- •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)
Point 3. Integration by substitution (change of variable)
Theorem 2. Let functions are continuous in corresponding intervals and the function has the continuously differentiable inverse function . In this case the next formula (formula of change of variable) is true
( 4 )
The formula implies returning to preceding variable x after integration with respect to the variable t. The word “differentiation” always means finding of differentials.
■The first method. The derivatives of the left and right sides of the formula (4) are equal because of
The second method. If is a primitive of the function , then the function is the primitive of the function on the strength of
.
Therefore by virtue of definition of the indefinite integral
■
Note 3. The formula (4) is often applied “from the right to the left”, and in this case it’s useful to write it in the next form
. ( 5 )
The formula (5) means: if an integrand is represented as a product of some function f of a function and the derivative of this latter then it’s well to put .
Note 4. We can use any letter instead t in the formula (5).
Ex. 9.
.
Ex. 10. Prove yourselves that .
Ex. 11.
Ex. 12. Evaluate the indefinite integral .
The integrand is a product of a function of , namely , and the derivative of (up to a constant factor , for ). On the base of the formula (5) we can put or better . We suppose that and so . Therefore
.
Ex. 13. Calculate the indefinite integral .
The integrand is a product of the function of and (up to a factor -20) the derivative of . So, putting , we reduce the given integral to a tabular one (see the tabular formula (12) where we must take )
Ex. 14. The case when an integrand is a fraction, the numerator of which is the derivative of the denominator. The integral is reduced to a tabular one (see 2)
( 6 )
For ex. a) ( 7 )
b) ( 8 )
Ex. 15. Prove that ( 9 )
Solution. Using the method of Ex. 12 we put . Hence, .
Ex.16. Indefinite integrals
( 10 )
are reduced to sums of integrals of the types (8), (9) and those tabular with the help of the substitution
. ( 11 )
For example: a) Using the substitution (11), the formula (8) and the tabular integral No. 13, we get
b) With the help of (11), (9) and the tabular integral No. 12 we obtain
Ex. 17. To evaluate the indefinite integral
we put , whence (with the help of the tabular integral No. 13)
Ex. 18. To prove the tabular formula No. 14 we’ll use so-called Euler’s substitution