd function.
• Function f is called one-to-one function if f(x1) = f(x2) for any pair x1, x2 ∈ D(f)
where x1 = x2. For example, y = x is a one-to-one function.
• Function f is called periodic function with period p ∈ R, p > 0 if for every x ∈ D(f)
also x + p ∈ D(f) and f(x + p) = f(x − p) = f(x). If p is a period of function
f, then also 2p, 3p, 4p, . . . are its periods. For example, trigonometric functions
(y = sin x, y = cos x, y = tan x, y = cotg x) are periodic.
Definition 2.1. Let f be a function f : D(f) → R and I ⊆ D(f). Function f is said to be
a) increasing on the set I if f(x1) < f(x2) for all x1, x2 ∈ I such that x1 < x2;
b) decreasing on the set I if f(x1) > f(x2) for all x1, x2 ∈ I such that x1 < x2.
7
Functions that are increasing or decreasing are called strictly monotonic.
Given two functions u : A → B and f : B → R, then the function F : A → R given by
y = f(u(x)) is called a composite function. Function u is called the inner component
and function f is called the outer component.
Example 2.1. For the functions below find its inner and outer component.
a) f : y = sin(x2
), b) g: y = x−1
x+1
.
Solution.
a) Here is y = sin u, u = x2
. That means that the outer component is the sinus function
and the inner component is quadratic function.
b) The outer component of given composite function is y =
√
u and the inner component
is
u =
x − 1
x + 1
,
where u must be nonnegative and therefore
x − 1
x + 1
≥ 0 → x ∈ (−∞, −1) ∪ [1, ∞).
The inverse function of a one-to-one function f is a function f−1
for which
1. D(f−1
) = H(f), H(f−1
) = D(f)
2. For every y ∈ D(f−1
) it holds f−1
(y) = x if and only if f(x) = y.
Note:
An inverse function can be defined only for one-to-one function.
The symbol f−1
does not mean 1
f
.
The graph of function f−1
is a reflection of the graph of function f in the line y = x.
Example 2.2. Find the inverses of the given functions
a) f : y = 2x−1
3x+5
, b) g: y = ln(5 − 2x), c) h: y = ex−3
.
Solution.
a) We will describe the procedure for finding the inverse function of a given function.
We first interchange x and y in equation y = f(x) and try to find y as a function of
variable x using equivalent operations. In this case, we get
x =
2y − 1
3y + 5
⇔ 3yx+5x = 2y−1 ⇔ y(3x−2) = −1−5x ⇔ y = −
1 + 5x
3x − 2
.
Thus the inverse function has a formula f−1
: y = −
1 + 5x
3x − 2
.
8
x
y
− 5
3
2
3
y = 2x−1
3x+5
x
y
2
3
− 5
3
y = − 1+5x
3x−2
Figure 2.1: Graphs of functions f and f−1
b) By interchanging x and y, we attain
x = ln(5 − 2y) ⇔ ex
= eln(5−2y)
⇔ ex
= 5 − 2y ⇔ y =
5 − ex
2
.
The inverse of the given function g is therefore g−1
: y =
5 − ex
2
.
c) After interchange x and y, we have
x = ey−3
⇔ ln x = ln ey−3
⇔ ln x = y − 3 ⇔ y = ln x + 3.
Thus the inverse function of h is the function h−1
: y = ln x + 3.
x
y
1
1
3
3
y = ex−3
y = ln x + 3
Figure 2.2: Graph of function h and its inverse function
Note:
Figures 2.1 and 2.2 illustrate an important property of inverse functions: Function f is
increasing on interval I if and only if its inverse function f−1
is increasing on interval J,
where J = f(I). Similarly for decreasing functions.
9
2.2 Polynomials
Definition 2.2. Function P : R → R given by P(x) = anxn
+ an−1xn−1
+ · · · + a0, where
a0, a1, . . . , an ∈ R is called polynomial. Numbers ai are called coefficients of the polyno-
mial.
If an = 0, then the degree of the polynomial is number n.
Number α ∈ C is called root of polynomial P if P(α) = 0.
Number α ∈ C is called k-tuple root of polynomial P if there exists polynomial Q such
that P(x) = (x − α)k
Q(x) and α is not a root of polynomial Q (that means Q(α) = 0).
Number k is the multiplicity of root α.
Examples of polynomials:
a) linear function y = ax + b, where a = 0 (polynomial of degree 1)
b) quadratic function y = ax2
+ bx + c, where a = 0 (polynomial of degree 2)
Theorem 2.1.
a) Any polynomial P of degree n (where n ≥ 0) has n complex roots (counted with the
multiplicity).
b) If complex number α is a k-tuple root of polynomial P, then the complex conjugate α
is also k-tuple root of polynomial P.
c) Let α1, α2, . . . , αr be all real roots of polynomial P with multiplicities k1, k2, . . . , kr
and (c1 ± id1), . . . , (cs ± ids) are all complex roots with multiplicities r1, . . . , rs, then
P(x) = an(x − α1)k1
· · · · · (x − αr)kr
· [(x − c1)2
+ d2
1]r1
· · · · · [(x − cs)2
+ d2
s]rs
.
d) Polynomials P, Q of degree n are equal if and only if all coefficients of corresponding
terms are equal.
Sign of Polynomials
When investigating the behavior of function, we need to find intervals on which a given
function assumes positive values (its graph is above the x-axis) and on which it assumes
negative values (its graph is below the x-axis).
This can be solved by factorizing the polynomial P in real domain. If x1 < x2 < · · · < xm
are all different real roots of P with odd multiplicities then polynomial P keeps the same
positive or negative sign on each interval (−∞, x1), (x1, x2), . . . , (xm, ∞). If we choose two
neighbouring intervals, P changes its sign from negative to positive or vice versa. Passing
over the root of even multiplicity polynomial P does not change its sign.
Example 2.3. Determine the sign of the polynomial P(x) = (x2
− x)(x − 2)2
and sketch
the graph.
10
Solution. We factorize the polynomial P(x) = x(x−1)(x−2)2
. Polynomial P has two simple
real roots (i.e. with odd multiplicities) x1 = 0 and x2 = 1 and one double root (i.e. even
multiplicity) x3,4 = 2. We mark numbers x1 = 0 and x2 = 1 on the number line. Polynomial
is positive for high values of x and it changes its sign when passing over the roots 0 a 1. For
x = 2, the polynomial is equal to zero.
Thus P(x) > 0 for x ∈ (−∞, 0) ∪ (1, 2) ∪ (2, ∞) (see fig. 2.3).
0 21 x
y
y = (x2 − x)(x − 2)2
Figure 2.3: Graph of polynomial P(x) = (x2
− x)(x − 2)2
2.3 Rational functions
Let P(x), Q(x) be polynomials. Function
R(x) =
P(x)
Q(x)
is called rational function.
a) If degree of P(x) is smaller than degree of Q(x), we call function R(x) proper rational
function.
b) If degree of P(x) is greater than or equal to degree of Q(x), we call function R(x)
improper rational function.
Examples of rational functions:
a) y = k
x
, where x = 0, k = 0,
b) y = ax+b
cx+d
, where x = −d
c
, c = 0, ad = bc.
Graph of both of these rational functions is a curve called hyperbola.
Sign of Rational Functions
The procedure of finding the sign of rational function is similar to finding the sign of polynomial.
A rational function R(x) = P(x)
Q(x)
where P and Q are polynomials, can change its
sign only at points, where P or Q change it. However, we have to remember that rational
11
functions are not defined at real roots of the denominator.
Consider R(x) to be a rational function whose numerator and denominator have no common
roots. Let x1 < · · · < xp be all different real roots of numerator and denominator
with odd multiplicities. Then P keeps the same positive or negative sign on each interval
(−∞, x1), (x1, x2), . . . , (xm, ∞) (where R is defined). If we choose two neighbouring intervals,
P changes its sign from negative to positive or vice versa.
Example 2.4. Determine the sign of R(x).
a) R(x) = (x−2)(x−5)
(x−3)(x+2)(x+5)
, b) R(x) = (x2−1)(x2−2x+1)(x−2)2
(x2+x−2)2(x+2)
.
