Добавил:
ivanov666
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз:
Предмет:
Файл:Mathematics (Математика). Учебное пособие
.pdf
41
Section 3. Differential calculation
x
x
x
D
x
x
¶
¶
-D+
¶
¢+¢
¶
¶
¶
of multi-variable functions
Topic 12. Multi-variable function. Derivatives
and differentials of multi-variable functions
Introduction. This lecture is devoted to the discussion of the following concepts: functions of several variables, the total differential of a
function of two variables.
When considering functions of several variables will restrict ourselves to a detailed description of functions of two variables, since all the
results will be fair for functions of arbitrary number of variables.
Definition: If each pair are independent of each other integers (x, y)
from some set on any rule is mapped to one or more values of the variable z, the variable z is called a function of two variables z = f(x, y).
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
You can write
x
=
D
z
D
Then the limit
lim
x
®D 0
x
is called a partial derivative of the function
z = f(x, y) on х . Designation:
z
-D+
D
¢
;; yxf
z
¶
Likewise is determined by the partial derivative functions on y.
z
=
lim
y
®D
y
¶
0
Definition: the total differential of a function z = f(x, y) is called the
following mathematical expression
=
For functions of arbitrary number of variables:
f
tzyxdf
dx
= ...),...,,,(
x
¶
yxfyxxf
),(),(
.
),(
yxf
¶
y
D
f
dy
+
y
¶
¢
xx
yxfyyxf
),(),(
f
dt
++
t
¶
).,(;
.
dyyxfdxyxfdz
),(),(
yx
42
Example. To find the total differential of a function
-
¶
x
D
.
yzxy
z
x
¶
dz
22
---
222
)(
yx
-
xy
2
-=
¢
z
¶
=
y
¶
-
2yxyx
=
-
)2()(
yyyxy
=
dx
22
yx
222
)(
222
2
yyx
+-
=
222
yx
-
22
yx
+
+
dy
222
yx
)()(
-
Reading questions.
z
1. How is this a limit
2. To find the total differential of a function
lim
x
®D 0
x
D
called?
=
y
z
=
+
yx
-
.
22
yx
-
22
yx
222
)()(
2
-
43
Section 4. Integral calculation
()
fxdx
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.
Newton and Leibniz had an absolutely brilliant intuition, and there is
no reason why we can't share it.
The geometry problem is to find the area under the curve.
The symbol ∫ was invented by Leibniz to represent the integral. It is a
stretched-out S, from the Latin word for sum. This symbol is a powerful
reminder of the whole construction: Sum approaches integral, S approaches I, and rectangular area approaches curved area: curved area
=
The "dx" indicates that Δx approaches zero. The sum of fi times Δx
approaches "the integral of f of x dx."
.
Figure 4
Fig.4 – graphic representation of the area under the curve.
44
We leave a side sums of rectangular areas, and their limits as Δх → 0.
ò
()
FxC
+
Antiderivative
Antideriv
a
tive
1
9
ò
ex
+ C 2
ò
10
ò
sinx + C
3
ò
a
ln
11
ò
-cosx + C
4
+22x
a
a
12
x2cos
tgx + C
5
-22a
x
13
x2sin
-ctgx + C
6
14
a
7
ò
1
15
x
cos
8
16
x
sin
x
x
x
x
1
x
¢
=
dxxc
=+
ò
0
c
¢
=
dxc
=
ò
Instead we search for an f(x).
Thus our goal is to find antiderivatives and use them.
there
+= ;)()( CxFdxxf
is the set of antiderivative functions.
Below I will give a table of common integrals, which can be used to
obtain the values of the indefinite integrals of various functions.
integral
tgxdx
ò
ctgxdx
x
dxa
dx
ò
dx
ò
dx
ò
±22ax
a
dxx
dx
ò
x
-ln½cosx½+C
ln½sinx½+ C
1
1
2
ln
x
x
a
C
+
arctga+
+
ln
a+-
1
+a
C
+a
Cx +ln
f(x)
x
ax
ax
C
C
22
Caxx +±+
1,
-¹a+
integral
x
dxe
xdxcos
xdxsin
1
ò
1
ò
dx
ò
1
ò
1
ò
f(x)
dx
dx
arcsin
-22xa
ln
dx
dx
x
x
æ
tg +
+
ç
è
x
tg +
ln
2
+ C
p
ö
÷
42
ø
C
C
You recognize that each integration formula came directly from a differentiation formula. The integral of the cosine is the sine, because the
derivative of the sine is the cosine. For emphasis we list three derivatives
above three integrals:
( )
x
ln =
1
¢
, and
( )
x
)ln( =-×-=
-
1
¢
1
dx
,
)1(
ò
ln
+= Cx
,
1
,
0
Rules for integrals.
