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Chapter 6
6.4 Integration of substitution and int6.4 Integration of substitution and integration by parts of the definite integregration by parts of the definite integralal
As we all known, the integration of substitution ( or
“change of variables”) and integration by parts are very
important tools to evaluate the indefinite integrals. In
this section we extend these methods to the definite
integrals.
1. Integration of substitution1. Integration of substitution
At first, the substitution technique extends to definite
integrals
Theorem 1Theorem 1 (Substitution in a definite integral)
b
abatt
t
tx
baxf
, and ,],[],,[ when )3(
0 and
,on derivative continuous a has 2
;,on continuous is 1
:satisfied are conditions following theIf
tttfxxf
txb
ad)()]([d)(Then
Substitution formula of definite integrals
][tLet ),()()(then
, of tiveantiderivaan is that Suppose
tFaFbFdxxf
xfxFb
a),()]([)()(
d
d
d
d)( ttftxf
t
x
x
Ft
b
axxfaFbFFF
tttf
d)]([)]([
)()(d)()]([
ttf ][ of tiveantiderivaan is t
ProofProof
NoticeNotice
One important is the necessary change in the limits of
integration. The following examples will apply this
technique.
Example 1Example 1 .dsincos Evaluate 2
0
5
xxx
SolutionSolution
,dsind then ,cosLet xxtxt
Thus 1. to0 from
goes cos ,2
to0 from goes as e,Furthermor xtx
.6
1
6ddsincos
1
0
60
1
5
0
52 t
ttxxx
Example 2Example 2
SolutionSolution
..dsinsin Evaluate0
53
xxx
02
3
0
53
2
353
dsincosdsinsin
sincossinsin)(
xxxxxx
xxxxxf
2
2
32
02
3
dsincosdsincos xxxxxx
Additivity over intervals
2
2
32
02
3
sindsinsindsin xxxx
.5
4sin
5
2sin
5
2
2
2
2
5
0
2
5
xx
Change of variables
Newton-Leibniz formula
Example 3Example 3
SolutionSolution
.)ln1(ln
d Find
4
3
e
e xxx
xI
43
)lnarcsin(2e
ex .6
4
3
4
3
2)ln(1
lnd2
)ln1(ln
lnd e
e
e
e x
x
xx
xI
Example 4Example 4
SolutionSolution
a
axxax
I0 22
)0(,d1
Find
,00,2
and ,dcosd then ,sinLet
txtax
ttaxtax
2
0
2
0
2
0 22
dcossin
sincos1
2
1d
cossin
cos
d)sin1(sin
cos
ttt
ttt
tt
t
ttata
taI
.4
cossinln2
1
22
1 20
tt
Example 5Example 5 4
1 if ,d2 Find xxf
SolutionSolution
01,cos1
10,
2
xx
xxexf
x
thus,,24
,11 and ,dd then ,2Let
tx
txtxtx
2
0
0
1
2
1
4
1 d d d d2 ttfttfttfxxf
2
0
0
1 de d
cos1
1 2
tttt
-t
2
1
2
1
2
1tan
2
1
2tan 4
2
0
0
1
2
eet t
2. Important simplification formulas2. Important simplification formulas
Theorem 2Theorem 2
,d)(2d)(
then function,even an is If 1
thatshow
,,on continuous is that Suppose
0
aa
axxfxxf
xf
aaxf
,d)(d)(d)(0
0
a
a
a
axxfxxfxxf
0d)( then function, oddan is If 2 a
axxfxf
SolutionSolution
then,let ,d)( evaluate order toIn 0
txxxfa
,d)(
d)(d)(d)(
0
0
00
a
a
aa
xxf
ttfttfxxf
aaa
axxfxxfxxf
00d)(d)( d)(
a
xxfxf0
d)(
,d)(2d)(
,2then
, then function,even an is If 1
0
aa
axxfxxf
xfxfxf
xfxfxf
0d)(
,0then
, then function, oddan is If 2
a
axxf
xfxf
xfxfxf
NoticeNotice
The properties of definite integrals for odd and even
functions provide a easy way to evaluate their
definite integrals. The following examples will apply
the property.
Example 6Example 6 .d11
cos2 Evaluate
1
1 2
2
x
x
xxx
SolutionSolution
1
1 2
2
d11
cos2x
x
xxx
1
1 2
2
d11
2x
x
x
even function
1
1 2d
11
cosx
x
xx
1
0 2
2
d11
4 xx
x
1
0 2
22
d)1(1
)11(4 x
x
xx
.4
odd function
1
0
21
0
2 d144d)11(4 xxxx
Area of unit circle
Example 7Example 7 .dsin1 Find 2
2
3202
xxx
2
2
3202 dsin1
xxx
SolutionSolution odd function
0
Example 8Example 8
? then ,dcossin
,dcossin
,dcos1
sin that Suppose
2
2
432
2
2
43
2
2
42
xxxxP
xxxN
xxx
xM
NMPDPMNC
NPMBMPNA
;
; ;
SolutionSolution
0dcos1
sin
2,
2on function oddan is cos
1
sin
2
2
42
42
xxx
xM
xx
xxf
0dcos2dcossin and 2
0
42
2
43
xxxxxN
0dcos2dcossin 2
0
42
2
432
xxxxxxP
Thus, we choice ( D )
Theorem 3Theorem 3
TTx
xttfttf
Txf
0dd that prove
, period offunction periodic a is that Suppose
ProofProof
Tx
T
T
x
Tx
xttfttfttfttf dddd
0
0
T
xxT
dttf
dssfdttfdttf
0
000
x
x
TdssTfdttfdttf
0
0
0 f (x+T)=f (x)
The integral properties of periodic functions provide
also a easy way to evaluate some definite integrals.