Solution.
a) We first find all real roots of numerator and denominator. These are −5, −2, 2, 3 and
5. We choose any one number that is not the root and determine the sign of R. For
example, for x = 0 is R(0) < 0. Hence R(x) has the negative sign on the interval
involving 0, that is the interval (−2, 2), and passing over the roots −5, −2, 2, 3, 5 function
R changes its sign from negative to positive or vice versa (see table below).
x (−∞, −5) (−5, −2) (−2, 2) (2, 3) (3, 5) (5, ∞)
R(x) − + − + − +
b) Numerator and denominator are not fully factorized. Therefore, we have to start with
finding all real roots
x2
− 1 = (x + 1)(x − 1), x2
− 2x + 1 = (x − 1)2
, x2
+ x − 2 = (x − 1)(x + 2)
and the complete factorization is
R(x) =
(x + 1)(x − 1)(x − 1)2
(x − 2)2
(x − 1)2(x + 2)2(x + 2)
=
(x − 1)3
(x + 1)(x − 2)2
(x − 1)2(x + 2)3
.
After cancellation we obtain
R(x) =
(x − 1)(x + 1)(x − 2)2
(x + 2)3
.
Real roots with odd multiplicities are −2, −1 and 1. These points separate the domain
of R into intervals. We calculate the value of R at an interior point of some interval,
for example x = 0, where R(0) < 0. The sign changes when passing over the roots −2,
−1 and 1.
x (−∞, −2) (−2, −1) (−1, 1) (1, ∞)
R(x) − + − +
12
2.4 Partial Fractions
Let R(x) = P(x)
Q(x)
be a proper rational function. We can decompose any of proper rational
functions into partial fractions:
a) If number α is a simple real root of polynomial Q, then function R can be written as
a partial fraction
A
(x − α)
.
b) If number α is a k-tuple real root of Q, then R can be written as a sum of k partial
fractions
A
(x − α)
+
B
(x − α)2
+ · · · +
M
(x − α)k
.
c) If α ± iβ are simple complex conjugate roots of Q, then R can be written as a partial
fraction
Ax + B
ax2 + bx + c
,
where ax2
+ bx + c has roots α ± iβ.
d) If α ± iβ are double complex conjugate roots of Q, then R can be written as a sum of
two partial fractions
Ax + B
ax2 + bx + c
+
Cx + D
(ax2 + bx + c)2
.
Analogously, to l-tuple pair of complex conjugate roots of the denominator there corresponds
a sum of l partial fractions.
To find numbers A, B, C, D we use a method of undetermined coefficients:
1. We decompose the denominator into the product of factors irreducible in the real
domain.
2. We write the assumed form of the decomposition into the sum of partial fractions with
undetermined coefficients.
3. We write an equation where the original proper rational function is equal to the sum
of the assumed form of partial fractions.
4. We multiply the equation by the denominator of the original rational function and get
the equality of two polynomials.
5. Two polynomials are equal if and only if coefficients of corresponding powers of x on
the two sides of the equation are equal. Thus we expand the polynomials on both
sides and equate coefficients of corresponding terms. That gives us a system of linear
equations.
6. We solve the system of equation and obtain the only solution.
7. We decompose the original rational function.
13
Example 2.5. Express
R(x) =
x2
+ 4x
x4 − 16
as a sum of partial fractions.
Solution. First of all, we have to factorize the denominator
x4
− 16 = (x2
)2
− (22
)2
= (x2
− 4)(x2
+ 4) = (x − 2)(x + 2)(x2
+ 4).
Therefore, the assumed form of the sum of partial fractions is
x2
+ 4x
x4 − 16
=
A
x − 2
+
B
x + 2
+
Cx + D
x2 + 4
,
and after multiplying by (x − 2)(x + 2)(x2
+ 4) we obtain
x2
+ 4x = A(x + 2)(x2
+ 4) + B(x − 2)(x2
+ 4) + (Cx + D)(x2
− 4).
Equating the coefficients of corresponding powers of x we get a system
x3
: 0 = A + B + C
x2
: 1 = 2A − 2B + D
x1
: 4 = 4A + 4B − 4C
x0
: 0 = 8A − 8B − 4D.
The solution of this system is A = 3
8
, B = 1
8
, C = −1
2
, D = 1
2
and the decomposition of
R(x) is
R(x) =
x2
+ 4x
x4 − 16
=
3
8(x − 2)
+
1
8(x + 2)
+
1 − x
2(x2 + 4)
.
14
Chapter 3
Limits, derivatives, minima and
maxima
3.1 Limits
A limit describes the behaviour of function in a neighbourhood of given point (apart from
the point).
Function y = f(x) has a limit L at point x0 if values of function f(x) approaches number
L as x approaches x0. We write
lim
x→x0
f(x) = L.
Note:
Point x0 may be any real number but also ∞ or −∞.
If L is a real number, we say that function f(x) has a finite limit.
If L is ∞ or −∞, we say that function f(x) has a infinite limit.
Function f has at most one limit at any point.
Limits, as x approaches a number x0 from left and right sides, may differ. We call the limit
of function f as x approaches x0 from the right the right-hand limit of f at point x0 and
the limit of function f as x approaches x0 from the left the left-hand limit of f at point
x0.
A function f(x) has a limit L at point x0 if and only if the right-hand limit and the left-hand
limit at x0 exist and they are equal, i.e.
lim
x→x−
0
f(x) = lim
x→x+
0
f(x) = L = lim
x→x0
f(x).
Example 3.1. a) Let f(x) be a function y = x2
and x0 = 0. Then
lim
x→0−
x2
= lim
x→0+
x2
= 0.
And therefore
lim
x→0
x2
= 0.
15
b) Let f(x) be a function y = 1
x
and x0 = 0. Then
lim
x→0−
1
x
= −∞ and lim
x→0+
1
x
= ∞.
Therefore,
lim
x→0
1
x
does not exist.
Rules for limits
If limx→x0 f(x) = L1 and limx→x0 g(x) = L2, where L1, L2 ∈ R, then
1. limx→x0 (f(x) ± g(x)) = L1 ± L2,
2. limx→x0 (f(x) · g(x)) = L1 · L2,
3. limx→x0
f(x)
g(x)
= L1
L2
if L2 = 0,
4. limx→x0 |f(x)| = | limx→x0 f(x)|.
Note:
When calculating the limit of quotient
lim
x→x0
f(x)
g(x)
,
where
lim
x→x0
f(x) = 1, lim
x→x0
g(x) = ±∞ or lim
x→x0
g(x) = 0,
symbolic expressions
1
±∞
= 0,
1
+0
= +∞,
1
−0
= −∞ (3.1)
holds.
If limx→x0 f(x) = c and limx→x0 g(x) = 0, then we have to consider sign of number c.
In case of limits in the form
∞ − ∞,
∞
∞
,
0
0
the operations are not defined in R. These cases are therefore called indeterminate forms.
We will show how to deal with such limits in section 3.4.
Definition 3.1. a) A function f(x) is continuous at point x0 if
lim
x→x0
f(x) = f(x0).
b) If the function f(x) is defined and continuous at every point of the closed interval [a, b],
we call the function continuous on interval [a, b].
16
Theorem 3.1. If the function f(x) is continuous on interval I = [a, b], then f(x) is bounded
and attains its minimal and maximal value on I.
If the function f(x) is continuous on interval I = [a, b], then f(x) attains all the values
between its maximum and minimum on the interval I.
Corollary 3.1. If the function f(x) is continuous on I = [a, b] and f(a) · f(b) < 0, then
there is at least one point c ∈ (a, b) where f(c) = 0.
3.2 Derivatives
Definition 3.2. Let f be a function of x and x0 ∈ D(f). If the limit
lim
x→x0
f(x) − f(x0)
x − x0
exists and is finite, we call this limit the derivative of f at point x0 and denote f′
(x0).
Similarly,
lim
x→x+
0
f(x) − f(x0)
x − x0
is called right derivative and denoted f′
+(x0);
lim
x→x−
0
f(x) − f(x0)
x − x0
is called left derivative and denoted f′
−(x0).
The derivative f′
(x0) equals to the slope of tangent line to the graph of f(x) at point x0.
Therefore,
y = f(x0) + f′
(x0)(x − x0)
is the equation of tangent line at point (x0, f(x0)).
A normal to the graph of f(x) at point x0 is a line which is perpendicular to the tangent
line at point x0. Such a line has an equation
a) y = f(x0) − 1
f′(x0)
(x − x0) if f′
(x0) = 0,
b) x = x0 if f′
(x0) = 0.
Theorem 3.2 (Rules for derivatives). Let functions f(x), g(x) have derivative on set M.