45
1.
(
)
¢
ò
(
)
ò
ò
òòò
ò
ò
ò
3
ò
ò
ò
323
¢
+=
);())(()( xfCxFdxxf =
2.
3.
4. the sum rule:
где u, v, w – some functions from x.
5. the constant rule:
Example.
ò ò ò ò
6. the substitution rule
Example.
Will make substitute t = sinx, dt = cosx dt.
ò ò
Reading questions
1. What techniques of integration do you know after reading this topic?
2. The integral of the cosine is the sine, because the derivative of the
sine is the …?
3. What do you know about the symbol ∫ ?
;)()( dxxfdxxfd =
+= ;)()( CxFxdF
-+=-+ ;)( wdxvdxudxdxwvu
×=× ;)()( dxxfCdxxfC
sin2)1sin2(
j= dtttfdxxf )())(()(
.
xdxx cossin
2
2/32/32/1
1
¢
j
+=+== .sin
CxCtdttdtt
322
+++=+-=+- ;cos2
Cxxxdxxdxdxxdxxx
Topic 14. Methods of integration
Introduction. This topic is about the most used techniques of integration (Integrals by substitution, integration by parts, Trigonometric Integrals).
Integrals by substitution.
We now present the most valuable technique in this sectionsubstitution.
There are two points to emphasize right away:
1. Constants are no problem – they can always be fixed.
46
2. Choosing the inside function g (or u) commits us to its derivative:
2
2cos
xx
2
sin
x
,/2
gxdgdxx
==
/
dudx
31
x
+
66
du
xxdxxxdxudx
dx
6565
cc
+=+
5
xxdxuduc
+==+
()
vudu
ò
244
xxdxxdxxdx
òòò
The integral of
is
+ C (
2
)
To substitute g for х2, we need its derivative. The trick is to spot an
inside function whose derivative is present.
Remark on writing down the steps. When the substitution is complicated, it is a good idea to get
where you need it. Here
2
needs 6x:
7(31)(31)6
òòò
Now integrate:
24244
+=+=
525
77(31)
ux
77
+
Check the derivative at the end. The exponent 5 cancels 5 in the denominator, 6x cancels 6, and 7x is what we started with.
Remark on differentials. In place of (du/dx) dx, many people just
write du:
5
244
(31)6
òò
u
This really shows how substitution works. We switch from x to u, and
we also switch from dx to du. The most common mistake is to confuse
dx with du.
Here are the four steps to substitute u for x:
1. Choose u(x) and compute du/dx
2. Locate v(u) times du/dx times dx, or v(u) times du
3. Integrate
to find f(u) + C
4. Substitute u(x) back into this antiderivative f
Integration by parts.
The method is based on the formula:
-= vduuvudv
òò
Example.
ln;;
uxdvxdx
==
ìü
lnln
222
xxxx
=-+=-+
ïï
==-×=-
íý
1
dudxv
ïï
îþ
ln
;;
==
x
CxC
2
x
2
(2ln1).
222
xxxx
2222
1ln1
x
47
Example.
32223
xxxdx
xxxxxx
òòò
xxdx
pxqxdx
sincossin()sin()
22
pxqxpxqxpxqx
sin8cos6(sin14sin2)
21422
xxdxxxdx
mxnxdxmnxmnxdx
mxnxdxmnxmnxdx
mxnxdxmnxmnxdx
ìü
ln;;
uxdvdx
==
lnln11ln1
=-+-+=--+
ïï
ïï
dxdx
==---×=-+
íý
11
ïï
dudxv
==-
ïï
xx
îþ
ln11ln1
xx
éù
-
222
22224
xxx
xCC
êú
ëû
1
3
x
;;
2
2
2
2222
.
Trigonometric Integrals
2
Find
2
p
sin8cos6
ò
0
. More generally find
p
sincos
ò
0
.
