Higher_Mathematics_Part_2

1.2. Use DeMoivre’s law to find the following complex numbers. Express your answer in both trigonometric and rectangular form.

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Module

INTEGRAL CALCULUS

General characteristic of the module. This chapter is designed to give a brief review of basic methods of indefinite and definite integrals application.

Text organization

Topic 1. Indefinite integrals. Table of integrals. Evaluating techniques. Topic 2. Polynomial functions. Rational functions.

Topic 3. Integrating of rational functions by partial fractions. Topic 4. Integrals involving powers of trigonometric functions. Topic 5. Integrating of irrational functions.

Topic 6. Definite integrals. Newton-Leibniz fundamental theorem. Properties of definite integrals. Evaluating techniques.

Topic 7. Improper integrals. Integrand unbounded. Topic 8. Application of the definite integrals

Basic concepts. 1. Antiderivative. 2. Indefinite integral. 3. Definite integral. 4. Improper integral. 5. Convergence of improper integrals.

Basic problems. 1. Finding indefinite integrals. 2. Definite integrals evaluation techniques. 3. Application of definite integrals. 4. Investigation to convergence of improper integrals.

Topic 1. Indefinite integral

Concepts antiderivative and the indefinite integral. The table of the integrals. The substitution technique. Integration by parts.

Literature: [1, section 6, ch. 6.1—6.3], [2, section 2, ch. 2.1], [3, section 7, § 1], [4, section 7, § 22], [6, section 8], [7, section 10, §§1—6], [9, §§29—30].

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1.1. Some basic indefinite integrals

There are two main concepts in calculus: the derivative and integral. The derivative provides information at a point or at a particular instant. The integral, the other major concept of calculus, does just the opposite. It is a tool for obtaining the numerical value of some overall quantity from local information.This section describes the substitution technique that changes the form of an integral, preferably to that of an easier integral. Before describing this technique, we collect some basic facts about integrals in order to have a supply of integrands.

Every formula for a derivative provides a corresponding formula for an antiderivative of integral. For instance, since (3x4+5)′=12x3 and (3x4−7)′=12x3 that does not imply that 3x4+5=3x4 − 7.

An antiderivative of a function f is a function F such that F′(x) = f(x),

or equivalently, in differential notation, dF = f(x)dx

For any constant C, (3x4+C)′=12x3.

Any two antiderivatives of a function differ only by a constant

Since 3x4 + C describes all antiderivatives of 12x3, we can refer to it as being the most general antiderivative of 12 x3, denoted by ∫12x3dx, which is read “the indefinite integral of 12x3, with respect to x”. Thus we write

∫12x3dx = 3x4+ C.

The symbol ∫ is called the integral sign, 12x3 is the integrand, and C is the constant of integration. The dx is a part of the integral notation and indicates the variable involved. Here x is the variable of integration.

More generally, the indefinite integral of any function f with respect to x is

written ∫f(x)dx and denotes the most general antiderivative of f. Since all antiderivatives of f differ only by a constant, if F is any antiderivative of f, then

∫f(x)dx = F(x) + C, where C is a constant.

To integrate f means to find ∫f(x)dx. In summary,

∫ f(x)dx = F(x) + C if and only if F′(x) = f(x).

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2. For two functions f(x) and g(x):

∫ ( f (x)) ± g(x)dx = ∫ f (x)dx ±∫ g(x)dx.

An antiderivative of the sum of functions can be found by adding

4.d ∫ f (x)dx = f (x)dx.

5.∫ F ′(x)dx = F(x) + C .

The following miniature integral table list has a few formulas that should be memorized. Each can be checked by differentiation.

1.3. Table of integrals

Let a function u(x) be continuous on an open interval [a, b] and have the continuous derivative u′(x) on [a, b]. Then on [a, b], the following formulas are accurated (tabl. 2.1).