Then
1. (cf(x))′
= cf′
(x),
2. (f(x) + g(x))′
= f′
(x) + g′
(x),
17
3. (f(x) · g(x))′
= f′
(x)g(x) + f(x)g′
(x).
4. If g(x) = 0, then f(x)
g(x)
′
= f′(x)g(x)−f(x)g′(x)
g2(x)
.
The second derivative of function f(x) is a function f′′
= (f′
)′
. In general, n-th order
derivative of function f(x) is a function f(n)
= (f(n−1)
)′
.
Theorem 3.3. Let function f has a derivative on an open interval I.
a) If f′
(x) > 0 for all x ∈ I, then f is increasing on I.
b) If f′
(x) < 0 for all x ∈ I, then f is decreasing on I.
c) If f′
(x) = 0 for all x ∈ I, then f is constant on I.
Derivatives in physics
Consider a function s(t) to be a displacement function. Then the first derivative s′
(t) is
a velocity function v(t) and the second derivative s′′
(t) (which is also the first derivative
of velocity function) is an instantaneous acceleration at time t.
3.3 Maxima and Minima
There are two different types of minima and maxima. The first one is a local minimum (or
maximum) and the second type is a global minimum (or maximum).
Definition 3.3. A neighborhood O(x0) of x0 ∈ R is an open interval (x0 − δ, x0 + δ), where
δ > 0.
Definition 3.4. A function f has at point x0
• a local minimum if there exists a neighborhood O(x0) of x0 such that f(x) ≥ f(x0)
for all x ∈ O(x0),
• a local maximum if there exists a neighborhood O(x0) of x0 such that f(x) ≤ f(x0)
for all x ∈ O(x0),
• a strict local minimum if there exists a neighborhood O(x0) such that f(x) > f(x0)
for all x ∈ O(x0) \ {x0},
• a strict local maximum if there exists a neighborhood O(x0) such that f(x) < f(x0)
for all x ∈ O(x0) \ {x0}.
Theorem 3.4. Suppose that a function f(x) has a local minimum or maximum at point x0
and that f′
(x) is defined at x0. Then
f′
(x0) = 0. (3.2)
The point satisfying (3.2) is called a stationary point.
The points where f′
= 0 or f′
fails to exist are commonly called the critical points of f.
None of these points is necessarily the location of a local maximum or minimum, but these
points are only points where local extreme may exist .
18
Theorem 3.5 (Sufficient Condition).
a) If the first derivative f′
of the function f changes its sign from positive to negative at
critical point x0, then the function f has a local minimum at point x0.
b) If the first derivative f′
of the function f changes its sign from negative to positive at
critical point x0, then the function f has a local maximum at point x0.
Theorem 3.6 (Second Derivative Test for Local Maxima and Minima).
a) If f′
(x0) = 0 and f′′
(x0) < 0, then function f has a strict local maximum at point x0.
b) If f′
(x0) = 0 and f′′
(x0) > 0, then function f has a strict local minimum at point x0.
Example 3.2. Find the local minimum and maximum values, if any, of the function
f(x) = x3
− 12x − 6.
Solution. To find all the stationary points we have to calculate the first derivative
f′
(x) = (x3
− 12x − 6)′
= 3x2
− 12.
The derivative is defined at every point of the function’s domain and is zero at x = 2 or
x = −2. The sign diagram of f′
is
interval (−∞, −2) (−2, 2) (2, +∞)
f′
+ − +
f ր ց ր
Since the derivative f′
changes its sign at both points, it has a local maximum at point
x = −2 and its value is 10. Local minimum value is - 22 and is located at x = 2.
Definition 3.5. Let f be a function on set M. If x0 ∈ M and
• f(x) ≥ f(x0) for all x ∈ M, we say that the function f has a global minimum at
point x0.
• f(x) ≤ f(x0) for all x ∈ M, we say that the function f has a global maximum at
point x0.
The only points where we can find a global minimum or maximum are critical points or
endpoints of the function’s domain (set M).
Example 3.3. Find the absolute minimum and maximum of
f(x) = x − x2
, x ∈ [0, 1].
19
Solution. Function f(x) is continuous and bounded on closed interval and therefore has
absolute minimum and maximum. The derivative is
f′
(x) = 1 − 2x
and we find stationary points where
f′
(x) = 0.
Therefore, there is a stationary point (which is also the only critical point) at x = 1
2
.
Finally, we have to compare the values at all critical points and endpoints of the interval
[0, 1]
f
1
2
=
1
4
, f(0) = 0, f(1) = 0.
Thus, function f has its absolute maximum at point x = 1
2
and absolute minimum at x = 0
and x = 1.
3.4 L’Hospital’s Rule
Theorem 3.7. Suppose x0 ∈ R ∪ {−∞, ∞}. If
lim
x→x0
f(x) = lim
x→x0
g(x) = 0 or lim
x→x0
|f(x)| = lim
x→x0
|g(x)| = ∞
and
lim
x→x0
f′
(x)
g′(x)
exists, then lim
x→x0
f(x)
g(x)
exists as well
and
lim
x→x0
f(x)
g(x)
= lim
x→x0
f′
(x)
g′(x)
.
Example 3.4. Find the limit
lim
x→0
xex
4 − 4ex
.
Solution. Substituting x = 0, we obtain
”
0
0
“. Therefore, we can use a L’Hospital Rule:
lim
x→0
xex
4 − 4ex
= lim
x→0
ex
+ xex
−4ex
= −
1
4
.
20
Chapter 4
Behaviour of a function,
Approximation of a function
4.1 Concavity and Points of Inflection
Definition 4.1. a) We say that function f is convex on interval I if the graph of
function f is located above the tangent at any point on I, i.e.
f(x) ≥ f(x0) + f′
(x0)(x − x0) where x, x0 ∈ I.
b) We say that function f is concave on interval I if the graph of function f is located
under the tangent at any point on I, i.e.
f(x) ≤ f(x0) + f′
(x0)(x − x0) where x, x0 ∈ I.
c) A point on the curve where the concavity changes is called a point of inflection.
Theorem 4.1. Let I be an open interval and let f have a second derivative f′′
on I.
a) If f′′
(x) > 0 for all x ∈ I, then f is convex.
b) If f′′
(x) < 0 for all x ∈ I, then f is concave.
c) If f′′
(x0) = 0 and the sign of f′′
(x0) changes on either side of x0, then there is a point
of inflection at x = x0.
Note:
Convex function is sometimes called concave up and concave function may be also called
concave down. (See e.g. [3] and [7]).
21
4.2 Asymptotes
If the distance between the graph of a function and some line approaches zero as a point on
the graph moves increasingly far from the origin, we say that the line is an asymptote of the
graph. For instance, the x-axis and y-axis are asymptotes of the curves y = 1
x
.
Definition 4.2. A line x = x0 is called a vertical asymptote of the graph of a function f
if
lim
x→x+
0
f(x) = ±∞ or lim
x→x−
0
f(x) = ±∞.
A function f can have a vertical asymptote only at point x = x0 where f is not defined.
Definition 4.3. A line y = ax + b, a, b ∈ R, is called an asymptote of f if
lim
x→−∞
(f(x) − (ax + b)) = 0 or lim
x→+∞
(f(x) − (ax + b)) = 0.
If a = 0, the asymptote is called horizontal, otherwise it is called oblique.
Theorem 4.2. A line y = ax + b is an asymptote of function f for x → +∞ if
a = lim
x→+∞
f(x)
x
, b = lim
x→+∞
(f(x) − ax)
(where both limits are proper limits). Analogically for x → −∞.
Example 4.1. Find all asymptotes of the graph of functions
a) f(x) =
x − 2
x + 1
, b) f(x) =
x3
x2 − 1
.
Solution.
a) Since D(f) = R \ {−1} and f is continuous on domain D(f), a vertical asymptote can
only occur at the point x = −1. We have
lim
x→−1−
x − 2
x + 1
= ∞.
Limit is improper thus there is a vertical asymptote at point x = −1.
Now, we find oblique and horizontal asymptotes. That means that we have to find
limits
a = lim
x→+∞
f(x)
x
= lim
x→+∞
x − 2
x(x + 1)
= 0
b = lim
x→+∞
(f(x) − ax) =
x − 2
x + 1
= 1.
Asymptote for x → ∞ is a line y = 1. Similarly for x → −∞ and therefore line y = 1
is an asymptote also for x → −∞.
