INTEGRAL AND APPLICATION OF INTEGRAL

THAI NGUYEN UNIVERSITY OF EDUCATION
MATHEMATICS
----------

PROBLEM

SEMINAR

PROBLEM:

INTEGRAL AND
INTEGRAL

APPLICATION OF

Subject:

Mathematical analysis I

Supervisors:

NGUYEN VAN THIN

Class:

English of mathematics (NO1)

Members:

NGUYEN NHU QUYNH

June, 2022

CONTENTS
INTRODUCTION......................................................................................................................3


CHAPTER 1: PRIMITIVE.......................................................................................................4
1. Primitive functions...........................................................................................................4
2. Basic primitive table.........................................................................................................4
CHAPTER 2: DEFINITE AND INDEFINITE INTERGRALS............................................6
I. INDEFINITE INTEGRALS:.............................................................................................6
1 . Some basic concepts and properties :..............................................................................6
2. Methods for Calculating Uncertainty Integral :................................................................8
3. Integral of rational fraction functions.............................................................................11
4. Integral of trigonometric functions:................................................................................14
5. Integral of some irrational functions..............................................................................16
II. DEFINITE INTEGRALS:.............................................................................................18
1. Definition, geometric meaning of definite integral........................................................18
2. Integralability criteria. Properties of definite integrals...................................................19
3. Derivative formula with bounds. Formula Newton-Leibniz..........................................22
4. Calculation methods.......................................................................................................24
III. APPLICATIONS OF DEFINITE INTEGRALS:.......................................................28
1. Compute the area of a plane figure.................................................................................29
2. Calculate the length of the curve....................................................................................31
3. Calculate the volume of an object..................................................................................34
4. Calculate the area of the rotating circle..........................................................................37
CHAPTER 3: IMPROPER INTEGRALS............................................................................39
I. IMPROPER INTEGRALS TYPE I:...............................................................................39
II. IMPROPER INTEGRALS TYPE II:............................................................................41
III. CONVERGENCE STANDARDS:...............................................................................43

IV. ABSOLUTELY CONVERGENT AND SEMI-CONVERGENT IMPROPER
INTEGRALS:.......................................................................................................................46
CONCLUSION........................................................................................................................49
REFERENCES.........................................................................................................................50

INTRODUCTION
2


Mathematics is the science that studies the spatial structure of numbers and
transformations. In other words, it is said that it is the subject of "figures and
numbers." Mathematics is the foundation for all other natural sciences. It can be
said that without mathematics, there would be no science at all. Mathematics is
divided into many sub-disciplines, including analysis. Calculus is a branch of
Mathematics that studies concepts: limits, derivatives, primitives, integrals... The
basic math of calculus is taking limits. Most learners are very confused and have
difficulty learning calculus in general and primes, integrals, and practical
problems that need to use integration in particular.
Integration of applications in some problems about finding limits, proving
inequality, or calculating area..
Besides, in the graduation exam of high school, university and college,
there are always problems related to integration.
With the desire to systematize knowledge about integrals and its
applications, we have chosen the topic "Integrals and applications" as the
research topic.

3


CHAPTER 1: PRIMITIVE

1. Primitive functions
a. Definition:
Let f ( x) be defined on K . The function F ( x) is called a primitive of f ( x) on K
if F '( x)  f ( x) , x �K .
- Theorem 1.1: If F ( x) is a primitive of the function f ( x) on K , then with
each of constant C , the function G ( x)  F ( x)  C is also a primitive of f ( x) on K .
- Theorem 1.2: Let f ( x ) be an integrable map on I and F a primitive. Then any
primitive of f is of the form F ( x)  c , with the constant c varying in �.
- Theorem 1.3: Every function f ( x) that is continuous on K has a primitive
on K .
b. Proposition:
- Proposition 1:

f '( x )dx  f ( x )  C
�

- Proposition 2:

kf ( x)dc  k �
f ( x)dx
�

( k is a constant, k �0 )

f ( x)dx ��
g ( x) dx
 f ( x) �g ( x) dx  �
- proposition 3: �

2. Basic primitive table.

1.
 1

1 ( ax  b)
 C ( �1)
  1
1
1
dx  ln ax  b  C
�
a
2. ax  b
1
e ax b dx  e ax b  C
�
a
3.


