课用平时第三章2节定积分

第二节定积分（二）一、微积分基本公式二、定积分的换元积分法三、定积分的分部积分法引例与速度函数在变速直线运动中,已知位置函数之间有关系:s

(t)

=

v(t)1221物体在时间间隔 内经过的路程为TTv(t)

d

t

=

s(T

)

-

s(T

)这种积分与原函数的关系在一定条件下具有普遍性。一、微积分基本公式babaf

(x)

dx

=

F

(x)

|

=

F

(b)

-

F

(a)定理.的一个原函数

,

则上式称为微积分基本公式（牛顿-莱布尼茨公式）。故牛顿-莱布尼茨公式架起了定积分与不定积分的桥梁。例1.221x

dx求7=3解:2

2x

dx1x3=323

132=

-1

3

3120dx

2

1-

x2求102=

2

arcsin

1

-

2

arcsin

0例2.解:

原式=

2

arcsin

x

2=

p3解:例3.计算正弦曲线的面积。π0sin

x

dxA

=π=

-cos

x

=

-(-1)

-(-1)

=

20Oyxy

=

sin

x例4.

求分段函数解:f

(x)

=

x

-1,x

£13

-

x,

x

>1在区间[0,

2]上的定积分.02

1201f

(x)f

(x)dxf

(x)dx

=dx

+(x

-1)(3

-

x)=

110x221x2=

(

-

x)

+

(3x

-

)2

2二、定积分的换元积分法1.

定积分的第一类换元法设

f

(u)

有原函数F

(u),

u=

g(x)

可导,

则有换元公式

f

(g(x))d

g(x)=

F

(g(x))

+

Ca=

F

(g(x))

|b=

F

(g(b))

-

F

(g(a))例5.

求01

1

dxex

+

e-x解:ex2

x

dx0

e+11原式=1012

xd

ex=e

+104=

arctan

ex

1

=

arctan

e-

p0psin3

x

-

sin5

xdx.解:

f

(

x)

=

sin3

x

-

sin5

x3=

cos

x

(sin

x)2p=03cos

x

(sin

x)2

dx(

)=p2023cos

x

sin

x(

)dx

-pp223dxcos

x

sin

x=p2023d

sin

x(sin

x)(pp-

223sin

x)

d

sin

x0)25p22=

5

(sin

xpp2)252-

5

(sin

x5=

4

.例6.计算原式40例7.40dx

x

-1

2x

+1解1.用换元法求不定积分

x

-1

dx令t

=22x

+1,

则x

=1(t

2

-1),2x

+1dx

=

tdt.dx

x

-1

2x

+1=tdt1

1

2

2

tt

2

-

-12=-

t

+

C1

3

1

3(

t

-

)dt

=

t2

2

6

23=

1

(2x

+1)32

-

3

(2x

+1)

12

+C6

2(替换t

=2x

+1)原函数40dx

x

-1

\3212x

+1

62123(2x

+1)

]=[

(2x

+1)

-3=

4例7(续).40dx

x

-1

2x

+1解2.

直接用换元法求定积分令t

=22x

+1,

则x

=1(t

2

-1),dx

=

tdt.40dx

x

-1

2x

+131t=tdt1

t

2

-

12

2-1321(1

32

2=t

-

)dt6

23(

t

-

t)=

1

3注意到当x

从0变化到4时，t

相应地从1变化到3;313=

4

.则有babaf

(

x)dx

=f

[j

(t

)]j

¢(t

)dt

.2.定积分的第二类换元积分法假设f

(x)在[a,b]上连续；函数x

=j

(t

)在[a

,b

]上是单值的且有连续导数；当t

在区间[a

,b

]上变化时，x

=j

(t

)的值在

[a,b]上变化，且j

(a

)=a、j

(b

)=b，f

(x)dx

=

f

[j

(t)]j¢(t)dt(令x

=j

(t))b=

a

f

(x)

d

xj

(t)

j

(t)j

(t)

j

(t)定积分的第二类换元法注意两点:也要换成相应于新变量t

的积分限[a

,b]；(2)求出f

(j(t))j

(t)的原函数后，不必变回原来变量x;此换元公式也可反过来使用,即2当x

=

0

时,

t

=

0;

x

=

a

时,

t

=

π

.∴

原式=2aπ20a2(1

+

cos

2

t)

d

t=2122a20=

(t

+

sin

2t

)

2ππ20cos

t

d

t2Oy

=

a2

-

x2xya例8.

计算解:

令

x

=

a

sin

t

,则

dx

=

a

cos

t

d

t

,

且证:若a-aa0f

(x)

dxf

(x)

dx

=

2则a-af

(x)

dx=a则-a

f

(x)

dx

=

00-af

(x)

dxa0f

(x)

dx

+f

(x)

dxa0+af

(-x)=f

(x)时f

(-x)=-f

(x)时偶倍奇零令x

=-t=

0

[

f

(-x)

+

f

(x)]dx=0-

f

(-t)

d

ta=a0f

(-t)

d

t证明:(1)(2)若结论1.4(1)

x

sin

xdx;p-p2-2(3)4

-

x2

dx20试求(2)54

2dx;x3

sin

2

x-5x

+

2x

+1解：x

sin

xdx(1)4p-p=0；dx54

2x3

sin

2

x(2)

-5x

+

2x

+1=0；2-22(3)4

-

x

dx=

244

-

x2

dx

=

2

·

p

·

22

=

2p2设x

=p

-t,dx

=

-dt,2当x

=0,t

=p

;2当x

=p

,t

=0,0p2f

(sin

x)dx=

-0

2p2

p

dtf

sin

-

t=0p2f

(cos

t)dt0=p2f

(cos

x)dx;证:200(1)f

(cos

x)dxp2f

(sin

x)dx

=设f

(x)在

[0,

1]

上连续，证明p00(2)nncos

xdxp2p2sin

xdx

=结论2.设

f

(x)

是以T为周期的连续函数,

证明证明:af

(x)

dx0=Tf

(x)

dx0+Tf

(x)

dxa+T+令

x

=

t

+T

,

当x

=

T

,

t

=

0;

当x

=

a

+T

,

t

=

a;Ta+T

a0f

(x)

dx

=a0af

(x)

d

x0f

(t)

d

t

=f

(x)

dxf

(t

+T

)

d

t

=0=Tf

(x)

dx0+af

(x)

dx0+aTf

(x)

dx0=结论3.——定积分的分部积分公式。u(x),v(x)在区间[a,b]上有连续的导数,udv

=定理.设则uvvduabba

-[

]ab三、定积分的分部积分法例9.

计算解:

原式

=e1e1例10.

求0解:

原式

=

x

arctan

x

1-

10xdx1+

x2102d

(1+

x2

)4

2

1+

x=

π

-

1

1420=

π

-

1

ln(1+

x2

)

142=

π

-

1

ln

2例11

求解:

令2x

=t

,

则2x

=

1

t

2

,

dx