22
0
y
x
1
−1
(a)
0
y
x−1 1
(b)
Figure 4.1: Graphs of functions and asymptotes
b) Function is not defined at points x = −1 and x = 1. We evaluate limits
lim
x→−1−
x3
x2 − 1
= ∞ and lim
x→1+
x3
x2 − 1
= ∞.
Both limits are improper. Therefore lines x = −1 a x = 1 are vertical asymptotes.
To find oblique and horizontal asymptotes we have to evaluate limits
lim
x→∞
x3
x3 − x
= 1
lim
x→∞
x3
x2 − 1
− x = lim
x→∞
x
x2 − 1
= 0.
Both limits are proper hence the graph of f has an oblique asymptote y = x as x → ∞.
Since all previous calculations are valid also when x → −∞, the line y = x is an oblique
asymptote as x → −∞.
4.3 Behaviour of Function
When studying a behaviour of a function, we go through the following steps:
1. Identify the domain D(f). Find (if possible) roots of f and intervals where function
f is positive and where it is negative (sign diagram). Determine if f is even, odd or
periodic.
2. Calculate f′
and find
23
a) intervals where f is increasing (i.e. f′
> 0),
b) intervals where f is decreasing (i.e. f′
< 0),
c) local minima and maxima.
3. Calculate f′′
and find
a) points of inflection,
b) intervals where f is convex (i.e. f′′
> 0),
c) intervals where f is concave (i.e. f′′
< 0).
4. Find all asymptotes of the graph of function f (vertical, oblique, horizontal).
5. Sketch the graph of function f.
Example 4.2. Investigate the behaviour of function
f : y = x3
− 3x2
+ 4.
Solution. 1. The function f is a polynomial so the domain D(f) = R. Since f(−x) =
(−x)3
− 3(−x)2
+ 4 = −x3
− 3x2
+ 4, function f is neither even nor odd. Function is
not periodic (nonconstant polynomials are not periodic).
Roots of f are x = −1 and x = 2. Therefore, a sign diagram is
interval (−∞, −1) (−1, 2) (2, +∞)
f − + +
2. We evaluate the first derivative
y′
= (x3
− 3x2
+ 4)′
= 3x2
− 6x.
Roots of polynomial 3x2
−6x = 3x(x−2) are x = 0 and x = 2. Those roots correspond
to stationary points.
The sign diagram for y′
is
interval (−∞, 0) (0, 2) (2, +∞)
f′
+ − +
f ր ց ր
Thus function f has a local maximum at x = 0 and a local minimum x = 2.
3. The second derivative is
y′′
= (3x2
− 6x)′
= 6x − 6.
We equate it with zero and get x = 1 which is the only potential point of inflection.
Sign diagram for y′′
:
interval (−∞, 1) (1, ∞)
f′′
+ −
f ∪ ∩
The function has a point of inflection at x = 1.
24
4. We find any asymptotes.
As f is continuous on R there is no vertical asymptote.
We identify an oblique or horizontal asymptote:
a = lim
x→±∞
f(x)
x
= lim
x→±∞
x3
− 6x2
+ 4
x
= ∞.
The limit is improper, therefore, there are no such asymptotes.
5. When calculating the values at several significant points we obtain: f(0) = 4, f(1) = 2,
f(2) = 0. Now, we can sketch the graph (see fig. 4.2).
0
y
x−1 1 2 3
−2
−1
1
2
3
4
Figure 4.2: Graph of function y = x3
− 3x2
+ 4
25
Chapter 5
Indefinite Integral
Chapter 3 deals with an important term derivative of function. In this chapter we will
investigate a converse problem. We are looking for a function F whose derivative f = F′
is
known. To find such functions F is part of integral calculus. To integrate function means
to find all the functions that have a given function as a derivative.
5.1 Basic Integrals
Definition 5.1. Let f and F be functions defined on interval I. Function F is called
an antiderivative of function f (or a primitive function of f) on interval I if
F′
(x) = f(x) for all x ∈ I.
The set of all antiderivatives of f is called an indefinite integral of function f and is
denoted
f(x) dx.
Note:
a) In case that interval I is not open, the derivatives at the end points are one-sided
derivatives.
b) Symbol originate from stretching letter S, which is the first letter of the word sum.
c) The function f is the integrand of the integral. The term dx is the differential of
variable x. It tells us that the variable of integration is x.
Theorem 5.1. If F is an antiderivative of f on interval I, then all other antiderivatives of
f on I has the form F + c, where c ∈ R.
We indicate this by writing
f(x) dx = F(x) + c, where c ∈ R.
Number c is the constant of integration.
26
Example 5.1. Find antiderivatives of following functions
a) y = 1, b) y = ex
, c) y = e−x
.
Solution. a) Antiderivative of function f(x) = 1 is F(x) = x because F′
(x) = x′
= 1.
b) Antiderivative of function f(x) = ex
is F(x) = ex
because when differentiating function
ex
it stays the same.
c) Similarly, (e−x
)
′
= −e−x
= −f(x) and thus the antiderivative of f is function F(x) =
−e−x
.
Using indefinite integrals we can write
dx = x + c, ex
dx = ex
+ c, e−x
dx = −e−x
+ c.
Example 5.2. Find an indefinite integral of
a) y = cos 2x, b) y =
1
x
.
Solution. a) We know that
(sin x)′
= cos x, therefore cos x dx = sin x.
Similarly,
(sin 2x)′
= 2 cos 2x, thus cos 2x dx =
1
2
sin 2x.
b) The antiderivative of function f(x) = 1
x
is F(x) = ln x on interval (0, ∞) and F(x) =
ln(−x) on interval (−∞, 0). Therefore function F(x) = ln |x| is a primitive function
on (−∞, 0) and (0, ∞) and we get
1
x
dx = ln |x| + c. (5.1)
Theorem 5.2. If a function f is continuous on the interval I, then it has an antiderivative
on I.
Theorem 5.3. Let integrals f(x) dx and g(x) dx exist on an interval I and let α be
a constant. Then the integrals (f(x) + g(x)) dx and αf(x) dx also exist on I and the
following equalities hold:
(f(x) + g(x)) dx = f(x) dx + g(x) dx, (5.2)
αf(x) dx = α f(x)dx. (5.3)
27
The following formulas give the indefinite integrals of several basic functions:
(1) 1 dx = x + c, (8) 1
x2+1
dx = arctg x + c
(2) xn
dx = xn+1
n+1
+ c, n = −1, (9) 1
(x−x0)2+a2 dx = 1
a
arctg x−x0
a
+ c,
(3) 1
x
dx = ln |x| + c, (10) 1√
a2−x2 dx = arcsin x
a
+ c,
(4) ex
dx = ex
+ c, (11) 1√
x2+a
dx = ln |x +
√
x2 + a| + c,
(5) ax
dx = ax
ln a
+ c, a > 0, a = 1, (12) 1
cos2 x
dx = tan x + c,
(6) sin x dx = − cos x + c, (13) 1
sin2 x
dx = −cotg x + c,
(7) cos x dx = sin x + c, (14) f′(x)
f(x)
dx = ln |f(x)| + c.
In the above table a means any non-zero number, apart from Formula (5). Number c ∈ R is
the integration constant. The formulas are valid on intervals on which both sides are defined
and x is an independent variable.
Example 5.3. Evaluate the indefinite integrals
a) x3
dx, b) 1
x2 dx,
c)
√
x + 1
x−1
+ 2 dx, d) 1
2x−5
dx,
e) tan x dx, f) tan2
x dx,
g) x+1
x2+2x+9
dx, h) 1
x2+9
dx,
i) 1√
9−x2 dx, j) x4
x2+9
dx.
Solution. a) Using formula (2) from the table above we obtain
x3
dx =
x4
4
+ c.
b) Rewriting the integral and using formula (2) we have
1
x2
dx = x−2
dx =
x−1
−1
+ c = −
1
x
+ c.