( ax  b)
�

dx 

10.

4.
5.

�

sin

2

e ax (a cos bx )  b sin bx
C
a 2  b2
1

cos(ax  b)dx  sin(ax  b)  C
�
a
11.

1

12.

1

sin(ax  b)dx   cos(ax  b)  C
�
a

e ax cos bxdx 
�

1

dx  tan( ax  b)  C

�
cos (ax  b)
a
2

1

tan( ax  b)dx   ln cos(ax  b)  C
�
a
13.

1
1
dx   cot(ax  b)  C
(ax  b)
a

dx

14.
4

�
a x
2

2



1
x
arctan  C
a
a


1

6.

cot(ax  b)dx  ln sin(ax  b)  C
�
a

7.

�
a x

dx

8.
9.

2

�a

2



dx
2

 x2

1
ax
ln
C
2a a  x

 arcsin

dx

15.

�x

16.

�
x x

a

2

2

dx

x
C
a

17.

� b�

ln( ax  b) dx  �x  �
ln( ax  b)  x  C
�
� a�

18.

5

2





 ln x  x 2  a 2  C

 a2



1
x
arccos  C
a
a

2
2
�a  x dx 

x a2  x2 a2
x
 arcsin  C
2
2
a

e ax sin bxdx 
�

e ax (a sin bx)  b cos bx
C
a2  b2

CHAPTER 2: DEFINITE AND INDEFINITE INTERGRALS
I. INDEFINITE INTEGRALS:
1 . Some basic concepts and properties :
a, b
Definition : Given a function f ( x ) defined on the interval   . A function
F ( x) that is differentiable on  a, b  is said to be a primitive of f ( x ) if F '( x )  f ( x)

It is easy to see that if F ( x) is a primitive of f ( x ) , then :
 Every function of the form F ( x)  C (with C �� ) is also a primitive of
f ( x) .

 All other primitives of f ( x) have the form F ( x )  C .

The family of all primitives of f ( x) , is called the uncertainty integral of f ( x )
, denoted by :
f ( x )dx  F ( x)  C
�

Theorem.
a, b
1. Every function f ( x ) that is continuous on the interval   has a primitive
on this interval.

2.
3.
4.

F '( x)dx  F ( x ).
�

 �f ( x)dx 

/

 f ( x)

.

f ( x )dx  b �
g ( x )dx  C
 af ( x)  bg ( x)  dx  a �
�

with a, b ��.

* Basic primitive table.
f ( x)

F ( x)
C

0

 x  a
1
xa



( x  a ) 1

( �1)
 1
ln x  a
6


sin  ax  b 
cos  ax  b 
1
cos 2 x
1
sin 2 x
1
sin x
1
cos x
1
x2  b

x2  b
1
a  x2
2

a2  x2
1
x  a2
eax b
2

1
 cos  ax  b 
a
1
sin  ax  b 
a
tan x

 cot x
ln tan

�x  �
ln tan �  �
�2 4 �
ln x  x 2  b
1
b
x x 2  b  ln x  x 2  b
2
2
x
arcsin , a  0
a

1
a2
�x �
x a 2  x 2  arcsin � �
2
2

�a �
1
x
arctan
a
a
1 ax  b
e
a

3

2�
�
� x  �dx
�
x� .
Example 1. Find the interval : �

Solution.
3

2�
�
x 3 2  6  12 x 3 2  8 x 3  dx

� x  �dx  �
�
x�
�

2
 x 5 2  x  24 x 1 2  4 x 2  C
5

Example 2. Find the interval :

x
2

sin 2 x cos 3 xdx
�

Solution.

