微積分第三章習題參考答案

第三章

一元函數(shù)積分學習題參考答案§3.1不定積分的概念及計算(51-52)33.

ln

|

x

|

arctan

x

c

.

2

4.

csc

x

cot

x

c

tan

x

c

.1x

c

.

2

xp51.一.1.

e

. 2.

c0

.x5. (

x

sin

x)

c

. 6.

e

27.

cot

x

tan

x

c

2csc

2

x

c

2cot2

2

x

c

.

1

2sin2

x

cot2

x

2sin2

x,sin2

x2p51.二.1.331x2

3x

5

2xdx

(2

5(

)x

)dx

2x2

x

5

c.3x

(ln

2

ln

3)2.

dx

1

tan

x

c.1

cos

2

x

2p52.三.1.

f

(sin2

x)

cos

2

x

cot2

xx

f

(

x)

1

2

x,

f

(

x)

ln

|

x

|

x2

c.p52.2.解:

設曲線方程為y

f

(

x),則f

(

x)

1

,

f

(

x)

ln

|

x

|

c,x曲線過點(e2

,3),

c

1,曲線方程為y

ln

|

x

|

1.3.用s(t

)表示t秒后物體離開出發(fā)點的距離.s(0)

0,

s(t

)

3t

2

,

s(t

)

t

3

,

s(3)

27(米);由s(t

)

t

3

360,得t

23

45(秒).p52.公式:shx

,chx

.221,22

x2

x22

xx

x

2

x2

1

e2

x,ex

shx,exchx都是e2

x的原函數(shù).ex

e

xex

e

x(ex

shx)

ex

(shx

chx)

e2

x

,(exchx)

ex

(chx

shx)

e2

x

,

1(

e

)

(e

shx)

(e

chx)

e

,四.證明:

(

e

)

ex一.1.

ee

c

,

ln

|

ln

x

|

c

.2.

ln

|

x

sin

x

|

sin

x

1

sin5

x

2

sin3

x

c

.5

3

ln

|

sin

x

cos

x

|

c n

2

2

x4n23.

I

.(sin

x

cos

x)

c n

2n

24.

I

x2

c

.§3.2不定積分的換元法(53-54)9

4

x2

9

4

x2二.1.4I

1

e2

x2

c;2.

1arcsin2342

x

9

4

x2

c

.I

dx

xdx

p53.3.I

cos4

x(1

cos2

x)2

d

cos

x

(cos4

x

2cos6

x

cos8

x)d

cos

x7

5

9

2

cos7

x

1

cos5

x

1

cos9

x

c.4.

I

(sec2

x

1)2

sec2

xd

sec

x

(sec6

x

2sec4

x

sec2

x)d

sec

x

1

sec7

x

2

sec5

x

1

sec3

x

c.7

5

32p54.sin

sin

1

[cos(

)

cos(

)]2cos

cos

1

[cos(

)

cos(

)]25.

I

1

(cos

2

x

cos12

x)dx

1

sin

2

x

1

sin12

x

c.4

2422dx6.(

x

3)dx

1

(2

x

2)dx

2

x

5

2

x

2

x

5

2

x

5

x2

x2

1

ln(

x2

2

x

5)

arctan

x

1

c.2

2x2dx

a2

sin2

tdta2

x22

ap54.三.1.

令x

a

sin

t,2

2

2a2

x22

2

a

(1

cos

2t

)dt

a

(t

1

sin

2t

)

carcsin

c.22x

xax2x2x2x

41

4p54.2.解法(1)

dx

dx

2

x

2

xx21

4

1

1

d

2

1

arcsin

2

c.解法(2)

令x

2sec

t,2

2x2x

4

dx

1

dt

1

t

c

1

arccos

2

c.2

xx

1

3

x

6

(t

2

t

1

t

1

2t

3

3t

2

6t

6ln(t

1

2

x

1

33

x

1

66

x

1

6ln(

6

x

1

1)

c.p5x432

x2

2

1

arctan

x2

c.p54.4.解法1:1x3

x41x

(

x

1)

3

4解法2:I

1

dx

1dx

1

x4x3

(

x4

1)xdx

x4

(

x4

1)xx1

dx

dx

1

x4

)dx

1

arctan

x2

c.

1 2

x2

2I

x3

(

x4

1)(

1

xdx

§3.3不定積分的分部積分法(55-56)2一.1.

2

x2e

x2

e

x2

23.

xf

(

x)

f

(

x)

c

. 4.

x

ln

x

x

c

.5.

x

arcsin

x

1

x2

c

.6.