28
c) Dividing the integral by the Theorem 5.3 and using the formulas (2), (14) and (1) we
get
√
x +
1
x − 1
+ 2 dx = x
1
2 dx +
1
x − 1
dx + 2 dx =
=
2
3
x
3
2 + ln |x − 1| + 2x + c.
d) We rewrite the integral and then use the formula (14)
1
2x − 5
dx =
1
2
2
2x − 5
dx =
1
2
ln |2x − 5| + c.
e) We apply the trigonometric identity tan x = sin x
cos x
to be able to use formula (14)
tan x dx =
sin x
cos x
dx = −
− sin x
cos x
dx = − ln | cos x| + c.
f) Using trigonometric identities and formulas (1) and (12) we have
tan2
x dx =
sin2
x
cos2 x
dx =
1 − cos2
x
cos2 x
dx =
=
1
cos2 x
dx − dx = tan x − x + c.
g) Since (x2
+ 2x + 9)′
= 2x + 2 we multiply the integral by 1
2
and rewrite the integrand.
Afterwards we use the formula (14)
x + 1
x2 + 2x + 9
dx =
1
2
2x + 2
x2 + 2x + 9
dx =
1
2
ln |x2
+ 2x + 9| + c .
h) We can use the formula (9) for x0 = 0 and a = 3
1
x2 + 9
dx =
1
3
arctg
x
3
+ c .
i) We use the formula (10) for a = 3
1
√
9 − x2
dx = arcsin
x
3
+ c .
j) We rewrite the given function as a sum of polynomial and proper rational function and
use formulas (1), (2) and (9)
x4
x2 + 9
dx = x2
− 9 +
81
x2 + 9
=
x3
3
− 9x + 27 arctg
x
3
+ c .
29
Example 5.4. Decide whether the following functions are the antiderivatives of the same
function.
f : y = sin2
x, g: y = −
1
2
cos 2x.
Solution. We show two different methods of solution.
a) Using the definition, i.e. we calculate the derivatives
(sin2
x)′
= 2 sin x cos x = sin 2x, −
1
2
cos 2x
′
= −
1
2
(− sin 2x)2 = sin 2x.
That means that given functions are antiderivatives of the same function.
b) We will show that given functions differ only in constant.
sin2
x − −
1
2
cos 2x = sin2
x +
1
2
(cos2
x − sin2
x) =
1
2
.
5.2 Basic Integration Methods
Two principal integration methods will be introduced in this chapter – integration by parts
and the substitution method.
Integration by Parts
Integration by parts is a technique for replacing difficult integrals by ones that are usually
easier to integrate. The formula for integration by parts comes from the product rule,
(uv)′
= u′
v + uv′
.
By integrating we get
uv = u′
v + uv′
.
Theorem 5.4. Let u(x) and v(x) be functions differentiable on the interval I. Then the
equality
u(x)v′
(x) dx = u(x)v(x) − u′
(x)v(x) dx. (5.4)
holds if at least one of the integrals involved exists.
Choosing u and v′
In general, we choose for u something that becomes simpler when differentiated and for v′
something whose integral is simple.
Most of the functions that are suitable for integration by parts can be divided into two
groups.
30
1. In the first group a polynomial is differentiated and the second factor is integrated.
The new integral is the product of the polynomial of degree smaller by one and of the
second function, which is of a type similar to that of the original integral (exponential
function eax
is kept, sine and cosine are exchanged).
Examples:
P(x)eax
dx, P(x) sin ax dx, P(x) cos ax dx.
2. In the second group a polynomial is integrated and the second factor is differentiated.
The reverse choice would not be possible, because we do not yet know how to integrate
a logarithm, inverse sine, etc.
Examples:
P(x) ln x dx, P(x) arctg ax dx, P(x) arcsin ax dx.
Example 5.5. Evaluate the indefinite integrals
a) x cos x dx, b) x2
ex
dx,
c) arctg x dx, d) ex
sin x dx.
Solution. a) The integral is of the first type, thus v′
= cos x will be integrated and the
polynomial function u = x will be differentiated. Then u′
= 1, v = sin x and by
substituting we get
x cos x dx = x sin x − sin x dx = x sin x + cos x + c.
b) Sometimes we have to use integration by parts more than once to obtain an answer.
Here is an example.
We choose u = x2
, v′
= ex
. Then u′
= 2x, v = ex
and we have
x2
ex
dx = x2
ex
− 2xex
dx.
The integral on the right side now requires another integration by parts: u = 2x,
v′
= ex
. We obtain u′
= 2, v = ex
and
x2
ex
dx = x2
ex
− 2xex
dx = x2
ex
− 2xex
+ 2 ex
dx
= x2
ex
− 2xex
+ 2ex
+ c = (x2
− 2x + 2)ex
+ c.
c) It seems that there is no product in this integral. However, we can rewrite it as
arctg x = 1 · arctg x.
31
Therefore u = arctg x, v′
= 1 and we get u′
= 1
1+x2 , v = x. Thus
arctgx dx = x arctg x −
x
1 + x2
dx
= x arctg x −
1
2
2x
1 + x2
dx
= x arctg x −
1
2
ln(1 + x2
) + c.
d) After using the integration-by-parts method, we can sometimes attain an expression
that again contains the original unknown integral. In this case, we can understand the
whole expression as an equation from which we find the unknown integral.
We choose u = sin x, v′
= ex
. Then u′
= cos x, v = ex
and
ex
sin x dx = ex
sin x − ex
cos x dx.
Similarly u = cos x and v′
= ex
. Then u′
= − sin x, v = ex
and
ex
sin x dx = ex
sin x − ex
cos x − ex
sin x dx
from which we find
ex
sin x dx =
1
2
(ex
sin x − ex
cos x) + c.
The Substitution Method
The substitution method of integration is derived from the derivative chain rule.
Theorem 5.5. Let function f have an antiderivative F on interval J. Let function t = ϕ(x)
be differentiable on I and ϕ(x) ∈ J for x ∈ I. Then composite function f(ϕ(x))ϕ′
(x) has an
antiderivative on interval I and the equality
f(ϕ(x))ϕ′
(x) dx = F(ϕ(x)) + c
holds.
To evaluate the integral f(ϕ(x))ϕ′
(x) dx we carry out the following steps:
1. Substitute t = ϕ(x) and dt = ϕ′
(x) dx to obtain the integral f(t) dt.
2. Integrate with respect to t.
3. Replace t by ϕ(x) in the result.
32
That means
f(ϕ(x))ϕ′
(x) dx = f(t) dt = F(t) + c = F(ϕ(x)) + c. (5.5)
Example 5.6. Evaluate
a) (3x − 4)7
dx, b) x
x2+1
dx,
c) 10x(x2
+ 13)12
dx, d) ln2
x
x
dx,
e) cos3
x dx, f) x
√
1 − x2 dx.
Solution. a) We substitute t = 3x − 4. To find dx we have to differentiate the equation
for substitution and we attain
dt = 3 dx.
Therefore,
dx =
1
3
dt.
Then
(3x − 4)7
dx = t7 dt
3
=
t8
24
+ c =
(3x − 4)8
24
+ c.
b) We use substitution t = x2
+ 1. Then dt = 2x dx, dx = dt
2x
, thus
x
x2 + 1
dx =
x
t
dt
2x
=
1
2
1
t
dt =
1
2
ln |t| + c =
1
2
ln |x2
+ 1| + c.
c) We substitute t = x2
+ 13. Then dt = 2x dx and dx = dt
2x
. Therefore,
10x(x2
+ 13)12
dx = 10x · t12
·
dt
2x
= 5 t12
dt =
5
13
t13
+ c =
5
13
(x2
+ 13)13
+ c.
d) We choose the substitution t = ln x. Then dt = 1
x
dx, dx = x dt and
ln2
x
x
dx =
t2
x
x dt = t2
dt =
1
3
t3
+ c =
1
3
ln3
x + c.
e) Substituting t = sin x we get dt = cos x dx and dx = dt
cos x
. We use the trigonometric
identity cos2
x = 1 − sin2
x and obtain
cos3
x dx = cos3
x
dt
cos x
= cos2
x dt = (1 − sin2
x) dt =
= (1 − t2
) dt = t −
t3
3
+ c = sin x −
sin3
x
3
+ c.
33
f) We substitute t = 1 − x2
. Then dt = −2x dx and dx = − dt
2x
. Thus
x
√
1 − x2 dx = x
√
t
dt
−2x
= −
1
2
t
1
2 dt = −
1
3
t
3
2 + c = −
1
3
(1 − x2
)
3
2 + c.
Note:
We can also use a ”converse” substitution, i.e. x = ψ(t). However, this substitution is not
so common. We will show the method for integral
√
a2 − x2 dx.
We use the substitution x = a sin t. This integral is used when calculating the surface of a
circle.
Example 5.7. Evaluate
√
1 − x2 dx.
