- •Міністерство освіти і науки україни донецький національний технічний університет
- •Integral calculus (інтеґральне числення)
- •Донецьк 2005
- •Integral calculus lecture no. 19. Primitive and indefinite integral
- •Point 1. Primitive
- •Properties of primitives
- •Point 2. Indefinite integral and its properties
- •Point 3. Integration by substitution (change of variable)
- •Point 4. Integration by parts
- •Lecture no.20. Classes of integrable functions
- •Point 1. Rational functions (rational fractions)
- •Point 2. Trigonometric functions
- •Universal trigonometrical substitution
- •Other substitutions
- •Point 3. Irrational functions
- •Quadratic irrationalities. Trigonometric substitutions
- •Quadratic irrationalities (general case)
- •Indefinite integral: Basic Terminology
- •Lecture no. 21. Definite integral
- •Point 1. Problems leading to the concept ofa definite integral
- •Point 2. Definite integral
- •Point 3. Properties of a definite integral
- •I ntegration of inequalities
- •Point 4. Definite integral as a function of its upper variable limit
- •Point 5. Newton-leibniz formula
- •Point 6. Main methods of evaluation a definite integral Change of a variable (substitution method)
- •Integration by parts
- •Lecture no.22. Applications of definite integral
- •Point 1. Problem – solving schemes. Areas
- •Additional remarks about the areas of plane figures
- •Point 3. Volumes
- •Volume of a body with known areas of its parallel cross-sections
- •Volume of a body of rotation
- •Point 4. Economic applications
- •Lecture no. 23. Definite integral: additional questions
- •Point 1. Approximate integration
- •Rectangular Formulas
- •Trapezium Formula
- •Simpson’s formula (parabolic formula)
- •Point 2. Improper integrals
- •Improper integrals of the first kind
- •Improper integrals of the second kind
- •Convergence tests
- •Point 3. Euler г- function
- •Definite integral: Basic Terminology
- •Lecture no. 24. Double integral
- •Point 1. Double integral
- •Point 2. Evaluation of a double integral in cartesian coordinates
- •Point 3. Improper double integrals. Poisson formula
- •Point 4. Double integral in polar coordinates
- •Double integral: Basic Terminology
- •Contents
- •Integral calculus 3
- •Integral calculus (Інтеґральне числення): Методичний посібник по вивченню розділу курсу ”Математичний аналіз” для студентів ДонНту (англійською мовою)
Lecture no. 23. Definite integral: additional questions
POINT 1. APPROXIMATE INTEGRATION
POINT 2. IMPROPER INTEGRALS
POINT 3. EULER Г-FUNCTION
Point 1. Approximate integration
We’ll study the case of nonnegative function when a definite integral
represents the area of a curvilinear trapezium bounded by straight lines , , the Ox-axis and a graph of the function . The same results remain valid in general case.
Rectangular Formulas
A. We divide the segment into n equal parts of the length by points
.
Straight lines divide the curve into n parts . Let’s denote by
Fig. 1
the values of the function at the division points (see fig. 1)
a) Substituting every part of the curve by the segments of straight lines we substitute the curvilinear trapezium by the set of rectangles with total area
Therefore
( 1 )
b) Similarly, substituting every part of the curve by the segments of straight lines , we get
( 2 )
Absolute error of the formulas (1), (2) has the order 1/n, that is
.
c) Dividing the segment into 2n equal parts of the length by points
(fig. 2),
we substitute the curvilinear trapezium by the set of rectan- Fig.2 gles with bases 2h, altitudes and total area
Hence
( 3 )
Absolute error of the formula (3) has the order 1/n2, that is
It means that the formula (3) is more exact than both (1) and (2).
Trapezium Formula
After dividing the segment into n equal parts
(fig. 3) we divide the graph of the function into n parts of the length by points
.
Fig. 3 Substituting every part of the graph by segments
we substitute the curvilinear trapezium by the set of trapeziums with total area
.
So we get the next approximate formula (trapezium formula) ,
( 4 )
Its absolute error has the order 1/n2, that is
.
It means that the formulas (3) and (4) have the same order of exactness.
Simpson’s formula (parabolic formula)
Let’s divide (by points ) the segment into an even number 2n of equal parts of the length and let
be points of the curve corresponding to the division points (see fig. 4 for the case 2n = 6).
At first we draw a quadratic parabola
through points (see fig. 4, 5). One can prove that the area of the figure between the arc of the parabola and the segment of the Ox-axis Fig. 4 equals
,
and we can write
.
Fig. 5 Carrying out the same procedure on the point triplets …, we get
.
Finally we get Simpson’s formula for approximate integration
(5)
Simpson’s formula (5) is the most exact in comparison with (1) – (4). Indeed, its absolute error has the order 1/n4 that is
.
For example in the case n = 3, 2n = 6 (fig. 4) Simpson’s formula has the next form
.
Ex. 1. Calculate approximately the integral .
Let’s form the next table of values of the argument and the function:
i |
|
|
|
0 |
0.0 |
0.00 |
0.0000 |
1 |
0.2 |
0.04 |
0.0400 |
2 |
0.4 |
0.16 |
0.1593 |
3 |
0.6 |
0.36 |
0.3523 |
4 |
0.8 |
0.64 |
0.5972 |
5 |
1.0 |
1.00 |
0.8415 |
6 |
1.2 |
1.44 |
0.9915 |
7 |
1.4 |
1.96 |
0.9249 |
8 |
1.6 |
2.56 |
0.5487 |
It corresponds to division of the segment into parts of the length
.
By the formula (1)
.
By the formula (2)
.
Using the formula (3) we take 2n = 8, n = 4, , and so
.
By the formula (4)
.
We’ll apply the formula (5) two times.
At first we divide the segment into 2n = 4 parts, , , , , , correspondingly , , , , , , and therefore
Dividing now the segment into 2n = 8 parts, , we have
It’s useful to compare all these results with known approximate value of the same integral up to , namely
.