二課堂9定積分計算

第八節(jié)定積分的計算定積分的換元法設單值函數(shù)滿足:φ(t

)

φ

(t

)則φ(α)

=

a

,

φ(

β)

=

b;在[α

,β]或[β

,α]上j(t

)?

C1[a

,b],換元公式定理1注11)

換元要換限,變量不代回.3)換元公式雙向使用：a=

b

f

(

x)d

x或配元φ(t

) d

φ(t

)φ(t

)

φ

(t

)2）x

=

j(t

),x

:

a

fi

b,

t

:

α

fi配元不換限φ(t

)

φ

(t

)φ(t

)

φ

(t

)β,

下限對應下限.令x

=φ(t

)定積分的分部積分法定理2

設u(

x),

v(

x)在[a,

b]上導數(shù)連續(xù)，則ba分部積分公式注2a-af

(

x)dx0，偶倍奇零=0af

(

x)d

x

,2注3設f

(x)在實軸上連續(xù)，且是以T為周期的周期函數(shù)，則對任意實數(shù)a成立Ta+Taf

(

x)dxf

(

x)dx

=

0一、絕對值積分例1122-2min{ ,

x

}dx.x求解112,

x

>

1

x

x2

,

x

￡

1,

x

}

=x

min{是偶函數(shù),x,

x

}dx1min{220原式=221102=

2dx1xx dx

+

223+

2ln

2.=解0p21

-

sin

2xdx.求p20sin

x

-

cos

x

dx原式=ppp40=

4

(cos

x

-

sin

x)dx

+

2(sin

x

-

cos

x)dx=

2 2

-

2.例210(2x2

+1)

x2

+1二、定積分換元

1

dt02sec2

t(2

tan

t

+1)sectp6令x

=tant=====cost02sin2

t

+cos2t=p6102d

(sint)1+

sin

tdt

=p60p=

arctan(sint)

|

62=

arctan

1例3

3

dx2x

,

x<

01

+

e1,

x

?

0

1

+

x1計算0

f

(

x

-1)dx.例4

設f

(

x)

=

1令u=x-1解:原式=====-1

f

(u)duu

=100-1du

+

1

du1+

e

1+

u1=

-ln(e-u

+1)

|0

+ln(1+

u)

|1-1

0=

-ln

2

+ln(e

+1)

+ln

2

=

ln(e

+1)

1ln

x

d

1-

x

e2三、定積分分部積分ln

xe2(1-

x)2

dx

=

ee11e2e2dxx(1-

x)ln

x

|

-e=2|11-

x2eex1-

e

1-

x-

-

ln1-

e2=1-

ln=e1+

e

e

+1例5

e2110x2e-t

dt

.計算

xf

(

x)

dx

,

其中f

(

x)

=

例61010f

(

x)

dx2xf

(

x)

dx

=解:212121002

11-

x2

f

￠(x)

dx=

x f

(x)

|01x22=

0

-

10|144

1

42xe-x

dx

=

e-x4=

1

(e-1

-1).40π4π2nn2n

-12n

-

2+I

n=

2n

-1In-1,

I0

=2n-1sec

x

dx

,證明降階遞推公式設I

=例70nsec2n-2p4證:

I

=0-(2n

-

2)

4

sec2n-2

x

tan2

x

dx=

sec2n-2

x

tan

x

|4px

d

(tan

x)psec

x

dx0402n2n-2n-1sec

x

dx

+(2n

-

2)=

2

-(2n

-

2)p40pn=

2n-1

-(2n

-

2)I

+(2n

-

2)I+2n-1In-11 2n

-12n

-

2n-1=

2n

-4000pp4sec

x

dx

=I

=四、含參量的變限積分例80設f

(

x)連續(xù),

F

(

x)

=

x

f

(

x

+

t

)

dt,

計算F

￠(

x).xxf

(u)

du2

x0令u

=x+tf

(x

+

t)

dt

====解:F

(x)=F

(x)

=

2

f

(2x)

-

f

(x)當被積函數(shù)中出現(xiàn)求導變量時,或利用代數(shù)方法將求導變量提出積分號;或利用換元積分法將求導變量放到積分的上,下限.例9)2022sin4

xxxfi

0+x

-

t

dt,arctan ln

1+(

x

-

t

)

(其中x

>

0,

計算極限lim

f

(x)

.設f

(x)=解:令x2

-t

=u,0arctan

u

ln(1+

u)

2uduf

(x)