7


1

1

1

sin 2 x cos 3 xdx  �
 sin 5 x  sin x  dx  cos x  cos 5 x  C.
�
2
2
10

2. Methods for Calculating Uncertainty Integral :
* Variation method:
Consider the integral

f ( x )dx  F ( x)  C
�

. Then, with the discriminant function

u ( x) , we have:
f (u ) du  F (u )  C.
�

Case 1 : Consider the integral

I �
f ( x)dx

. Set t  u ( x) , if we get

f ( x)dx  g (u )du , then:
f ( x)dx  �
g  u ( x)  du ( x)  G (u )  C ,
�
(when G

is a primitive of g )

The main idea of the above method is that we put a factor of f ( x) into d (.) ,
so that the integral expression can be easily computed .

Example 1. Find the interval :

e
�

sin x

cos xdx

Solution.
e
�

sin x

cos xdx  �
esin x d sin x  esin x  C.

ln 2 x
�x dx
Example 2. Find the interval :

Solution.
ln 2 x
1
ln 2 xd ln x  ln 3 x  C.
�x dx  �
3
x

Example 3. Find the interval :

dx
�
x 1
4

Solution.
x
1
dx 2
1
dx

dx  arctan  x 2   C.
2
4
�
�
x 1
2  x2   1
2
8


e

Example 4. Find the interval :

x

�x dx

Solution.
e

x

e d
�x dx  2�
x

x  2e

x

 C.

Example 5. Find the interval :

x 2 x  3dx
�
.

Solution.
t2  3
x
, dx  tdt.
2
Set t = t  2 x  3 . Hence:

1

1

x 2 x  3dx  �
 t  3 t dt  10 t
�
2


1
10

2

2

 2 x  3

5



1
2

Case 2 : Consider the integral

5

1
 t3  C
2

 2 x  3

3

C

I �
f ( x)dx

. Set x  x(t ) , hence:

f ( x)dx  �
f  x (t )  x '(t )dt ,
�

Where x  x(t ) is a function with continuous derivative and inverse function.
Setting x  x(t ) needs to satisfy the requirement that the integral expression after
setting must be simpler and easier to calculate than the original expression. For
2
2
example, if in the expression of f ( x) there is a  x , we put x  a sin t ; and in

1
2
2
f

(
x
)
case the expression
contains x  a , we need to set x  a tan t .

dx

Example 1. Find the interval :

�1  x



2 32

Solution.
�  �
x  sin t , t ��
 ; �
.
2
2
�
�
Set

9

dx

�1  x



2 32

dt
 � 2  tan t  C, t  arcsin x.
cos t

Example 2. Find the interval :

�x


1

2

 4

2

dx

Solution.
Set x  2 tan t. We have:

�x



1

2

 4

2

dx 

1
1� 1
x
�
cos 2 tdt  �
t  sin 2t � C , t  arctan .
�
8
16 � 2
2
�

* Integral by parts method:
uv '   uv  ' vu '
With two differentiable functions u ( x), v( x) , we have
. From

that, we infer:
f ( x)dx  �
u ( x)v '( x)dx  �
u ( x)dv( x)  u ( x)v ( x)  �
v( x )u '( x )dx.
�

(2.1)

The meaning of the above method is: instead of integrating the whole
expression f ( x) (complex), we only need to integrate one of its factors. In the
above formula, we analyzed f ( x)  u( x)v '( x) and integrated the second factor .
This analysis must ensure that the requirements of the resulting integral are
simpler than the original integral expression.
* Integral types calculated by the method of partial integration:
Integral form
P ( x)e
�
n

ax

dx

How to apply the integral part formula
ax
Let e into in d (.) , apply (2.1) n times