Solution. First we substitute x = sin t, dx = cos t dt
√
1 − x2 dx = 1 − sin2
t · cos t dt =
√
cos2 t cos tdt = cos2
t dt.
Then we apply the trigonometric identity cos2
t = 1+cos 2t
2
and get
cos2
t dt =
1 + cos 2t
2
dt =
1
2
t +
1
2
sin 2t + c =
1
2
(t + sin t cos t) + c =
=
1
2
(t + sin t 1 − sin2
t) + c =
1
2
arcsin x + x
√
1 − x2 + c.
5.3 Integration of Rational Functions
Rational function is a function in the form
R(x) =
P(x)
Q(x)
,
where P, Q are polynomials.
The basic method for preparing rational functions for integration is the method of partial
fractions (see chapter 2.3 ).
In general,
34
• if R(x) is an improper rational function, we divide the polynomial P(x) by Q(x) and
rewrite it as a sum of polynomial and proper rational function.
To integrate a polynomial we use formulas (1) and (2) from the table on page 28.
• if R(x) is a proper rational function, we decompose it into partial fractions, which
might be of two types:
M
(x − α)k
, or
Ax + B
(ax2 + bx + c)k
depending on the type of roots of polynomial Q.
Integration of Partial Fractions with Real Roots in the Denominator
If number α is a simple real root of polynomial Q, then the rational function R can be
written in the form
A
x − α
. (5.6)
We use the formula (14) to integrate the fraction (5.6) and obtain
A
x − α
dx = A ln |x − α| + c.
If number α is a k-tuple real root of Q, the rational function R can be written as a sum
of k partial fractions
A
x − α
+
B
(x − α)2
+ · · · +
K
(x − α)k
.
To integrate these fractions we use (5.6) and for k ≥ 2 formula (2) on page 28, by which
M
(x − α)k
dx = M (x − α)−k
dx = M
(x − α)−k+1
−k + 1
+ c =
M
(1 − k)(x − α)k−1
+ c.
Integration of Partial Fractions with Complex Roots in the Denominator
If numbers α ± iβ are complex conjugate simple roots of polynomial Q then the partial
fraction is
Ax + B
ax2 + bx + c
,
where ax2
+bx+c has roots α±iβ. We use formulas (9) and (14) on page 28 for integration.
If numbers α ± iβ are complex conjugate double roots of Q, then R contains two partial
fractions
Ax + B
ax2 + bx + c
+
Cx + D
(ax2 + bx + c)2
.
35
Example 5.8. Evaluate the integrals:
a) 2x2+6x−2
x(x+2)(x−1)
dx, b) x2+2x+6
(x−1)3 dx,
c) x−1
x4+3x2+2
dx, d) 1
x3+1
dx.
Solution. a) We decompose the function into partial fractions and we use the formula
f′(x)
f(x)
dx = ln |f(x)| + c:
2x2
+ 6x − 2
x(x + 2)(x − 1)
dx =
1
x
+
2
x − 1
−
1
x + 2
dx =
= ln |x| + 2 ln |x − 1| − ln |x + 2| + c.
b) We decompose the function and substitute t = x − 1. Then dx = dt and
x2
+ 2x + 6
(x − 1)3
dx =
9
(x − 1)3
−
4
(x − 1)2
+
1
x − 1
dx =
9
t3
−
5
t2
+
6
t
dt =
= −
9
2t2
+
5
t
+ 6 ln |t| + c = −
9
2(x − 1)2
+
5
x − 1
+ 6 ln |x − 1| + c.
c) We decompose the given function into partial fractions
x − 1
x4 + 3x2 + 2
dx =
x − 1
x2 + 1
dx −
x − 1
x2 + 2
dx.
We evaluate each of the integrals separately. We first multiply each integrand by a
constant so that the numerator would be equal to the derivative of denominator
x − 1
x2 + 1
dx =
1
2
2x − 2
x2 + 1
dx,
x − 1
x2 + 2
dx =
1
2
2x − 2
x2 + 2
dx
and divide into two fractions
x − 1
x2 + 1
dx =
1
2
2x
x2 + 1
dx − 2
1
x2 + 1
dx ,
x − 1
x2 + 2
dx =
1
2
2x
x2 + 2
dx − 2
1
x2 + 2
dx .
We use formula f′(x)
f(x)
dx = ln |f(x)| + c for the first integrals and formula (9) for the
second ones, i.e. dx
(x−x0)2+a2 = 1
a
arctg x−x0
a
+ c:
x − 1
x2 + 1
dx =
1
2
ln |x2
+ 1| − arctg x,
x − 1
x2 + 2
dx =
1
2
ln |x2
+ 2| +
1
√
2
arctg
x
√
2
.
Altogether
x − 1
x4 + 3x2 + 2
dx =
1
2
ln |x2
+ 1| − arctg x −
1
2
ln |x2
+ 2| −
1
√
2
arctg
x
√
2
+ c.
36
d) We decompose integrand into partial fractions
1
x3 + 1
dx =
1
3
1
x + 1
dx −
1
3
x − 2
x2 − x + 1
dx.
And then
1
3
1
x + 1
dx =
1
3
ln |x + 1|
x − 2
x2 − x + 1
dx =
1
2
2x − 4
x2 − x + 1
dx =
1
2
2x − 1 − 3
x2 − x + 1
dx =
=
1
2
2x − 1
x2 − x + 1
dx −
3
2
1
x2 − x + 1
dx =
=
1
2
2x − 1
x2 − x + 1
dx −
3
2
1
(x − 1
2
)2 + 3
4
dx =
=
1
2
ln |x2
− x + 1| +
3
√
3
arctg
x − 1
2
√
3
2
.
Altogether
1
x3 + 1
dx =
1
3
ln |x + 1| −
1
6
ln |x2
− x + 1| −
√
3
3
arctg
2
√
3
x −
1
√
3
+ c .
37
Chapter 6
Definite Integral
The previous chapter deals with the concept of the indefinite integral, which assign a set of
functions (varying in constant) to a function. The definite integral, on the contrary, assigns
a number to a function.
The definition of a definite integral is based on the requirement of the meaning of the area of
the region between the graph of a nonnegative continuous function y = f(x) and an interval
a ≤ x ≤ b.
6.1 Definition and Basic Properties of Definite Integral
When we define the definite integral, we can use either antiderivative (Newton-Leibniz formula)
or summation definition (Riemann definite integral). Since both definitions coincide
for continuous functions, we will show only the Newton-Leibniz formula.
Definition 6.1. Let function f be continuous on the interval [a; b]. The definite integral
of function f over [a; b] is a number
b
a
f(x) dx = F(b) − F(a), (6.1)
where F is the antiderivative of function f on interval [a, b].
Note:
i) We often write F(x)
b
a
instead of F(b) − F(a), i.e.
b
a
f(x) dx = F(x)
b
a
.
ii) The existence of the antiderivative F is assured for continuous function by Theorem
5.2. The number
b
a
f(x) dx is unique which can be proved by using Theorem 5.1. The
theorem says that any two functions that have f as a derivative on [a; b] must differ
by some fixed constant throughout [a; b]. Therefore,
b
a
f(x) dx = F(b) + c − (F(a) + c) = F(b) − F(a).
38
iii) The number a is a lower limit, the number b is a upper limit, the interval [a; b] is
an integration domain, and the function f is an integrand. The common name for
lower and upper limits is limits of integration.
Geometrical meaning
Let a function f be nonnegative and continuous function on interval [a; b]. The geometrical
interpretation of definite integral is the area P of the region beneath the graph of f from a
to b (see fig. 6.1.). This region is called a subgraph of function f.
y = f(x)
P
a b
y
x
Figure 6.1: Geometrical meaning of definite integral
The definition is based on approximation of the region with inscribed rectangles.
The technique is:
a) We divide the interval [a, b] into n intervals [xi−1, xi] (called intervals of partition)
of the same length so that x0 = a and xn = b. The norm of partition is
xi − xi−1 = (b − a)/n.
b) We approximate each interval with an inscribed rectangle with sides of the length
(xi − xi−1) and f(ci) where ci lies inside the interval of partition. The area of the
rectangle, its height times its base length, is
f(ci)(xi − xi−1).
The sum of all these areas provides the estimate of the area P, i.e.
P ≈
n
i=1
f(ci)(xi − xi−1) =
n
i=1
f(ci)
b − a
n
.