=

-xx0u

arctan

u

ln(1+

u)du=

2402xlim

f

(x)sin4

xxu

arctan

u

ln(1+

u)du=

limxfi0+xfi

0+型

004x3=

lim

2x

arctan

x

ln(1+

x)xfi

0+=

1210202f

(

x)dx.(

x

-

1)x-

y

+2

ye

dy,求設f

(x)=解3

0原式=1

1

f

(x)d

(x

-1)311030

313=

1dx(

x

-

1)

ee dy]0

-[

(

x

-

1)3

-

x2

+2

xx-

y2

+2

y=

-102162d[(

x

-

1)

](

x

-

1)

e2

-(

x-1)

+1令(x

-1)2

=u-01ue

due6-u1=

6

(e

-

2).例0-

2(x5

+

4cos5

x)cos3

x

dx

.p2p計算cos8

x

dx-

2-

2p2pp2p解:原式=cos

x

dx208=

0

+

8x5

cos3

x

dx

+

4p8

6

4

2

2=

8

7

5

3

1

p32=

35

p五、利用被積函數(shù)的奇偶性證明或計算例11-2計算定積分2

x

ln(1

+

ex

)

dx例122x)(-dt

)-2-t令x=-t解法一:

-2

x

ln(1

+

e

)

dx

====2

-

t

ln(1

+

e22

t-2[t

-

t

ln(e

+1)]

dt=t222-2-2t

ln(e

+1)

dtt dt

-=t

dt2-2212原式=83=22

x-2-2x

ln(1+

ex

)

dx-

x

x

2

+

e

2

dx=

x

ln

e

2

+

ln

e解法二:222x-2-2-

x

x

x

ln

e

2

+

e

2

dx

=

xdx

+

202=x

dx

+

0=

83xx

lnt1111+

t證法一:

f

( )

=uudt

====12

du11+

1x

-

ln

u

-1令u

=tx=1duu(u

+1)ln

u11+

t

t(t

+1)

1

x

lnt

lnt+f

(x)

+

f

( )

=xt

dt

=12x

ln

tdt

=

1

ln2

x六、積分等式的證明例13

設f

(

x)

=

x

ln

t

dt

,

證明f

(

x)

+

f

(

1

)

=

1

ln2

x.1

1

+

t

x

22x

21

1f

(

)

-

ln

x證法二:令g(x)=f

(x)+xxxln

1x21+

11+

x則g￠(x)

=

ln

x

+-1

-

ln

x

1

=

0于是g(x)為一常數(shù),又g(1)

=

0,

故g(x)

”

0.解p0

sin

x

+

cos

xsin

x求

2

dx.psin

x由I

=

2pdx,0

sin

x

+

cos

xcos

xdx,

設J

=

2,20p20

sin

x

+

cos

xpdx

=則I

+J

=0

sin

x

+

cos

xp

sin

x

-

cos

xI

-

J

=

2dx

=

-20sin

x

+

cos

xp

d

(cos

x

+

sin

x)=

0.2故得2I

=p,4即I

=p.例144

000dx

.a

a1+sin

2xf

(a

-

x)

dx

,f

(x)

dx

=p

1

-

sin

2x并利用此式計算證明設f

(x)在[0,a](a

>0)上連續(xù),例15a0證:0af

(t)(-dt)令t

=a-xf

(a

-

x)

dx

====a0=af

(x)

dx0f

(t)

dt

=dx4

01+sin

2xp

1

-

sin

2x解:dx0

2

2

=p41+sin

p

-

2x

1-

sin

p

-

2x

4

0=p

1-

cos

2xdx1+cos

2x

2cos

x402dx

=p

2sin2

x0=p0p4

(sec2

x

-1)

dx

=

(tan

x

-

x)

|

44=1-

pdt

=

0.1 ln

tt

tf

(t

+

)x1x例16

f

(x)在(0,+￥

)上連續(xù),試證明x

>0,有dt,1

lntt

tx1xf

(t

+)證法一:令g(x)=111=

0,-1x2xlnx則

g￠(

x)

=

f

(x

+

1

)

ln

x

-

f

(

+

x)x

x

xg(x)為一常數(shù),g(x)

=

g(1)=

0.duu1ln

111xxu2-1f

(

+

u)uu====令t

=1dtt

t1

lntx1xf

(t

+

)證法二:

g(

x)

=u

u1

lnuf

(x1xu+

u)

du

=

-g(x)=

-\

g

(x)

=

0.七、積分不等式的證明試證明02psin(x2

)

dx

>

0證明:原式tsin

t2

0====令t

=x2

12p

dtdttt=

2

01

2p

sin

tdt

+

2

p1

p

sin

tdudt

+=ppu

+p00dtsinttsintt00dt

-=ppsin

u

+p2

sint

t

+p)

dtsint

(120=pt

(t

+p

)

t

+p

-

t

sin

t

?