P ( x )sin(ax  b)dx
�

Let sin( ax  b) into in d (.) , apply (2.1) n times

P ( x ) cos(ax  b)dx
�
P ( x) lnQ ( x)dx
�

Let cos(ax  b) into in d (.) , apply (2.1) n times

n

n

n

m

e cos( ax  b)dx
�
ax

Let Pn ( x) into in d (.) , apply (2.1) once
Put one of the two factors in d (.) , apply (2.1)
twice
10


e cos( ax  b)dx
�

ax

Put one of the two factors in d (.) , apply (2.1)
twice

Example 1. Find the interval :

ln xdx
�

Solution.
ln xdx  x ln x  �
1dx  x ln x  x  C .
�

Example 2. Find the interval :

arctan xdx
�

Solution.
x

1

arctan xdx  x arctan x  � dx  x arctan x  ln( x
�
x 1
2

2

2

Example 3. Find the interval :

e x
�
x

2

 2 x  3 dx

 1)  C .

.

Solution.
Applying the integral by parts formula twice, we have:
e x
�
x

2

 2 x  3  dx  �
 x 2  2 x  3 de x

 e x  x 2  2 x  3  2 �

e x ( x  1)dx
 e x  x 2  2 x  3  2 �
( x  1)de x
 e x ( x 2  3)  C.

Example 4. Find the interval :

I �
arccos 2 xdx

Solution.
2
Set u  arccos x. We have

du 

2 arccos xdx
1  x2

11

.

This implies that


2 xar cos x
I  x arccos 2 x  �
dx
1  x2

 x arccos 2 x  2 �
arccos xd 1  x 2
 x arccos 2 x  2 1  x 2 arccos x  2 x  C

3. Integral of rational fraction functions
In this category, we consider integrals of the following form:
Pn ( x )

dx,
�
Q ( x)
m

Where Pn ( x), Qm ( x) are polynomials of degree n, m variables x . Without loss
of generality , we can assume n  m , because in the other case we only need to
divide the numerator by the denominator and get the remainder . The general
idea for calculating rational integrals is that we decompose the expression under
the integral sign into the sum of simplified rational integrals. There are two types
of simplified fractions:
1
1
q
2
(a  x ) p and  x  ax  b 
2
When a  4b  0 (i.e. it is not possible to decompose the quadratic trinomial

x 2  ax  b into a monomial ) , p, q ��.

Example 1. Find the interval :

1
I1  �
dx
p
 a  x

and

1
I2  �
dx
q
 x 2  ax  b 

Solution.
Applying transformation 1 , we easily get:
1

�
 a  x

p

�ln x  a  C
if p  1
�
 p 1
dx  � x  a 
 C if p �1
�

�  p 1

The second integral can be reduced to the form:
12


�
(u

2

1
du
  2 )q

a
a2
u  x  ;  b  .
2
4 ). Set u   tan t , we have:
( when

�u


1
2





2 q

du   1 2 q �
cos 2q  2 tdt.

The final integral can be calculated by applying the lower order formula to
the cos function.
dx

�
x  2x  2
2

Example 2. Find the interval :
Solution.

d ( x  1)
 arctan( x  1)  C.
2
1

dx


�
x  2x  2 �
( x  1)
2

dx

Example 3. Find the interval:

�
x  3x  2
2

Solution.
dx

�1

1 �

 �

� ln
�
x  3x  2 �
�x  2 x  1 �
2

x2
 C.
x 1

4 x3  8 x2  6 x  4

Example 4. Find the interval:

� 1  x 

2

(1  x 2 )

dx

Solution.
4 x3  8 x 2  6 x  4

Analysis:

 1 x

2

 1 x 
2



A
B
Cx  D

 2
.

2
x  1 ( x  1)
x 1

4 x 3  8 x 2  6 x  4  A  x  1  x 2  1  B  x 2  1   Cx  D   x  1 x ��.
2

That implies:

Identifying the coefficients of x on both sides, we get the system of equations:

13


� AC  4
�A  B  2C  D  8
�
�
� A  C  2D  6
�
� A B  D  4

Solving the above system of equations, we get: A  2, B  1, C  2, D  1 .
This implies that:
�2
1
2x
1 �
I �


 2
 2 �
dx
�
2
�x  1  x  1
x 1 x 1 �
�
�
 2 ln x  1 

1
 ln  x 2  1  arctan x  C
x 1
.

