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Файл:Mathematics for Foreign Students integrals. Study Guide
.pdf
41
=
+
+
+
=
1
2
1
22
t
dt
t
tdt
=+++++ Cttt 1ln21
22
Cxxxxx ++−+−++−= 542ln254
22
.
Do the following exercises and test Yourself
1.
+
++
dx
x
xx
3
2
1
1
2.
+
5
2
xxx
dx
3.
( )
++
−
dx
xx
x
22
13
2
4.
( )
−+
−
2
25
118
xx
dxx
Answers.
1.
( ) ( ) ( ) ( )
C
xxxx
+
+
+
+
+
+
−
+
7
16
2
13
5
16
8
13
6
7
6
4
6
10
6
16
2.
++−++−+− Cxx
x
xxx
1lnln
1
2
1
3
1
4
1
10
1010
10
10
2
10
3
10
4
3.
( ) ( )
Cxxx +++++−++ 111ln4113
22
4.
( )
;
6
1
arcsin3168
2
C
x
x +
−
−−−−
42
To integrate the functions containing a radical of the form:
•
x
22
)dxaR(x,
−
, we make a replacement х=asint, dx=accost.
•
)dx+xaR(x,
22
, we make a replacement x=atgt,
t
adt
dx
2
cos
=
.
•
)dxaxR(x,
−
22
, we make a replacement
t
a
x
cos
=
,
t
dta
dx
2
cos
sin
=
.
Example. To solve
dxx
−
2
7
.
Solution.
=−=
=
=
=
=− tdtt
x
t
tdtdx
tx
dxx cos7sin77
7
arcsin
cos7
sin7
7
22
=+
+=
+
== C
t
tdt
t
tdt
2
2sin
2
7
2
2cos1
7cos7
2
C
x
x
x
+
−+=
2
7
1
2
7
7
arcsin
2
7
Cхx
x
+−+=
2
7
2
1
7
arcsin
2
7
.
And that Because of
ttt cossin22sin =
, and
7
sinxt =
,
2
7
1cos
−=xt
.
43
Example. To solve
dx
x
x
+10
2
2
.
Solution.
−+=
+
−+
=
+
dxxdx
x
x
dx
x
x
10
10
1010
10
2
2
2
2
2
2
10
10
dx
x
=
+
The second integral is tabular
=
+
dx
x 10
10
2
=
+
dx
x 10
1
10
2
Сxx +++ 10ln10
2
.
To find the first integral we will use the method of replacement as
a following
=+
dxx 10
2
=
=
=
=
10
cos
10
10
2
x
arctgt
t
dt
dx
tgtx
=
+
=
t
dt
ttg
2
2
cos
10
1010
1
==
t
dt
t
2
cos
10
cos
10
1
=
=
=
=
tdtdz
tz
dt
t
t
cos
sin
cos
cos
2
=
−
=
2
1 z
dz
=+
+
−
− C
z
z
1
1
ln
2
1
=+
+
−
− C
t
t
1sin
1sin
ln
2
1
C
x
arctg
x
arctg
+
+
−
−=
1
10
sin
1
10
sin
ln
2
1
.
=
+
dx
xx10
2
2
10ln10
2
++− xx
C
x
arctg
x
arctg
+
+
−
−
1
10
sin
1
10
sin
ln
2
1
.
44
Do the following exercises and test Yourself
1.
+
++
dx
x
xx
3
2
1
1
;
2.
+
5
2
xxx
dx
;
3.
( )
++
−
dx
xx
x
22
13
2
;
4.
( )
−+
−
2
25
118
xx
dxx
;
5.
dxx
−
2
4
;
6.
dxx
+
2
1
.
Answers.
1.
( ) ( ) ( ) ( )
C
xxxx
+
+
+
+
+
+
−
+
7
16
2
13
5
16
8
13
6
7
6
4
6
10
6
16
2.
++−++−+− Cxx
x
xxx
1lnln
1
2
1
3
1
4
1
10
1010
10
10
2
10
3
10
4
3.
( ) ( )
Cxxx +++++−++ 111ln4113
22
4.
( )
;
6
1
arcsin3168
2
C
x
x +
−
−−−−
5.
C
x
x
x
+
−+
2
2
1
2
arcsin2
Cxxx
x
+++++
22
1ln
2
1
1
2
;
6.
.
45
46
D E F I N I T E I N T E G R A L , M E T H O D S O F I T S
C A L C U L A T I O N
A definite integral as a limit integral sum
Let the function f(x) be defined on the segment [a,b]. The segment is
divided into n parts by points a=x0<x1<x2<…<xn=b. Let
i
an arbitrary
point is taken inside the segment [x
i-1,xi
], and let’s write the following amount
,)(
1
=
=
n
i
iin
xfS
(
1−
−=
iii
xxx
), which is called the Riemann sum
f(x) on the segment [a,b].
i
x
and
|)(|
i
f
.
0xnx1x1−nx2x12n
x
y
Then the definite integral is defined by the following limit
b
a
dxxf )(
=
,)(lim
1
)(
0max
=
→
→
n
i
ii
n
x
i
and the numbers a and b are called the lower and upper limits of the integral,
respectively.