1

x

cos

2

x

1

sin

2

x

c

1

x

sin2

x

x

sin

2

x

c

.4

8

2

4

8

x

sin

x

cos

xdx

1

x

sin

2

xdx

1

xd

cos

2

x2

4

1

x

cos

2

x

1

cos

2

xdx

1

x

cos

2

x

1

sin

2

x

c.4

4

4

8

2

4

c

.2.

x

ln

x

x

c

.x31

x2

1

x3

arctan

x

1

x2

1

ln(1

x2

)

c.3

6

6p55.二1.3

x2

cos

xdx

x2d

sin

x

x2

sin

x

2

x

sin

xdx

x2sin

x

2

xd

cos

x

x2

sin

x

2

x

cos

x

2sin

x

c.2.

x2

arctan

xdx

1

arctan

xdx313

1

x3

arctan

x

13

dxp56.3.I

3x

sin

xdx

3x

d

cos

x

3x

cos

x

cos

x3x

ln

3dx

3x

cos

x

3x

ln

3d

sin

x

3x

cos

x

3x

sin

x

ln

3

3x

sin

x

ln2

3dx

3x

cos

x

3x

sin

x

ln

3

ln2

3I

,1(3x

sin

x

ln

3

3x

cos

x)

c.1

ln2

3

I

p56.

4.令u

x

,則x

u2

,dx

2udu.

sin2

xdx

u(1

cos

2u)du2

u2

u2

2

1

ud

sin

2u

1

usin

2u

1

sin

2udu2

2

2442

u

1

1usin

2u

cos

2u

c2

2

x

12

2x

sin

2

x

1

cos

2

x

c.p56.6.

當

1時,2

x

ln

xdx

x1

ln

xdx

1

ln2

x

c;當

1時,ln

xdx

1

c.x

dxx

ln

xdx

1

1

1x

ln

x

1

1

1

1

x

ln

x

x

1

1 (

1)2

e

xp56.

5.

xe

xdx

xe

x

cp56.

7.I0

x

c,

I1

x

ln

x

x

c,In

x

ln

x

n

ln

xdx

x

ln

x

nIn1n

n1

n

x

lnn

x

nx

lnn1

x

n(n

1)In2

x

lnn

x

nx

lnn1

x

n(n

1)x

lnn2

x

(1)n2

n(n

1) 3

x

ln2

x

(1)n1

n(n

1)(n

2)

2I

,(接下頁)1

x

lnn

x

nx

lnn1

x

n(n

1)x

lnn2

x

(1)n2

n(n

1)

(1)n1

n(n

1)3x

ln2

x2(

x

ln

x

x)

c,k

0nnx

lnnk

x

c.(1)k

n!n

k

!

I

§3.4

有理函數(shù)的不定積分(57-58)p57.一1

ln4

x

12.23d

(

x

1)

(

x

1)2

4

1

arctan

(

x

2

23.

2

x

5

x2dxdx

1

arctan

x

arctan

x

c.

(

x2

1)(

x2

3)

212

3x3dx

273

2

x

3

x67(4

x

6)dx

(

1

)dx

5

x2

6

x

x x

3

x

2

ln

x

6ln(

x

3)

7

ln(

x

2)

c.

x32.x

3

x

3

(

x2

3

x

9

)dx

9

x

27

ln

|

x

3

|

c.3

21

1

arctan

x

c.3.2(

x

1)

2xdx

(

x

1)2

(

x2

1)

p57.二.1.注意

(

x2

1)

(

x

1)2

2

x.p58.三.1.3

cos

x22

2cos2

x2222d

tan

x22tan

x

1

arctan2

c.22

tan2

x

dx

dx

d

xxxcos2

(1

sec2

)

1u

x

432x

4

xp58.2.

dx

cos2

2

222d

tan

x1

tan

x

2

ln

|

1

tan

x

|

c.4u

du

4

(u

1

13.)duu

1x

x1

sin

x

cos

x2cos2

x

2sin

dx

dxu

u1

x2

arcsin

x

1

x21

x2

c.1

xdx

4.1

x

1

dx

§習題課(59-60)5.

x

ln(

x

1

x2

)

1

x2

c.3.

c

.

4.

tan

t

.f

(

x)

4

x

(6.

2e x

1)

c

.

7.

1

e2

x2

c

.

2

2

8.

1

x2

f

(

x2

)

1

f

(

x2

)

c

.

x

一.1. 2

x(ln

x

1)

c

.

2. cos

x

2sin

x

c.xp59.二.1.

esin

x

sin

2

xdx

2

esin

x

sin

xd

sin

x

2

sin

xdesin

x

2esin

x

sin

x

2

esin

xd

sin

x

2esin

x

sin

x

2esin

x

c.x

1|

x

|

1x

1

x2dx2.max{

x2

,1}dx

dx

x2dxx333

1

x3(接上頁)

33

2

cx

1|

x

|

1.

c x

1

max{|

x

|,1}dx23x

c

1

x

3p59.3.

e

2

x3

dx

u

2

x

3

euudu

udeu22

x3

ueu

eu

( 2

x

3

1)e

c.p60.4.

sin(ln

x)dx

x

sin(ln

x)

cos(ln

x)dx

x

sin(ln

x)

x

cos(ln

x)

sin(ln

x)dx,

sin(ln

x)dx

1

[

x

sin(ln

x)

x

cos(ln

x)]

c.earctan

xdx2

31

x2u

arctan

xp60.