Methods of integration:

1. Integration by tables.

2. The substitution technique.

3. Integration by parts.

1.4. Integration by tables

Certain forms of integrals that occur frequently may be found in standard tables of integration formulas. A shot table appears in table 2.1 and its use will be illustrated in this section.

A given integral may have to be replaced by an equivalent form before it fits a formula in the table. The equivalent form must match the formula exactly. Consequently, the steps that you perform should not be done mentally. Failure to do this can easily lead to incorrect results.

1.5. The substitution technique

This method involves the introduction of a function that changes the form of the integrand, hopefully to that of a simpler integrand. Several examples will illustrate the mechanics of the method, known as substitution. The proof that it works will be given after these examples. Substitution is the most important tool in computing integrals.

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Table 2.1

There are two cases in order to apply the substitution technique to find

∫ f (x)dx :

Here φ(t) and ω (x) are both continuous and differentiable functions. In the first case: dx = φ′(t)dt and

∫ f (x)dx = ∫ f (φ (t)) φ′(t)dt .

The second substitution is used if a function f has the form f (x)dx = g(ω (x)) ω′(x)dx ,

then

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In this case a function ω′ (x) can be written:

ω′(x)dx = d (ω (x)) = dt

and is called basic integration form.

It is important to keep in mind that there is no simple routine method for antidifferentiation of elementary functions. This is in contrast with the routine that exists for differentiation. Practice in integration pays off in the quick recognition of which technique is most promising.

1.6. Integration by parts

Integration by parts is used not only to obtain integrals, but also to establish properties of functions.

Just as the chain rule is the basis for integration by substitution, the formula for the derivative of a product is the basis for integration by parts. Let u = u(x)

and v = v(x) be differentiable functions on some interval. Then

∫udv = uv − ∫vdu .

This differential form is the most useful and is called the integration by parts formula.

The key to applying integration by parts is the labeling of u and dv. Usually three conditions should be met.

First: v can be found by integrating and should not be too messy. Second: du should not be messier than u.

Third: ∫vdu should be easier than the original ∫udv .

Some types of integrals where need that use of integration by parts are: 1) integrals in forms

∫ P(x)e kx dx , ∫ P(x) sin kxdx , ∫ P(x) cos kxdx ,

where P(x) is polynomial. In this case we consider the factor P(x) as u

and dv is the remainder; 2) integrals in forms

∫ P(x) ln xdx , ∫ P(x) arcsin xdx , ∫ P(x) arccos xdx , ∫ P(x) arctg xdx , where P(x) is polynomial. In this case dv = P(x)dx and u is remainder;

3)integrals in forms

∫eαx sin βxdx , ∫eαx cosβxdx ,

where α, β are real numbers. In this case integration by parts is applied twice and provides an equation for the unknown integral

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Т.1 Typical problems

І. Using a table of integrals 1. Find ∫1dx.

Solution. By formula 2 with k = 1, ∫1dx = 1x + C. Usually we write ∫1dx as

∫dx. Thus ∫dx = x + C.

2. Find ∫ x3dx .

3. Find ∫5xdx .

Solution. By property 1 with k = 5, f(x) = x and by formula 3 with n = 1 we

One constant of integration is enough. 5. Find ∫ xdx .

Solution. Here x is variable of integration. We rewrite the integrand so that a

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7. Find ∫(x − x3 + 2)x2dx .

Solution. Sometimes, in order to apply the basic integration formulas it is necessary to first perform algebraic manipulations on an integrand, as is shown:

In Example 7 we first multiplied the factors in the integrand. We point out that ∫(x − x3 + 2)x2dx ≠ ∫(x − x3 + 2)dx ∫ x2dx .

More generally, ∫ f (x)g(x)dx ≠ ∫ f (x)dx ∫g(x)dx .

The integral of a product is not the product of the integrals.

8. Find ∫ x3 − 2 x + 3dx. x2

Solution. We can break up the integrand into fractions by dividing each term in the numerator by the denominator:

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