5. Integral of trigonometric functions:
* The function under the integral is even/odd for sin/cos :
Integral form
Odd for sin x

Solution

Set t  cos x
Set t  sin x

Odd for cos x
Even for both sin x and cos x Set t  tan x or applying the downgrade formula
Table 3 : How to solve for even/odd function case for sin/cos
Example 1. Find the interval :

I �
cos3 x sin 8 xdx

.

Solution.
Set t  sin x , we have:
I �
cos 2 x sin 8 x  cos xdx 

�
 1  sin 2 x  sin 8 x  cos xdx 
�
 1  t 2  t 8dt


t 9 t 11
  C , t  sin x.
9 11

 2sin x  3cos x  dx
I �2
sin x cos x  9 cos3 x .
Example 2. Find the interval :
Solution.
14


3

Dividing both numerator and denominator by cos x and set t  tan x , we have:

 2 tan x  3 d  tan x 
I �
tan 2 x  9
2t  3
�
dt
t2  9
2t
3
�
dt  �
dt
2
2
t 9
t  32
t
 ln  t 2  9   arctan  C , t  tan x.
3
* The function under the integral sign is a rational function for sin and
cos:
I �
R  sin x, cos x  dx

. Set

t  tan

x
2 , applying the formula:

2t
1 t2
2dt
sin x 
, cos x 
, dx 
2
2
1 t
1 t
1 t 2

to return the integral of the rational analytic function .
dx
I �
3sin x  4 cos x  5
Example 1. Find the interval :

Solution.
x
�  �
t  tan , x ��
 , �
2
2 2�
�
Set

. We have:

dt
I  2�
6t  4(1  t 2 )  5  1  t 2 
dt
2
x
2
 2�
 2�
 C , t  tan .
 t  3 d  t  3 
2
t  6t  9
t 3
2

 2sin x  cos x  3 dx
I �
3sin x  4 cos x  5
Example 2. Find the interval:
Solution.
Analysis: 2sin x  cos x  3  A  3sin x  4 cos x  5   B  3sin x  4 cos x  5  ' C.
15


That implies , we have a system of equations:
3 A  4B  2
�

�
�4 A  3B  1
�5 A  C  3
�

Solution the system of equations , we get that : A  2 5, B  1 5, C  1.
Hence:
d  3sin x  4 cos x  5 
Cdx
I  A�
dx  B �
�
3sin x  4 cos x  5
3sin x  4 cos x  5
2
1
 x  ln  3sin x  4 cos x  5   I1.
5
5

where I1 is the integral in the previous example .
5. Integral of some irrational functions
In this section, we consider integrals containing roots. In the program, we
encounter mainly the following two types of integrals:
x2
I �
dx
x2  2x  2
2x  2
I  �x 2  2 x  2dx  �

dx
2
x  2x  2

d  x2  2 x  2

 � x  1  1d  x  1  �
2





x2  2x  2



1
1
 x  1 x 2  2 x  2  x  1  x 2  2 x  2  2 x 2  2 x  2  C.
2
2

� ax  b
I �
R�
�x, cx  d
�
Form 1:
tm

�
�
�
�, when R( , ) is a rational function

ax  b
cx  d , represent x in terms of t and then substitute

In this case, we set
the original integral to return to the rational function integral form.
dx
I �
2 x 1  4 2 x 1
Example 1. Find the interval :
16


Solution.
4
3
Set 2 x  1  t , t  0. This implies that 2dx  4t dt. We have:

2t 3dt
t � 2
�
I �
 2�
t 1
dt  t  2t  2 ln t  1  C , t  4 2 x  1

�
�
2
t t
t 1 �
�
.
x 1
I �
dx
x2
Example 2. Find the interval :

Solution.