0t

+p

-

t

>

002p\

sin(x2

)

dx

>

0(令u

=t

-p

)例17f

(

x)dxf

(

x)dxab

ba?

(b

-

a)2

.證明設f

(x)在區(qū)間[a,b]上連續(xù)，且f

(x)>0.證

作輔助函數(shù)f

(t

)dtf

(t

)dtxaxa-

(

x

-

a)2

,F

(

x)

=11f

(

x)f

(t

)dtf

(t

)xaxa-

2(

x

-

a)dt

+

F

￠(

x)

=

f

(

x)=xaxaxa2dt,dt

-f

(

x)f

(t

)dt+f

(t

)f

(

x)例18(+

-

2)dt

?

0f

(t

)

f

(

x)f

(

x)

f

(t

)xa即F

￠(x)=\f

(

x)

+

f

(t

)

?

2f

(t

)

f

(

x)

f

(

x)

>

0,F

(x)單調(diào)增加.又

F

(a)=0,\

F

(b)

?

F

(a)

=

0,f

(

x)dxf

(

x)dxbaba?

(b

-

a)2

.即e

sin

tdt

>

0xsin

tx+2p證明F

(x)=例192p0sin

te

sin

tdt證:F

(x)==pp2psin

t0sin

te

sin

tdte

sin

tdt

+(在第二個積分中,令u

=t

-p

)00sin

te

(-sinu)

due

sintdt

+=-sin

up

p0-1)dt=e-sin

t

sint(e2sin

tp當0

<

t

<

p時,

sin

t

>

0,

e2sin

t

-1

>

0,

\

F

(x)

>

0.10310f

(

x)

dx

.f

(x)

dx

?2￠0

<f

(x)￡1,證明設f

(x)在[0,1]上連續(xù),在(0,1)內(nèi)可導,且f

(0)=0,例20030xxf

(t)

dt

-

2證:令F

(x)=f

(t)dt,則F

(0)=0,30f

(t)

dt2

f

(x)x-

f

(

x)F

(x)

=￠20-

f

(

x)f

(t)

dt=

f

(x)

2x20(

x),xf

(t)

dt

-

f令G(x)=2則G(0)=0,G

(x)

=

2

f

(x)

-

2

f

(x)

f

(x)

=

2

f

(x)[1-

f

(x)]由f

(0)

=

0及0

<

f

(x)

￡1,

得f

(x)

>

f

(0)

=

0,有G

(x)

?

0,

G(x)

?,F

(x)

?

0,F

(x)

?,G(x)

?

G(0)

=

0,F

(x)

?

F

(0)

=

0,故當0

￡

x

￡1時,有F

(x)?0.特別有F

(1)?0,10312

0f

(

x)

dx

?

0

.f

(

x)

dx

-此即2

[11-2+

ln2

(1

-

x)]dx.sin

x解2x8

+

11-21ln(1

-

x)dx原式=0

+=1200-21x)dxln(1

-ln(1

-

x)dx

-=

3

ln

3

+

ln

1

.2

2

2備例1

求102f

(x)

dx.￠-(

x-1),f

(0)=0,計算備例設2

f

(x)

=

ex0f

￠(x)

dx-

f

(0)

=解:

f

(

x)

=

f

(x)dxex=0-(

x-1)2無法積出￠101010xf

(x)

dxf

(x)

dx

=

xf

(

x)

|

-dxxe10-(

x-1)2=

f

(1)

-仍無法解出1010f

(

x)

d

(

x

-1)f

(x)

dx

=(x)

dx1010￠-

(x

-1)

f=

(x

-1)

f

(x)

|dx10-(

x-1)2(x

-1)e=

0

-02=

1

e-(

x-1)2

|12=

1

(1-

e-1

)baba12￠-

b)

f

(

x)dx(

x

-

a)(

x[

f

(a)

+

f

(b)]+b

-

a2f

(

x)dx

=證明:ba(

x

-

a)(

x(

)ba-

b)

f

"(

x)dx

=-

b)df'

x(

x

-

a)(

x-

baba(

)(

-

a

-

b)dxf'

x

2

x(

)

-

a)(

x

-

b)=

f' x

(

x=

-ba(2

x

-

a

-

b

df

x)

(

)

(

)(+baba-

a

-

b)f

(x)

2dx=

-

f

x

2

x(

)(

)

(

)(

)ba-

b

-

f

b

b

-

a

+=

f

a

af

(x)