The higher number n is, the more accurate the area value we get.
c) When n → ∞, we get an exact value of the area
P = lim
n→∞
n
i=1
f(ci)
b − a
n
. (6.2)
This number is called a definite integral and is denoted
b
a
f(x) dx.
39
f(c1)
f(c2)
f(c3)
f(c4)
f(c5)
f(c6)
a = x1 x2 x3 x4 x5 x6 b = x7
Figure 6.2: Approximating the area with rectangles
Example 6.1. Find the area bounded by the parabola y = x2
, x-axis and line x = a. What
are odds of S1 : S2 (see fig. 6.3)?
0
y
x
S1
S2
×
a
Figure 6.3: Parabola and rectangular
Solution. To find the area S1 under the parabola, we calculate the definite integral
S1 =
a
0
x2
dx =
x3
3
a
0
=
a3
3
.
The area of rectangle is a · a2
= a3
and therefore the area above the parabola is S2 = 2a3
3
.
The odds are S1 : S2 = 1 : 2.
40
Example 6.2. Evaluate the definite integrals
a)
π
0
sin x dx, b)
1
0
1
x2 + 1
.
Solution. From (6.1) we attain
π
0
sin x dx = − cos x
π
0
= − cos π − (cos 0) = −(−1) + 1 = 2.
Similarly,
1
−1
1
x2 + 1
dx = arctg x]1
−1 = arctg 1 − arctg(−1) =
π
4
+
π
4
=
π
2
.
Theorem 6.1 (Properties of the Definite Integral). If f and g are continuous functions on
the interval [a, b], then
a)
b
a
[f(x) + g(x)] dx =
b
a
f(x) dx +
b
a
g(x) dx;
b)
b
a
cf(x) dx = c
b
a
f(x) dx;
c)
b
a
f(x) dx =
c
a
f(x) dx +
b
c
f(x) dx, where a < c < b;
d)
b
a
f(x) dx ≥ 0, if f(x) ≥ 0 on interval [a, b];
e)
b
a
f(x) dx ≥
b
a
g(x) dx, if f(x) ≥ g(x) on interval [a, b].
Note:
i) Formula c) can be used in case when a function f is continuous on [a, c] and [c, b] and
is not continuous at point c. For example, function sgn x is not continuous at x = 0,
however, we can define the definite integral
1
−2
sgn x dx =
0
−2
sgn x dx +
1
0
sgn x dx = −2 + 1 = −1.
ii) For a > b we define
b
a
f(x) dx = −
a
b
f(x) dx,
and integral
a
a
f(x) dx is
a
a
f(x) dx = 0.
41
6.2 Integration Methods for the Definite Integral
Integration by Parts
Theorem 6.2 (Integration by Parts for the Definite Integrals). Let u(x) and v(x) have
continuous derivatives on interval [a, b]. Then
b
a
u(x)v′
(x) dx = u(x)v(x)
b
a
−
b
a
u′
(x)v(x) dx.
Example 6.3. Evaluate the integral
e
1
x3
ln x dx.
Solution. We integrate by parts and choose u = ln x, v′
= x3
. Then
u′
=
1
x
, v =
x4
4
and we attain
e
1
x3
ln x dx =
x4
4
ln x
e
1
−
1
4
e
1
x3
dx =
e4
4
−
1
4
x4
4
e
1
=
3e4
+ 1
16
.
Substitution Method
Theorem 6.3 (Substitution Method for the Definite Integral). Let function f(t) be continuous
on [a, b]. Let function ϕ(x) have the continuous derivative on the interval [α, β] and
ϕ(x) maps the interval [α, β] into interval [a, b]. Then
β
α
f(ϕ(x))ϕ′
(x) dx =
ϕ(β)
ϕ(α)
f(t) dt.
Note: When using a substitution method for the definite integral, we have to find new limits
of integration.
Example 6.4. Evaluate
5
0
x
√
1 + 3x
dx.
Solution.
We substitute
t =
√
1 + 3x.
Then t2
= 1 + 3x. Hence x = 1
3
(t2
− 1) and dx = 2
3
t dt. To obtain new limits for variable t,
we substitute x = 0 and x = 5 into the equation t =
√
1 + 3x. Thus we get 0 1, 5 4.
Finally, we evaluate the definite integral with new limits of integration
5
0
x
√
1 + 3x
dx =
2
9
4
1
(t2
− 1)dt =
2
9
t3
3
− t
4
1
= 4.
42
6.3 Applications of the Definite Integral
In this part we present several applications of the definite integral in geometry.
Area of a Plane Figure
Let function f be a continuous and nonnegative on the interval [a, b]. Then the area of
subgraph of function f (see fig. 6.1) is given by the formula
P =
b
a
f(x) dx. (6.3)
Example 6.5. Evaluate the area of disc with radius r.
Solution. The area of disc P is equal to the area of one quarter of the disc multiplied by
four. That is
P = 4
r
0
√
r2 − x2 dx.
We substitute x = r sin t. Then dx = r cos t dt and new limits of integration are: 0 0,
r π/2. We obtain
P = 4
r
0
√
r2 − x2 dx = 4r2
π
2
0
1 − sin2
t cos t dt = 4r2
π
2
0
√
cos2 t cos t dt = 4r2
π
2
0
cos2
t dt.
Applying the formula cos2
t = 1+cos 2t
2
we get
P = 4r2
π
2
0
1 + cos 2t
2
dt =
4r2
2
t +
1
2
sin 2t
π
2
0
= πr2
.
Example 6.6. Find the area of the region bounded by y = 6 − x2
and x + y − 4 = 0.
0
y
x
y = 4 − x
y = 6 − x2
−1 2
43
Solution. We first find limits of integration, i.e. x-coordinates of points of intersection. We
solve the equation
6 − x2
= 4 − x,
and therefore x = −1 and x = 2. Thus the area of the region is
2
−1
[6 − x2
− (4 − x)] dx =
2
−1
(2 + x − x2
) dx = 2x +
x2
2
−
x3
3
2
−1
=
9
2
.
Example 6.7. Evaluate the area of the region bounded by the graphs of the functions
y = 2x3
, y = 2
x
and y = x − 1 for x ≥ 0.
0
y
x
U2
U1
1 2
1
2
y = 2x3
y = 2
x
y = x − 1
Solution. We have to divide the region U into two parts U1 and U2, where U1 ∪ U2 = U (see
figure above). We find points of intersection of boundary curves
2x3
=
2
x
⇒ x = 1 (x = −1 does not satisfy geometrically)
and
x − 1 =
2
x
⇒ x = 2.
Therefore, the area of the given region S(U) is
S(U) = S(U1) + S(U2) =
1
0
2x3
dx +
2
1
2
x
− (x − 1) dx =
=
x4
2
1
0
+ 2 ln x −
x2
2
+ x
2
1
=
=
1
2
+ 2 ln 2 − 2 + 2 +
1
2
− 1 = 2 ln 2.
44
Volume of a Solid of Revolution
Let function y = f(x) be continuous nonnegative on interval [a, b]. The volume of a solid
made by revolving a subgraph of function f
P = {(x, y) : a ≤ x ≤ b, 0 ≤ y ≤ f(x)}
about an x-axis is given by the formula
V = π
b
a
f2
(x) dx. (6.4)
Example 6.8. Evaluate the volume of the cone with height v and a base radius r.
Solution. We get a cone by revolving a triangle bounded by the lines
y = 0, y = kx + q and x = v
about the x-axis. Where y = kx + q is equation for a line passing points [0, 0] and [v, r].
Therefore the gradient of the line is r
v
, q = 0 and the volume of the cone is
V = π
v
0
r
v
x
2
dx = π
r2
v2
v
0
x2
dx =
= π
r2
v2
x3
3
v
0
=
1
3
πr2
v.
Example 6.9. (Revolving of catenary curve) Find the volume generated by revolving the
subgraph of function y = 1
2
(ex
+ e−x
) about the x-axis for x ∈ [0, a].
Solution. The volume of the solid is given by the formula (6.4) and therefore
V = π
a
0
1
4
(ex
+ e−x
)2
dx =
π
4
a
0
(e2x
+ 2 + e−2x
) dx =
=
π
4
e2x
2
+ 2x −
e−2x
2
a
0
=
π
4
e2a
2
+ 2a −
e−2a
2
.
Example 6.10. (Volume of Torus) Find the volume of solid generated by revolving the
circle
x2
+ (y − 3)2
= 1
about the x-axis.