Set

t

x 1
x  2 , we have

2t 2  1
2t
x 2
, dx 
dt
2
2
t 1

 t  1

and

t2
I  2�
dt
2
 t 2  1


1 �1
1
1
1 �



dt
�
2
2�
2�
t  1 t  1  t  1
 t  1 �
�
�
�

1 t 1 1 �1

1 �
 ln
 � 
� C , t 
2 t  1 2 �t  1 t  1 �





I �
R x, ax 2  bx  c dx

Form 2:

x 1
.
x2

, when R  ,  is a rational function

In this case, we find a way to return to one of the basic formulas shown in
Table 1 or convert to integral trigonometric functions.
dx

Example 1. Find the interval :

�2 x  x

2

Solution.
dx

�2 x  x

2

d  x  1
�
 arcsin  x  1  C.
2
1   x  1

Example 2. Find the interval :

�x

dx
2

 2x  2

Solution.
17


�x

d  x  1

�
 ln x  1  x 2  2 x  2  C.
2
 2x  2
 x  1  1





dx

2

x2
I �
dx
2
x

2
x

2
Example 3. Find the interval :
.

Solution.
2x  2
I  �x 2  2 x  2dx  �

dx
2
x  2x  2

d  x2  2x  2

 � x  1  1d  x  1  �
2





x2  2x  2



1
1
 x  1 x 2  2 x  2  x  1  x 2  2 x  2  2 x 2  2 x  2  C
2
2

II. DEFINITE INTEGRALS:
1. Definition, geometric meaning of definite integral

Let the function f ( x) be defined on the segment  
. Consider the
problem of finding the area of a curved trapezoid, which is the plane part
a, b ��

bounded by the graph y  f ( x ) , the line x  a, x  b and the Ox axis. Since the
area of a curved trapezoid cannot be calculated using an explicit formula, the
most natural approach is to subdivide the area to be calculated and approximate
it by the sum of the rectangular areas. More specifically, we proceed to divide
a, b
segment   into n segments by dividing points x0  a  x1  x2  ...  xn  b . On

each segment dividing i , we choose a representative point i and construct a
rectangle with height f  i  . Then the sum of the areas of the rectangles is

� f   
n

i 1

i

i

. Set

 : max   i :1 �i �n

f   
. The value of �
n

i 1

i

i

is called the

a, b
sum of the integrals of the function f over the segment   . Obviously, this

sum depends not only on the function f , but also on how the segment  a, b is
divided and how the representative point i is chosen. Intuitively, it can be seen
that, as the length of each division approaches zero, the sum of the areas of the
rectangles will gradually approach the area of the curved trapezoid.
18


lim �0 �i 1 f  i   i
n

Definition. If the limit

exists, regardless of how the

a, b
segment   is divided and how the representative point i is chosen, then we
a, b
say the function f ( x) integral on the segment   . Also, the value of that limit
a, b
is called the definite integral of the function f ( x) over the segment   ,
b

denoted by

�f ( x)dx .
a

Example. Consider the integrability of the Dirichlet function, given by the
�1
f ( x)  �
0
�
formula

if x ��
if x ��\ � on the closed interval  0,1 .

Solution.
0,1
Divide segment   into n segments of equal length, on the i-th division,

choose two representative points i �� and i ��\ � . The respective sums of
integrals are
Sn 

