Mathematics (Математика). Учебное пособие

131

8. Finally, let’s draw a graph of the function, using all this infor-

(

)

mation.

Fig.5

Fig.5 – graphic representation of the function

=

y

2

+

x

( )

14

-

x

3

2

Note:

When drawing the graph, the scales of OX and OY axes may differ.

Find by yourself:

1. the highest and lowest values of у = 2х – х2 at segment [0; 1].

2. the highest and lowest values of у = 3х – х3 at segment [0; 1].

132

Section 3. Differential calculation

x

D

x

D

x

¶

22

.

xy

222

(4)

zyx

xxy

¶-

2222222

222222222

(4)(4)(4)

zyxyyyxyyxy

yxyxyxy

dzdxdy

23

.

xy

.

xy

of multi-variable functions

Topic 12. Multi-variable function. Derivatives

and differentials of multi-variable functions

Introduction. This workshop is devoted to the discussion of the following concepts: functions of several variables, the total differential of a function of two variables.

Definition. Let in a certain field is set to the function z = f(x, y). Let us take an arbitrary point M(x, y), and ask increment the Dх to the variable х.. Then the value is Dxz = f( x + Dx, y) – f(x, y) is called the partial

increment function on х.

z

D

lim

x

®D 0

x

=

z

x

is called a partial derivative of the function

D

z

You can write

Then the limit

z = f(x, y) on х . Designation:

¶

Example. To find the total differential of a function

=

¶-

¢

¶----++

(4)(2)424

===

¶---

84

=-+

xyxy

22222

(4)(4)

xyxy

--

Find by yourself:

1. the total differential of a function

2. the total differential of a function

yxfyxxf

-D+

),(),(

¢

;; yxf

z

8

z

=

3

z

=

5

.

),(

yxf

¶

22

+

y

-

2

y

3

-

¢

).,(;

xx

z

y

=

4

-

133

Section 4. Integral calculation

1

k

x

a

ò

of single-variable functions

Topic 13. Indefinite integral

Introduction. This topic is about the idea of integration, and also about the technique of integration. We explain how it is done in principle, and then how it is done in practice.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

ò

k

dxx

ò ò

x

dxa

ò ò

ò

+

ò

-

ò

dx

ò

2

cos

dx

ò

2

sin

;Cxdx +=

1

+

k

x

=

+

;ln1Cxdx

+=

x

a

;lnC

+=

4а.

;1,

-¹+

kC

xx

;Cedxe

+=

;cossin Cxdxx +-=

;sincos Cxdxx +=

11

dx

22

ax

1

-

22

ax

a

dx

22

xa

11

dx

=

2

1

22

ax

±

x

+=

x

; arctg

C

+=

a

x

;arcsin

C

+=

a

ax

-

a

+

;tg

Cx

;ctg

Cx

+-=

;ln

C

+

ax

22

;ln

Cxaxdx

++±=

134

Exercise 1 : Calculate the integral:

ò

æ

5

ç

x 7

а)

ò

ç

x

è

г) ;3

72ò-dxx

x

ò

ж)

4

-

к)

× dxxx

н)

р)

ò

( )

4

3

dx

4

x

2

;tg

;arccosòdxx

dx

ö

2

÷

--+ dxx

÷ ø

б)

;

д)

;

з)

л)

-+

sin2cos2sin xxx

о)

;

с)

ò

4

ò

1

ò

dx

( )

21

+ x

arctg

x

dx

2

+

x

e

2

x

-

e

x

3

dx

x

49

+

2

23

3

;

3

;

dx

7

;

++

63

xx

;

dx

+-

dxx

65

xxx

;

4

22

2323

-+- xx

в)

е)

и)

м)

п)

т)

x

ò

sin

× ;sin dxee

ò

-

2

ò

2

;

dx

52

x

xx

5sin

x

;

dx

2

5cos4

x

× dxxx

x

;cos

6

;

dx

1

+-

xx

;5cos3cos

dxxx

4

у)

4

;sin

dxx

ф)

dx

;

ò

e

2x

-

1

Solution:

а) let’s find the integral by using the properties of an indefinite integral and formulas (1) and (2) of the integration table:

4

æ

5

x

ç

3

x

è

4

=

×+

15

+

ö

3

2

--+

1315

+-+

-

13

+-

æ

÷

ç è

ø

2

1

+

3

xxx

7

2

+

3

Cx

1

2

ö

3

--×+=

÷

ò ò ò òòò

ø

5

26

-

3

xxx

4

×+=+-

6

2

-

7

5 3

3535

--

6

x

Cx

6

2 3

3

32

2

x

Integrals (б – л) are solved by changing the variable.

