- •Донецьк 2006
- •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 a 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 2. Arс length
- •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
- •Simpson10 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
- •Differential equations lecture no.25. First and second order differential equations
- •Point 1. General notions
- •Point 2. Integrable types of the first order differential equations (of de - 1)
- •1. Separated de-1 (de-1 with separated variables)
- •2. Separable de-1 (de-1 with separable variables)
- •3. Homogeneous de-1
- •4. Linear de-1
- •5. Bernoulli de-1
- •Point 3. Order reducing second order differential equations
- •Lecture no.26. Second order linear differential equations
- •Point 1. General notions
- •Point 2. Linear dependence and independence
- •Point 3. Homogeneous equations Structure of the general solution of so lhde
- •So lhde with constant coefficients
- •Point 4. Nonhomogeneous equations Structure of the general solution of so lnde
- •Method of variation of arbitrary constants
- •Method of undetermined coefficients for so lnde with constant coefficients
- •Lecture no. 27. Systems of differential equations. Approximate integration of differential equations
- •Point 1. Normal systems of differential equations
- •Point 2. Approximate integration of differential equations Successive approximations method
- •Euler method
- •Differential equations: Basic Terminology
- •Bibliography textbooks
- •Problem books
- •Contents
Point 5. Newton-leibniz formula
Theorem 3. If a function is continuous one on a segment a, b, and F(x) is one of its primitives, then Newton4-Leibniz5 formula for evaluation of the definite integral of the function over the segment a, b is true
( 23 )
■ We have two primitives: and the integral (21) with upper variable limit x. By corresponding property of primitives the difference
.
To find the value of the constant C we put . So
,
and therefore
.
Substituting x by b and t by x we obtain the formula (23).■
Note 1. The expression
,
which means the action , is often called the double substitution.
Ex. 5. Calculate the definite integral
A primitive of is and therefore by Newton-Leibniz formula
Ex. 6. Find the area of a figure bounded by the next lines , (fig. 5).
The figure in question is a curvilinear trapezium, and so its area by the formula (10) equals the definite integral
.
Fig. 5 Ex. 7. A particle moves in a straight line and t sec. after passing a point O the velocity of the particle is ft. per sec. Find the distance of the particle from O after 2 sec.
On the base of the formula (12) the distance in question equals
ft.
Ex. 8. Find the mean value of the function on the segment .
On the base of the formula (20)
.
Ex. 9. Find the mean value of the velocity of the particle of the example 7 during the time interval from to .
By (20) (and taking into account the result of integration in ex. 7) we get
.
Point 6. Main methods of evaluation a definite integral Change of a variable (substitution method)
Theorem 4. Let: 1) a function is continuous on a segment [a, b]; 2) a function is continuous with its derivative on a segment [α, β]; 3) φ (α) = a, φ (β) = b. Then the next formula (formula of change of a variable) is true
. ( 24 )
■ Let is some primitive of a function . Then is the primitive of the function . By Newton-Leibniz formula
;
.■
Note 2. As distinct from an indefinite integral it isn’t necessary to return to the preceding variable after integration by the formula (24).
Ex. 9. Calculate the definite integral .
Let’s put . Then , so
Ex. 10. Find the area of a figure bounded by an ellipse (fig. 5).
It’s sufficient to find the quadruplicated area of the Fig. 5 part OAB of the figure.
The first way. From the equation of the ellipse , and so
The second way. It’s better to pass to parametric equations of the ellipse, na-mely 6. In this case
.
Integration by parts
Theorem 5. If functions are continuous with their derivatives on a segment , then the next formula (formula of integration by parts) is true
( 25 )
■To prove this formula it’s sufficient to integrate from a to b both parts of the identity
and apply Newton-Leibniz formula for the integral of the expression ■
Ex. 11. Evaluate the definite integral .
Ex. 12. Find the area of a figure boun-ded by two curves (see fig. 6).
The curves inter-sect at the points and form a given figure . Its area is equal to the difference of the areas of two curvilinear tra- Fig. 6 peziums .
.
Ex. 13. Let . Prove that
.
■
■
For example
.
Ex. 14. Find the remainder of Taylor formula in Lagrange form.
Let, for example, , and
.
By Lagrange formula
.
Taking we get
,
and after integration over the segment
Therefore
To get for any n we write
,
then we put and integrate n times over . As the result we’ll get