1 n
�f  i   1, n
n i 1

And
S 'n 

1 n
�f  i   0, n
n i 1

The limits of two integral sums are different, depending on how the
representative points are chosen. So function f ( x) is not integrable
In the problem of calculating the area of a curved trapezoid, we have divided
a, b
segment   into segments of very small length and considered the function
f ( x ) to be constant on each of these divisions. Thus, we can approximate the

area of a curved trapezoid by the sum of the areas of the rectangles. We will use
this idea again in the path length problem. Assume an object moving in a straight
line has a velocity equal to v(t ) at time t . Calculate the length of the object's
t ,t
path between t1 and t2 . To solve this problem, we divide segment  1 2  into
19


segments of very small length i . Consider the body moving uniformly on each
f  
of these segments. Then, the distance traveled by the object in time i is  i  i
t ,t
with i � i . Therefore, the distance covered by the object in time  1 2  is exactly
t2

equal to the integral

�v(t )dt

t1

2. Integralability criteria. Properties of definite integrals.

Theorem 2.1. A function f ( x) defined on a segment   will be integrable
on that segment if one of the following conditions is satisfied:
a, b

a, b
 The function f ( x) is continuous on the interval   .

 The function f ( x) is bounded and has a finite number of discontinuities on

 a, b  .
a, b
 The function f ( x) is monotonous and bounded on the interval   .
1

Example 1. Find the integral

�x dx
2

0

by definition.

Solution.
The function f ( x )  x is continuous so it is integrable on the interval  0,1 .
From this, we infer that the value of the limit of the sum of the integrals does not

2

0,1
depend on the division of segment   and the choice of representative point in
0,1
each division. We can equally divide   into n equal segments, each of which
1
is equal to n . On each division, we choose a representative point that coincides

with the right end of that segment, i.e.
obtained to be

i 

2

n

1
n . Then, the total integration is

�i �1 1 n
S n  �� �  3 �i 2
n �n n i 1
i 1 �

We have:
20

n( n  1)(n  2) 1

n ��
6n3
3

lim S n  lim
n ��

1

Therefore, we conclude

1

�x dx  6 .
2

0

a, b
Theorem 2.2. Let f ( x ), g ( x) be two integrable functions on the interval   .
Then

1. For all  ,  �� , we have:
b

b

b

a

a

a

�  f ( x)   g ( x)  dx   �f ( x)dx   �g ( x)dx.
2. For all

c � a, b 

, we have:
b

c

b

a

a

c

�f ( x)dx  �f ( x)dx  �f ( x)dx.
b

a

a

b

Note

�f ( x)dx :  �f ( x)dx

3. If

f ( x) �g ( x) x � a, b 

4. The function

f ( x)

if a  b .
b

b

a

a

f ( x)dx ��
g ( x) dx.
then �

a, b

is also integrable on the interval   and
b

b

a

a

�f ( x)dx ��f ( x) dx .
 a, a 
5. If f ( x ) is an even function and integrable over the interval 
, then
a

a

a

0

�f ( x)dx  2�f ( x)dx.
 a, a 
6. If f ( x ) is an odd function and integrable over the interval 
, then
a

�f ( x)dx  0.
a

7. If f ( x ) is a periodic function with period T and integrable over every
segment in �, then

21


a T

T

a

0

� f ( x)dx  �f ( x)dx
8. (First integral mean theorem) If

a ��.

m �f ( x ) �M , x � a, b 

, then

b

m(b  a ) ��f ( x )dx �M (b  a ).
a

a, b
In particular, if f ( x ) is continuous on the segment   , then there exists

c � a, b 

b

f ( x )dx  f (c)(b  a ).
such that �
a

9. (Second integral mean theorem). Suppose
(i) m �f ( x) �M , x � a, b  .

(ii ) g ( x) does not change sign on  a, b  .

We have
b

b

a

a

�f ( x) g ( x)dx   �g ( x)dx, m � �M .
a, b
c � a, b 
In particular, if f ( x ) is continuous on   , then there exists
such
that
b

b

a

a

�f ( x) g ( x)dx  f (c) �g ( x)dx.
5
8
3 sin x cos x
I �
dx
4

cos
x

1
Example2. Calculating the integral

Solution.
Since the function under the integral sign is a periodic function with period
2 , we have
5
8
 sin x cos x
I �
dx

cos 4 x  1

On the other hand, the function under the integral sign is an odd function, the
 ;  
integral is signed on the symmetry segment 
, so we have I  0 .
22


2 sin x
I �
dx
0
x
Example 3. Determine the sign of the integral
.