+=

xt

;21

б)

dx

4

( )

+

1

dt

=

3

x

21

dxdt

1

=

dtdx

2

;2

ò òò

4

3

t

3

1

-

4

===

dtt

2

74747

dxdxxdxxdxxdxxxxdxx

=--+=

2

x

5

;7

Cx

+---=+--

135

{to solve this, let’s use formula (2)}

5

3

ln3713

ln37

212

1

в)

4

1

t

1

2

4

x

sin

( )

4

dx

52

x

1

=

xt

=

4

5

4

dxxdt

5

1

=

dtdxx

sin

++=++=+×=

1

dt

5

{ to solve this, let’s use formula (12)}

1

ctg

1

5

;ctg

CxCt +-=+-=

-=

xt

г)

,72

-

72

=

dx

3

-=

dxdt

,7

1

-=

dtdx

1

æ

-×=

3

ç

7

è

7

{ to solve this, let’s use formula (4)}

72CCxt

1

-

;

+×-=+×-=

д)

arctg

+

1

x

=

dx

2

x

dt

xt

arctg

1

=

1

dx

2

+

x

tdt

òò

{ to solve this, let’s use formula (2)}

2

t

2

;arctg

CxC

+=+=

x

=

xx

е)

ò ò

dxee

et

=×

x

=

dxedt

sinsin

{ to solve this, let’s use formula (5)}

x

;coscos CeCt

+-=+-=

CxxC

ö ÷

ø

tdt

;212212

dt

1

òòò

5

t

1

7

==

sin

òòò

===

22

t

ttx

=-=

dtdt

3

136

2

(

)

2

=

ж)

x

ò ò ò

4 t

dx

4

-

x

=

2

xdxdt

1

=

dtxdx

xt

1

dt

2

=

-

4

t

2

{ to solve this, let’s use formula (8)}

1

arcsin

2

x

;

C

+=

з)

x

e

x

e

=

dx

-

7

x

( )

e

x

e

22

( )

-

=

dx

2

7

{ to solve this, let’s use formula (10)}

2

=

5cos

и)

ò ò òò

dxx

2

-

5cos9

x

-=

=

5sin

5sin5

-=

5sin

xt

dxxdt

1

-

=

9

dtdxx

5

{ to solve this, let’s use formula (9)}

1

×=

3215

×

ln

3

t

-

3

t

+

1

C

ln

×=+

3215

×

x

-

x

+

1

=

x

=

et

=

x

xx

++-=++-=

dt

2

,

x

=

-

2

=

,

dxedt

òòò

dtdxe

;7ln7ln

CeeCtt

=

222

dt

2

-

t

( )

7

=

2

1

dt

1

5

=

5

-

t

35cos

t

1

C

=+

30

dt

1

dt

=

5

-

9

ln

t

x

=

2222

-

3

35cos

+

;

C

+

35cos

2

xx

к)

tg

2

dxxx

sin

cos

==×

dx

2

x

{ to solve this, let’s use formula (3)}

1

ln

1

2

CxCt +-=+-=

=

sin

;cosln

cos

2

xt

-=

2

dxxxdt

sin2

1

-=

dtdxxx

2

1

dt

2

ò òò ò

t

dt

1

2

=-=-=

t

137

x

2

3ln212

213

ln

ò

=

3

t

1

x

=

dxdx

2

x

=

3

=

3ln3

dxdt

1

dtdx

3ln

dt

=

3ln

22

+

2

t

x

( )

x

3

2

+

23

л)

ò ò ò

x

=

+

49

x

3

{ to solve this, let’s use formula (7)}

x

1

arctg

t

arctg

3

;

CC

+=+×=

Find the integrals, that can decompose into basic table constants:

а)

б)

в)

г)

д)

е)

ж)

+ dxxx62)34( ;

ò

2sin2 ;

ò

3

x

dx

8

ò

x

+

9

- dxx)25cos(

ò

x

2sin

cos

2

+

cos1

2

x

32

x

ò

;

xxdxsin

xx

dx

;

dx

x

;

dx

Topic 14. Methods of integration

Introduction. This topic is about the most used techniques of integration (Integrals by substitution, integration by parts, Trigonometric Integrals).