A better approach, which applies to all angles px and qx, is to use the
identity
11
=++-
Separated like that, sines are easy to integrate:
22
pp
òò
00
1cos141cos2
()0
=--I=
xx
11
=+
22
2
p
0
With two sines or two cosines (also of sine times cosine), we go back
to the addition formulas
coscoscos()cos()
òò
1sin()sin()
mnxmnx
éù
=+
êú
2
ëû
sincossin()sin()
òò
1cos()cos()
éù
=--
êú
2
ëû
sinsincos()cos()
òò
1sin()sin()
=-+
2
+-
mnmn
+-
mnxmnx
mnmn
+-
éù
êú
ëû
mnmn
mnmn
+-
1
=++-
[ ]
2
1
=++-
[ ]
2
+-
1
=-++-
[ ]
2
+-
48
Example.
18
10
2
2
sin10cos7cos4sin10[cos7cos4]
sin7cos21coscos13
28
xxxdxxxxdx
xdxxxx
+=----
òò
1
2sin7sin Cxxxdxxdxxdxx +-=-=
Example.
11
sin10cos11sin10cos3
=+=
òò
22
111
=-+
òòò
444
1111
ò
484452
1
-+
Reading questions
1. What techniques of integration do you know after reading this topic?
2. What the formula is based method of integration by parts?
3. In what integrals do use the addition formulas?
xxdxxxdx
sin21sinsin13
xdxxdxxdx
cos7.
xC
5cos
=
1
9cos
òòò
1
1
5sin
.9sin
Topic 15. Definite integral.
Calculation of a definite integral
Introduction. This topic is about the definite integral and the methods to calculate definite integrals.
The Definite integral.
The indefinite integral contains "+ C." The constant is not settled because f(x) + C has the same slope for every C. When we care only about
the derivative, C makes no difference. When the goal is a number – a
definite integral-C can be assigned a definite value at the starting point.
Theorem: (Theorem of Newton – Leibniz)
If the function F(x) is any antiderivative of the continuous function
f(x), then
49
b
()()()()
vxdxvtdtFbFa
cos(1cos2)sin2sin.
222444
pp
==+=+=+=
()0
=
-=
ò
a
aFbFdxxf )()()(
this expression is known as the formula of Newton – Leibniz.
Don't pay attention to t or x, pay attention to the great formula of integral calculus:
bb
òò
aa
==-
With regard to the methods to calculate definite integrals, they are no
different from all those techniques and methods that have been discussed
above in finding indefinite integrals.
Feature is only then that the application of these techniques it is necessary to extend the transformation not only on the integrand function,
but also on the limits of integration. Replacing the variable of integration, do not forget to change respectively the limits of integration.
Example.
1/2
22
11sincos
xdxttdt
-==-
òò
00
/2/2
pp
2
tdttdttt
òò
00
sin;
xt
=
ìü
íý
0;/2
abp
==
îþ
1111
p
/2
æö
ç÷
èø
p
0
p
Properties of the Integral and Average Value.
All the normal rules for rectangular areas are obeyed in the limit by
integrals.
Property1. Areas over neighboring intervals add to the area over the
combined interval:
Property 2.
b
ò
b
b
ò òò
a
vxdx
c
a
b
+=
c
dxxfdxxfdxxf )()()(
That comes from Property 1 when c = b. Equation (1) has two identical integrals, so the one from b to b must be zero.
What happens when an integral goes backward? The "lower limit" is
now the larger under b. The "upper limit" a is smaller. Going backward
reverses the sign:
50
()()
ba
vxdxvxdx
xdx
(()())
fxgxdx
Property 3.
Property 4.
Property 5.
òò
ab
b
a
b
a
=-
=
òò
b
a
;)()(
dxxfAdxxAf
b
±=±
a
b
òòò
a
dxxfdxxfdxxfxf )()())()((
2121
Reading questions
1. What is the difference in techniques of computation of the definite
integral of an indefinite integral?
2. What are the properties of the definite integral do you know?
3. Find
3
2
5
ò
3
Topic 16. Applications
of a definite integral. Improper integrals
Introduction. This lecture 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.
We are experts in one application of the integral – to find the area under a curve. The curve is the graph of y = v(x), extending from x = a at
the left to x = b at the right. The area between the curve and the x axis is
the definite integral.
The area between two curves.
Area between two curves =
Example. To find the area between two lines y = x, y = x2, x = 2.
b
ò
a
-
Соседние файлы в предмете [НЕСОРТИРОВАННОЕ]