Solution.
 sin x
2 sin x
I  � dx  �
dx
0

x
x
 sin x
 sin( y   )
�
dx  �
dy ( set y    x)

0
0
x
y 
 sin x
 sin x
 � dx  �
dx
0
0
x
x 
  sin x
�
dx  0
0 x( x   )

3. Derivative formula with bounds. Formula Newton-Leibniz
Up to now, there does not seem to be a relationship between the indefinite
integral and the definite integral. To compute a definite integral, we need to
calculate the limit of the sum of the integrals. This is cumbersome and difficult to
do. In this section, we will give a formula for the definite integral of a function
through the primitive of that function.
a, b
Theorem 3.1. Let the function f ( x) be integrable on the segment   . For

all

x � a, b 

x

, set

F ( x) : �f (t )dt.
a

Then, we have
a, b
1. If f ( x ) is integrable on the interval   then F ( x) is continuous on that
interval.

2. If f ( x ) is continuous on the segment  a, b then F ( x) is a primitive of f ( x) ,
that is



x

�f (t )dt
a



/

 f ( x),

(2.3)

At the same time we have:
b

�f (t )dt  F (b)  F (a)
a

23

(2.4)


Equation (2.4) is called the Newton-Leibniz formula. It is also true when F ( x)
is any primitive of f ( x) . Furthermore, the formula (2.3) can be extended to the
case that both bounds of the integral depend on x:

�

 ( x)

 ( x)



/

f (t ) dt  f   ( x)   '( x)  f   ( x)   '( x).

d sin x dt
�
cos x

1 t4 .
Example 1. Calculating dx

Solution.
d sin x dt
cos x
sin x


.
�
4
4
cos
x
dx
1 t
1  sin x
1  cos 4 x
arcsin x

lim

x �0

Example 2. Calculate the limit:

� arctan tdt

0

arctan x

.

� arcsin tdt
0

Solution.
0
It is easy to see that the limit has the form 0 . Applying L'Hospital's rule, we

have
/

arcsin x
�
�
1
arctan(arcsin x).
� � arctan tdt �
1  x 2  1.
� lim
lim � 0
/

arctan x
x �0
1
�
� x �0

arcsin(arctan x). 2
arcsin
tdt
��
�
x 1
�0
�

Example 3. Calculate limit
1� 1
2
n�
L  lim �
1


1


...

1

.
�
�
n �� n �
n
n

n
�
�

Solution.
We will write the expression to compute the limit as the sum of the integrals
0,1
of a function f ( x ) on the interval   . Then, the limit to be calculated is exactly
24


1

equal to the definite integral

�f ( x)dx . We divide the segment  0,1
0

into n

1
segments, each of length n . On each division, choose a representative point i

that coincides with the right end of the division, i.e.

i 

i
n . Then, the sum of the

0,1
integrals of the function f ( x) over the segment   is equal to
1 n �i �
�f � �.
n i 1 �n �

Compared with the expression to calculate the limit, we need to choose the
function f ( x )  1  x . Then we have
1

L  �1  xdx 
0





2
2 2 1 .
3

The Newton-Leibniz formula acts as a bridge between primitives and definite
integrals. Thanks to this formula, to calculate the definite integral of a function,
we just need to calculate the primitive of that function and then substitute the
bound for the found primitive. In the next section, we present methods for
definite integrals, which are a direct consequence of the primitive methods and
the Newton-Leibniz formula.
4. Calculation methods
* Variable method:
b

�f ( x)dx . Set

Suppose we need to integrate
have

x   (t ),  ( )  a,  ( )  b . Then we

a

b



a



�f ( x)dx  �f   (t )   '(t )dt ,

(2.5)

where  '(t ) is a continuous function.

 1 x 
Example 1. Calculating the integral �
1

0

Solution.
25

2

3

dx.