Exercise 1: solve the integrals:

в)

4

x

ò

sin

;

dx

52

x

е)

xx

× ;sin dxee

и)

5sin

x

ò

-

dx

2

5cos4

x

;

138

н)

ò

=¢=

xxdx

ò

2

++

63

;arccosòdxx

о)

ò

xx

23

+-

;

dx

65

xxx

п)

6

x

ò

2

;

dx

1

+-

xx

р)

у)

ò

( )

4

dxx

dx

-+

;sin

;

с)

sin2cos2sin xxx

ò

ф)

ò

3

dxx

4

22

dx

;

2x

-

1

e

т)

;

2323

-+- xx

2

м)

× dxxx

dxxx

;cos

Let’s find integrals (м – н) using the partial integration method and

formula

2

м)

=

cos

ò ò

¢

× VdxUVUdxVU

= xdxxxx

xdxx sin2sin

-×=

2

=

¢

UxU

2;2

-==

cos;sin

2

xVxV

(13):

¢

×

¢

=

xUxU

2;

==

sin;cos

2

=×-=

xVxV

( ) ( )

( )

cos2cos2sin

ò

=-×--×-×==

dxxxxxx

{ to solve this, let’s use formula (6)}

2

=

= dx

arccos

ò ò

н)

dxx

¢

;sin2cos2sin

Cxxxxx +-+=

-

¢

UxU

;arccos

;1

1

=

==

2

-

x

1

xVV

arccos

-

-×=

1

xxx

2

-

x

1

{the second summand is calculated by changing variables, using formula (2)}

;5cos3cos

2

xt

-=

x

-

dx

2

x

-

1

At the end, we get

1

dxxdt

-=

=

2

1

=-

2

ò

1

dt

2

òò

t

dtxdx

Try to solve this integral by yourself 2cos

1 2

t

1

2

CxC

+-=+×==

1

2

;1arccosarccos

Cxxxdxx +--×=

.

139

Topic 15. Definite integral.

(44)(4)

xxxdxx

4

1

xxdx

1.

edxe

=-

24.

pp

45

(33).

2

33.

=

()35.

=+

.

44

p

=

327.

Calculation of a definite integral

Introduction. This topic is about the definite integral and the methods to calculate definite integrals.

Example. Calculate a definite article.

Solution. Because when solving a definite integral we use tabulated integrals and rules the same way we do with indefinite integrals, the result of our calculations is the difference between the primitive, taken at the highest and lowest limits of the integral.

4

2

-+-=-+-

ò

1

Solve these definite integrals by yourself:

2

1

ò

0

+

1

x

xx

dx

23

++

)13(

Find the upper limit of integration, if

Find the lower limit of integration, if

Find the distance travelled by a body in 3 sec. from the start, if its speed is

vtt

Find the upper limit of integration, if

32

xx

32

1

;

(43)

+

ò

0

5

26

|

;

*

ò

1

2

ò

*

3

ò

*

2

ò

*

ò

2

*

ò

0

x

xdx

edx

xdx

= 4,5

dx

=

+=

x

dx

2

x

+

2

=

,

140

Find the upper limit of integration, if

8

(3).

3

+=

98

(3).

3

(45)4,5.

9.

=

()321

vttt

=++

()41

=+

2

;

51

4

dx

3;

tgxdx

Find the upper limit of integration, if

Find the lower limit of integration, if

*

ò

3

-

*

ò

0

*

ò

1

-

3

ò

*

2

xdx

2

xdx

+=

xdx

-=

2

xdx

Find the distance travelled by a body in 1 sec. from the start, if its speed is

2

(m/sec).

Find the distance travelled by a body in 2 sec. from the start, if its speed is

vtt

3

(m/sec).

Solve:

1

в)

е)

и)

xdx

ò

2

x +

0

( )

1

-

e

x

-

ò

e

+

0

p

4

x

ò

0

Topic 16. Applications of a definite integral.

Improper integrals

Introduction. This workshop is devoted to the discussion of the following concepts: The area between two curves, volumes by slices, the volume of solid of revolution, length of a Plane Curve and improper integrals.

Exercise: Calculate improper integrals or establish their divergence.