5.(1

x

)

eu

cos

udu2

1

eu

(sin

u

cos

u)

c1

x

c.2

1

earctan

xp60.26.

cos4

xdx

(1

cos

2

x

)2dx4

1

(1

2cos

2

x

cos2

2

x)dx

1

[1

2cos

2

x

1

(1

cos

4

x)]dx4

2

3

x

1

sin

2

x

1

sin

4

x

c.8

4

32p60.三.

當x

1時,

F

(

x)

ch(

x

1)

c1

,當x

1時,F

(x)

1

x2

ln

x

1

x2

c,2

414由F

(1

)

F

(1

),得c

c

5

,f

(

x)dx

14x2

1

x2

ln

x

4

c

ch(

x

1)

c

5

x

1.x

1

2習題課(課外作業(yè))(61-62)

x1

34.

2

x(1

x)e2

x

ln(

x

1

x2

)

c

.1

x2一.1.2.

3e

c

.

2

. 5.

xx2

x2

a2

ln

|

x

a2

|

c

.1

16.,sin

x

sin2

xdx

. 7.

ln(4

x4

)

,

xdx

.t

1

ln

x

.2sin

x[

f

(sin

x)

f

(sin

x)]

2[

f

(sin

x)

f

(sin

x)]

c

.xx

a

3.

F

(arcsin

)

c

.p61.二.1.

x21

x2

dx

u

1

x2

(u4

u2

)du3(1

x2

)2

c.22155(1

x2

)2

135

3u5

u3d

(u

1)2.

x

x

2

x

1 1

2u

uu

1

x

d

(u

1)

arcsin

u

1

c22

x2

(u

1)2

arcsin

1

x

c.

c

dxx21

x21

x2p62.3.

arctanx

1dx1

x

arctan4x

1

1

x

arctan34.

ud

(cot

u)

ucot

u

ln

|

sin

u

|

carcsin

x

ln

|

x

|

c.dxx

1

1

(x

1x

2)

x

1

c.arcsin

x

dx

u

arcsin

x

ucsc2

udux

n

x

n

xn11

I0

e

,

I

xe dx

xe

e

c.x

x

x

x1p62.5.In

x

e dx

x

e

nI

,

c.6.令u

ln

x,nnxnkk

0(1)k

n!n

k

!I

ex

uu

0,f

(u)

1x

f

(

x)

x

c1x

0,

f

(0

)

f

(0),xx

0.

c

1

c,

f

(

x)

x

1

ce

0

ue

c x

0e

c x

0§3.5

定積分(63-64)一.1.充分.0. 2.(幾何意義)

4

2.

.3.(1).

,

2

.4.

.

.

9

3

(2).

,

2

.1xa5.

I

lim

f

(

x)

f

(a).10k

2n

1

n2101

x2

)

|1

ln(1

2

).2011

1.

11n2

k

21

ln(

x

1

xnp63.二.

1.(1).

limnn

k

1p64.(2).limnn

k

1

lim

n

1

2

nnn

1

nndx2

n

nn

1

n

lim

[(

)

(

)

(

)

(

)

]x dx

1201np64.2.(1)

|12(2)

limdx

lim

0.1

xn

1

2(0

1

)n

pnsin

xxsin

xxnn

pnnlimxnn

pdx

|dx

0.

nn

pndx

2e§3.6

微積分基本定理(65-66)00cos

x一.1.0,eab

sin

b

..sin

x

1(3.

1

.

12

3

4.

2xf

(

x2

)dx

.

5.

0

. 6.

2

.yxyye

ydy

dx2.

dy

cos

x

dxe

dt

cos

tdt

0,

e

cos

x

0,

.)02032200502502p65.sin

x

|

cos

x

|

dxsin3

x

sin5

xdx34525sin

2

x

45

2sin

xd

sin

x

.2.|

2

x

4

|

dx(4

2

x)dx

(2

x

4)dx

4

9

13.二.1.20122220111010

1

xp65.3.2min(

x

,

x)dx

xn

,1,n

1dx

0.xnxnn

2

1

3

1

.0

1

x1

x3

2

61

n4.lim

x

dx

0. 0

nxn0

1

x

0

n

limxdx

x dx

xdxdx

x dx

212001

3

e2

e.p66.