45
Solution. We first divide the circle into two functions f1 : y = 3 +
√
1 − x2 and f2 : y =
3 −
√
1 − x2. The volume of generated solid is equal to the difference in volumes of solids
generated by revolving the subgraphs of f1 and f2. Furthermore, both functions are symmetrical
about the y-axis. Therefore,
V = V1 − V2 = 2π
1
0
(3 +
√
1 − x2)2
dx − 2π
1
0
(3 −
√
1 − x2)2
dx =
= 2π
1
0
(3 +
√
1 − x2)2
− (3 −
√
1 − x2)2
dx =
= 2π
1
0
[9 + 6
√
1 − x2 + 1 − x2
− (9 − 6
√
1 − x2) + 1 − x2
] dx =
= 24π
1
0
√
1 − x2 dx = 6π2
.
Length of a Curve
Let function f have a continuous derivative on interval [a, b]. The length of its graph is given
by
L =
b
a
1 + (f′(x))2 dx. (6.5)
Example 6.11. Evaluate the length of the curve y = ln x on interval [
√
3,
√
8].
Solution. Since (ln x)′
= 1
x
and the length of the curve is given by (6.5), we have
√
8
√
3
1 +
1
x2
dx =
√
8
√
3
√
x2 + 1
x
dx.
To solve the obtained integral, we substitute t2
= x2
+ 1. Then 2x dx = 2t dt and therefore
dx = t
x
dt. New limits of integration are
√
3 2 and
√
8 3. We attain
3
2
√
t2
x
t
x
dt =
3
2
t2
t2 − 1
dt.
Finally, we decompose the integrand into partial fractions and evaluate the integral:
3
2
1 +
1
2
1
t − 1
−
1
2
1
t + 1
dt = t +
1
2
ln
t − 1
t + 1
3
2
= 1 +
1
2
ln
3
2
.
46
6.4 Improper Integrals
In the definition of the infinite integrals that we dealt with in the previous chapter, functions
to be integrated had to satisfy two restrictions:
• the integration domain was a bounded closed interval,
• the function was bounded on this interval.
This section deals with improper integrals, whose integration domain is a closed interval
unbounded from one side or whose integrand is an unbounded function.
Improper Integral on Unbounded Intervals
Definition 6.2. Let function f be continuous on interval [a, ∞). If there exists a proper
limit
lim
b→∞
b
a
f(x) dx, (6.6)
we say that improper integral
∞
a
f(x) dx
converges and assumes the value
∞
a
f(x) dx = lim
b→∞
b
a
f(x) dx. (6.7)
On the contrary, if the limit 6.6 is infinite or does not exist, the improper integral di-
verges.
Example 6.12. Evaluate the improper integrals
a)
∞
0
1
x2+1
dx, b)
∞
0
e−x
dx,
c)
∞
1
1
x
dx, d)
∞
1
1
x3 dx,
e)
∞
1
1
3√
x
dx, f)
∞
0
sin x dx.
Solution.
a) First we evaluate
F(b) =
b
0
1
x2 + 1
dx = [arctg x]b
0 = arctg b.
Then
lim
b→∞
F(b) = lim
b→∞
arctg b =
π
2
.
Therefore, the improper integral converges and its value is
∞
0
1
x2 + 1
dx =
π
2
.
47
b) Similarly, we evaluate
lim
b→∞
b
0
e−x
dx = lim
b→∞
−e−x b
0
= lim
b→∞
(−e−b
+ 1) = 1.
Thus the integral converges and assumes the value
∞
0
e−x
dx = 1.
c) To find the value of improper integral, we find the limit
lim
b→∞
b
1
1
x
dx = lim
b→∞
[ln x]b
1 = lim
b→∞
ln b = ∞.
Therefore, the integral diverges.
d) This time we have
lim
b→∞
b
1
1
x3
dx = lim
b→∞
−
1
2x2
b
1
= lim
b→∞
−
1
2b2
+
1
2
=
1
2
.
e) We evaluate
lim
b→∞
b
1
1
3
√
x
dx = lim
b→∞
3
2
x
2
3
b
1
=
3
2
lim
b→∞
b
2
3 − 1 = ∞.
f) We first find
F(b) =
b
0
sin x dx = − cos x
b
0
= − cos b + 1.
Then
lim
b→∞
F(b) = 1 − lim
b→∞
cos b
does not exist and the given integral diverges.
Note:
In general, integral
∞
1
1
xk
dx
converges to 1
k−1
for k > 1 and diverges for k ≤ 1.
48
Improper Integral of Unbounded Functions
Definition 6.3. Let function f be continuous and unbounded on interval (a, b]. Then the
point a is called a singular point of function f and the improper integral is defined by
b
a
f(x) dx = lim
c→a+
b
c
f(x) dx. (6.8)
If the limit (6.8) is a proper limit, we say that the integral converges.
If the limit is infinite or does not exist, the improper integral diverges.
Example 6.13. Find the singular points and evaluate the improper integrals
a)
1
0
1
x
dx, b)
2
0
1
2−x
dx,
c)
1
0
1√
1−x
dx, d)
1
0
1√
1−x2 dx.
Solution.
a) As we approach zero from the right side, function 1
x
is increasing to infinity and x = 0
is a singular point. Using (6.8) we get
1
0
1
x
dx = lim
c→0+
1
c
1
x
dx = lim
c→0+
[ln x]1
c = lim
c→0+
(ln 1 − ln c) = ∞,
That means that integral diverges.
b) Singular is x = 2. From (6.8) we obtain
2
0
1
2 − x
dx = lim
c→2−
c
0
1
2 − x
dx = lim
c→2−
[− ln(2 − x)]c
0 = lim
c→2−
[− ln(2 − c) + ln 2] = ∞,
Integral diverges.
c) Since the singular point is x = 1, we evaluate the limit
lim
c→1−
c
0
1
√
1 − x
dx = lim
c→1−
−2
√
1 − x
c
0
= lim
c→1−
(−2
√
1 − c + 2
√
1) = 2.
Therefore
1
0
1√
1−x
dx = 2.
d) Singular point is x = 1. We have
lim
c→1−
c
0
1
√
1 − x2
dx = lim
c→1−
[arcsin x]c
0 = lim
c→1−
(arcsin c − arcsin 0) =
π
2
.
Thus
1
0
1√
1−x2 dx = π
2
.
49
Note:
Generally, the integral
1
0
1
xk
dx
diverges when k ≥ 1 and converges to 1
1−k
when k < 1.
Example 6.14. Evaluate
a)
1
0
x ln x dx, b)
1
−1
1
x2 − 1
dx.
Solution. a) Singular point is x = 0. First we evaluate the definite integral
F(b) =
1
b
x ln x dx
using integration by parts, where u = ln x and v′
= x. Therefore u′
= 1
x
, v = x2
2
and
we get
1
b
x ln x dx =
x2
2
ln x
1
b
−
1
b
x
2
dx =
x2
2
ln x
1
b
−
x2
4
1
b
=
=
b2
4
−
1
4
−
1
2
b2
ln b.
Then we evaluate the limit
lim
b→0+
F(b) = lim
b→0+
b2
4
−
1
4
−
1
2
b2
ln b = −
1
4
−
1
2
lim
b→0+
b2
ln b.
Applying the L’Hospital’s rule we attain
lim
b→0+
b2
ln b = lim
b→0+
ln b
1
b2
= lim
b→0+
1
b
−2
b3
= lim
b→0+
(
1
−2
b2
) = 0.
Altogether
1
0
x ln x dx = −
1
4
.
b) Since both limits of integration are singular points, we divide the given integral into
two improper integrals whose one limit of integration is not a singular point:
1
−1
1
x2 − 1
dx =
0
−1
1
x2 − 1
dx +
1
0
1
x2 − 1
dx.
50
We evaluate the integrals separately and start with the second one. Using partial
fractions we attain
lim
b→1−
b
0
1
x2 − 1
dx = lim
b→1−
1
2
b
0
1
x − 1
−
1
x + 1
dx =
1
2
lim
b→1−
[ln |x − 1| − ln |x + 1|]b
0 =
=
1
2
lim
b→1−
ln
b − 1
b + 1
− ln | − 1| =
1
2
lim
b→1−
ln
b − 1
b + 1
= −∞.
As this particular integral diverges, the given integral diverges as well and the behaviour
of the first integral is not important.
51
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