Next example will apply the property.
Example 9Example 9
k
kxxxI dsinsin Evaluate 53
SolutionSolution
2
2
3
2
0
2
3
0
2
3
dcossindcossin
dcossin
xxxxxx
xxxI
period offunction
period a isit ,cossincossin
and ,cossinsinsin
2
3
2
3
2
353
xxxx
xxxx
5
4sindsinsindsin
2
2
3
2
0
2
3
xxxx
ProofProof( 1) Let tx 2
,dtdx
0x ,2 t
2x ,0 t
0 2
00
00
dcos1
sin find and
d)(sin2
d)(sin 2
;d)(cosd)(sin 1
thatprove [0,1],on continous is If
22
xx
xx
xxfxxxf
xxfxxf
xf
Example 10Example 10
2
0d)(sin
xxf
,0 and ,0
,ddLet 2
txtx
txtx
0
2
d2
sin
ttf
2
0d)(cos
ttf ;d)(cos2
0
xxf
0
0d)][sin()(d)(sin
ttftxxxf
,d)(sin)(0 ttft
00
d)(sind)(sin tttfttf
,d)(sind)(sin00
xxxfxxf
.d)(sin2
d)(sin00
xxfxxxf
00 2
0 20 2
)arctan(cos2
)(cosdcos1
1
2
dcos1
sin
2d
cos1
sin
formula, above theapplyingBy
xxx
xx
xx
x
xx
.4
)44
(2
2
3 Integration by parts3 Integration by parts
Theorem 4Theorem 4
Next, the integration by parts extends to definite
integrals
b
a
b
a
b
auvuvvu
ba
xvxu
dd
then,,on derivative
and continuous are and that Suppose Integration by parts for definite integrals
,d)( and ,b
a
b
auvxuvvuvuuv
,dd
equation,
above of sidesboth integral definite theTaking
b
a
b
a
ba xvuxvuuv
.dd b
a
b
a
b
auvuvvu
ProofProof
This completes the proof
SolutionSolution
Example 11Example 11 .darcsin Evaluate 21
0 xx
,,1
dd
then,dd ,arcsin Setting
2xv
x
xu
xvxu
)1(d1
1
2
1
62
11
darcsindarcsin
2
0 2
0 200
21
21
21
21
xx
x
xxxxxx
.12
3
121
1221
02
x
Example 12Example 12 ..2cos1
d Evaluate 4
0
x
xx
SolutionSolution
,cos22cos1 2 xx
xx
x
xx
x
xxtand
2cos2
d
2cos1
d 444
00 20
xxxx dtan2
1tan
2
1 44
00
.4
2ln
8secln
2
1
840
x
Example 13Example 13 ...d)2(
)1ln( Evaluate
1
0 2
xx
x
SolutionSolution
1
0
1
0 2 2
1d)1ln(d
)2(
)1ln(
xxx
x
x
1
0
1
0
)1ln(d2
1
2
)1ln(x
xx
x
xx
21
11x
xxd
1
1
2
1
3
2ln 1
0
.3ln2ln3
5)2ln()1ln(
3
2ln 10 xx
Example 14Example 14
constant anot 0;
constant; negative a constant; positive a
? is then ,dsineLet 2 sin
DC
BA
xFttxFx
x
t
SolutionSolution
A
2
0
sin2 sin
sin
dsine dsine
,2 periodith function w
periodic a is sine integrand theSince
ttttxF
xxf
tx
x
t
x
2
0
sin2
0
sin
2
0
sin
edcosecos
cosde
tt
t
tt
t
0dcose2
0
2sin
ttt
Thus, we choose (A)
.d)( find ,dsin
)( that Suppose1
01
2
xxxftt
txf
x
Example 15Example 15
SolutionSolution
directly, find way tono tive,antideriva
elementaryan no has sin
function theSince
xft
t
1
0
21
0
21
02
1
0
21
0
d)(2
1)1(
2
1)(d
2
1)(
2
1
)(d)(2
1d)(But
xxfxfxfxxfx
xxfxxxf
,sin2
2sin
)(
,0dsin
)1( ,dsin
)( and
2
2
2
1
11
2
x
xx
x
xxf
tt
tft
t
txf
x
1
0
21
0d)(
2
1)1(
2
1d)( xxfxfxxxf
).11(cos2
1cos
2
1
dsin2
1dsin2
2
1
1
02
1
0
221
0
2
x
xxxxx
number oddan is 1 ,3
2
5
4
2
31
numbereven an is ,22
1
4
3
2
31
dcosdsin that Prove 22
00
nn
n
n
n
nn
n
n
n
xxxxI nnn
Example 16Example 16
SolutionSolution
,cos ,dcossin)1(d
then,dsind ,sin Setting2
1
xvxxxnu
xxvxun
n
xxxnxxI nnn dcossin)1(cossin 22
0
220
1
x2sin1 0
nn
nnn
InIn
xxnxxnI
)1()1(
dsin)1(dsin)1(
2
00
2 22
,2
3 and ,
1 Thus 422
nnnn I
n
nII
n
nI
nI of formulareduction The
,2,1 where
,3
2
5
4
7
6
12
22
12
2
,2
1
4
3
6
5
22
32
2
12
112
02
m
Im
m
m
mI
Im
m
m
mI
m
m
thus,,1dsin,2
d and 22
0100
xxIxI
.3
2
5
4
7
6
12
22
12
2
,22
1
4
3
6
5
22
32
2
12
12
2
m
m
m
mI
m
m
m
mI
m
m