47
Some properties of definite integrals
1.
b
a
dxxf )(
= -
a
b
dxxf )(
.
2.
a
a
dxxf )(
=0.
3.
b
a
dxxf )(
=
c
a
dxxf )(
+
b
c
dxxf )(
, с ϵ [a,b].
4.
dxxfxfxf
b
a
−+ ))()()((
321
=
b
a
dxxf )(
1
+
b
a
dxxf )(
2
-
b
a
dxxf )(
3
.
5.
b
a
dxxAf )(
=А
b
a
dxxf )(
, where A=const
6.
−
−
=
−
нечетная. если ,0
четная, если ,)(2
)(
0
f(x)
f(x)dxxf
dxxf
a
a
a
The Newton-Leibniz formula
+= CxFdxxf )()(
, when F(x) – one of the primitive functions of f(x),
then a definite integral can be calculated by the Newton-Leibniz formula
),()()()( aFbFxFdxxf
b
a
b
a
−==
Example:
+−
2
1
2
)123( dxxx
.
Solution:
( )
=+−=
+−=+−
2
1
23
2
1
23
2
1
2
2
2
3
3)123( xxxx
xx
dxxx
( )
( )
5111222
23
=+−−+−=
.
48
Example:
−+
1
0
2
23 xx
dx
.
Solution:
3+2х–х2=–(х2–2х–3) =–(х2–2х+1–1–3) =–((х–1)2–4) = 22–(х–1)
2
.
dx=d(x–1),
−+
1
0
2
23 xx
dx
=
=
−
=
−−
−
1
0
1
0
22
2
1
arcsin
)1(2
)1( x
x
xd
.
62
1
arcsin
2
1
arcsin0arcsin
=
=
−−=
Example:
−
+
1
0
xx
ee
dx
.
Solution:
=
+
−xx
ee
1
=
+
=
+
x
x
x
x
e
e
e
e
1
1
1
1
2
1
2
+
x
x
e
e
.
==
+
=
+
=
+
−
1
0
1
0
2
1
0
1
0
2
1
)(
1
x
x
x
x
x
xx
arctge
e
ed
e
dxe
ee
dx
=
.
4
1
0
−=−=− arctgearctgarctgearctgearctge
Changing the variabl
e for the definite integral
Often to calculate the integral
b
a
dxxf )(
it is useful to replace the in-
tegration variable x by a new variable t.
)(tx
=
,
)(1xt−=
, where
)(1x
−
- inverse function to
)(tx=
.
49
)(
1
1
at−=
,
)(
1
2
bt−=
and then
=
b
a
t
t
dtttfdxxf
2
1
)())(()(
Example:
dxxx
−
−
3
3
22
9
.
Solution:
x=3sint, dx=3costdt.
x=3, 3=3sint, sint=1,
2
=t
; x=-3, -3=3sint, sint=-1,
2
−=t
.
=−=−
−
−
dttttdxxx
2
2
22
3
3
22
cos3sin99sin99
=
=−
−
dtttt
2
2
22
cossin1sin81
=
dttt
−
2
2
22
cossin81
=
+
−
=
dt
tt
2
0
2
2cos1
2
2cos1
281
=
dtt
−
2
0
2
)2cos1(
2
81
==
dtt
2
0
2
2sin
2
81
=
−
dt
t
2
0
2
4cos1
2
81
=
8
81
4sin
4
1
4
81
2
0
=
− tt
.
Example:
+
2
0
cos2
x
dx
.
Solution:
=−
−
dtttt
2
2
22
cossin13sin27
50
t
x
tg =
2
.
+
2
0
cos2
x
dx
=
+
−
=
+
==
1
2
0
0
1
1
cos
1
2
2
2
2
2
t
x
t
t
x
t
dt
dxt
x
tg
=
+
−
+
+
1
0
2
2
2
1
1
2
1
2
t
t
t
dt
=
=
+
−++
+
1
0
2
22
2
1
122
1
2
t
tt
t
dt
=
+
1
0
2
3
2
t
dt
=
+
1
0
22
)3(
2
t
dt
1
0
33
2 t
arctg=
=
9
3
6
3
2
0
3
1
3
2
==
−= arctgarctg
.
Example:
−
2ln2
2ln
1
x
e
dx
.
Solution:
te
x
=−1
,
1
2
+= te
x
,
)1ln(
2
+= tx
,
dt
t
t
dx
1
2
2
+
=
.
x=ln2,
1
2ln
1 te =−
,
1
12 t=−
, t1=1,
x=2ln2,
2
2ln2
1 te =−
,
2
4ln
1 te =−
,
2
14 t=−
, t2=
3
.
−
2ln2
2ln
1
x
e
dx
=
+
3
1
2
1
2
t
dt
t
t
=
3
1
3
1
2
2
1
2 arctgt
t
dt
=
+
=
=2(arctg
3
-arctg1) = 2
643
=
−
.
Integration by parts for the definite integrals
−=
b
a
b
a
b
a
vduuvudv
,
where
)()()()( avaubvbuuv
b
a
−=
.
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