2dx備例3

f

￠(

x)

在

[a,

b]連續(xù),證明baba122(

x

-

a)(

x

-

b)

f

￠(

x)dxf

(b)]

+[

f

(a)

+b

-

af

(

x)dx

=\021/

2

20x

f

(

)dx.2￠

xf

(2x)dx=1/32,計算備例4

設f

(x)連續(xù),

且已知f

(1)

=1/

8,

f

(1)

=1/

4,202202￠

xx

f

(

)dx

=

2

x

df

(

)2

2￠

x解:￠

x￠

x202

2xf

(

)dx2-42=

2x

f

(

)

|020xxdf

(

)2￠=

8

f

(1)

-8xxf

(

)dx2202=

2

-8xf

(

)

|0

+8(令x

=2u)120f

(2u)

4du2=

2

-16

f

(1)

+82=110

p4

2<dx

<p

1

6

4

-

x2

-

x3備例5

證明證:當0

<x

<1時,111<<4

-

x2

4

-

x2

-

x3

4

-

2x2關(guān)于x從0

到1

積分dxdx

<101014

-

x2

-

x3

1

4

-

x2|0

=x

1

2arcsin6p

=2

4

22111010p<xdx

=

arcsin

|

=4

-

2x231-1f

(x)

dx

￡

1

.f

￠(x)￡1,證明備例6

設f

(-1)

=

f

(1)

=

0,

f

(-1)

=

f

(1)

=

0,￠1-1-11-1f

(x)dx

=

xf

(x)

|1

-

xf

(x)dx證:112-1f

￠(x)dx2=

0

-￠=

--11-122

+

1x

f

(x)

|12-12

1

1

1x2

f

￠(x)dx

=

0

+

x2

f

￠(x)dxx

f

(x)

dx12121-1-1f

(x)

dx

=121-1x2

f

￠(x)

dx￠

￡x

dx￡1-121213=備例7+￠2000

0

02!(

x

-

x

)f

(

x

)f

(

x)

=

f

(

x

)

+

f

(

x

)(

x

-

x

)

+[稱此式為帶積分形式余項的泰勒公式]設函數(shù)f

(x)在x0

點的某個鄰域內(nèi)有n

+1階連續(xù)導數(shù)試證明x1n!

n!x0f

(

n)

(

x

)+

0

(

x

-

x )n

+0f

(

n+1)

(t

)(

x

-

t

)ndt證明:n

(n)x

xx0x0(

x

-

t

)

df

(t

)f

(t

)(

x

-

t

)

dt

=(

n+1)

nxx0x0(-1)dtf

(n)(t

)]x

-=

[(

x

-

t

)nf

(n)(t

)

n(

x

-

t

)n-1xx0df

(t

)(

x

-

t

)f

(n)(

x

)

+

n=

-(

x

-

x

)n(n-1

)n-100[]x(n)x000n-1

(n-1

)-

t

)

f

(t

)(

x

)

+

n(

x=

-(

x

-

x

)n

fxx0(-1)dt-

nf

(n-1

)

(t

)(n

-

1)(

x

-

t

)n-2f

(

n)

(

x

)

-

n(

x

-

x

)n-1

f

(

n-1)

(

x

)0

0

0n=

-(

x

-

x0

)xx0f

(

n-1)

(t

)(

x

-

t

)n-2

dt+

n(n

-

1)n

(

n)

n-1

(

n-1)=

-(

x

-

x0

)

f

(

x0

)

-

n(

x

-

x0

)

f

(

x0

)n-2

(

n-2)-

n(n

-

1)(

x

-

x0

)

f

(

x0

)

-xx0f

'

(t

)dt+￠20000

02!(

x

-

x

)f

(

x

)-

x

)

+\

f

(

x)

=

f

(

x

)

+

f

(

x

)(

xxn!1n!x0f

(

n)

(

x

)+

0

(

x

-

x )n

+0f

(

n+1)

(t

)(

x

-

t

)ndt-

n(n

-

1)(n

-

2)

2(

x

-

x

)

f

'

(

x

)

+

n(n

-

1)

20

0=

-(x

-

x

)n

f

(n)

(x

)

-

n(x

-

x

)n-1

f(n-1)

(x

)0

0

0

0-

n(n

-1)(x

-

x

)n-2

f

(n-2)

(x

)

-0

0-

n!(x

-

x

)

f

'

(x

)

+

n!

f

(x)

-

n!

f

(x

)0

0

0￠2000002!(

x

-

x

)

+f

(

x

)-

x

)

+(

x

)(x\

f

(

x)