5.f

(

x)dx

(

x

1)dx

2三.證明:

1.(1)F

(

x)

f

(

x)

1

2

0.f

(

x)

1

f

(

x)

1

f

(

x)(2)

F

(a)

dx

0,F

(b)

f

(

x)dx

0,

F

(x)

0在(a,b)內(nèi)至少有一實根,xabbabae

dxdx

1p66.三.證明:

1.(1)F

(

x)

f

(

x)

2

0.f

(

x)

1

f

(

x)

1

f

(

x)(2)

F

(a)

dx

0,F

(b)

f

(

x)dx

0,abbabadx

F

(x)

0在(a,b)內(nèi)至少有一實根,由(1),F

(x)在[a,b]上是嚴格增加的,

F

(x)

0在[a,b]上至多有一個實根,

F

(x)

0在(a,b)內(nèi)恰有一個實根.22p66.

2.

f

(

x)

f

(

x)

f

(

x)

M

(

x

a),f

(a)

f

(t

)dt(b

a)f

(

x)dx

M.2證法(2).x

(a,b],應用微分中值定理,

(a,x),使得f

(

x)

f

(

x)

f

(0)

f

(

)(

x

a),

f

(

x)

M

(

x

a),(b

a)f

(

x)dx

M.2xababa(

x

a)2(

x

a)2p66.證法(3).令F

(

x)

M2F

(

x)

M

(

x

a)

f

(

x),F

(

x)

M

f

(

x)

0,

F

(

x)

0,

F

(

x)

f

(a)

0,F

(a)

0,f

(

x)dx

M.2xa

f

(x)dx

(a

x

b),則xa即§3.7定積分的換元法與分部積分法(67-68)2

10

7.

2

6e2

;4.

1

; 5.

2

;

6.

f

(1)

f

(0)

;8.

2

2e1

.p67.,(5)

2

;

3

1.(1)

0

,(2)

0

,(3)

0

,(4)

21

3

41

2

2 2

n

2.

e

e

,

; 3.

2( 3

1) ,

1

2ln

;1100

2

2

2

2u

ln

xp67.

2.sin(ln

x)dx2

1

eu

(sin

u

cos

u)

|1

1

(e

sin1

e

cos1

1).2注意公式:eusin

ue

du

c

c(

sin

x

cos

x)e

x

e

x

sin

xdx

(

cos

x

sin

x)e

x

e

x

cos

xdx

1201001

x

ln(

x

x2

1)

1x2

1021010110p67.

3.ln(

x

x

1)dx

ln(1

2

)

ln(1

2

)

2

1.u

x

1p68.

4.f

(

x

1)dxf

(u)duxxx2

1dxdx

dx1

e

1

x

0

1

ex

ex110010

1

[

x

ln(1

ex

)]

ln(1

x)

ln(1

e).10011010u

1

xxm

(1

x)n

dx(1

u)m

und

(1

u)(1

u)m

undum

n(1

x)

x

dxdxdx

1

ex

1

xp68.三.1.200202220200nsin

xdxnsin

xdx

nsin

xdxnsin

udunnnsin

xdx,sin

xdx

2x

u

sinn

xdxp68.三.2.sin

xdx.(公式)習題課(69-70)22

ye

y0111x

x2000

d

x

1

(

x

)210

4

5.a

1,

b

1;p69.一.1.

a

;(積分中值定理)2.

;(定積分定義)3.lncos

x;4.a

2;6.

xf

(

x)

f

(t

)dt;

1

7.

2

dx

x

1

x

2arcsin

x

.xdx

03421222201

2

2.2.Max(1,

x

)dx

2[x

dx]

20

.3dx

p69.二.1.01

sin

2

xdx

0

|

sin

x

cos

x

|

dx

4

(sin

x

cos

x)dx

3

(sin

x

cos

x)dx223

x2

3

3e

x26

6e

xx450x

2e

t

dt

)x303

x5x042x0x03.lim(

13

x2x0

1

1p69.

3

x

3

lim

lim

lim15

x60

x120

x

10212

xe

x

1x0

lim

.(洛必達法則)xxe

dt

tx2

sin

tx

sin2

t12x

x2002sinx2p69.

4.解:2

n

1

n

33

1

2

.

n

1n

n

2

4

2

2n

3

4

2dt,,

f

(1)

0.nx

2nsi

n

tdtdxn為正偶數(shù)tx

n

n

2

5

3f

(

x)

1

f

(

x)

1

n為正奇數(shù)p70.

5.解:x21100112001200xf

(

x)dx

f

(

x)d212x

f

(

x)dx2x2

f

(

x)

f

(1)

x

sin

x dx

21cos

x22

1

(cos1

1).p70.(接上頁)5.

MaxA0f

(

x)

x(1

x)sin2n

x

0

(

x

0),

x1

1,

x2

kk

1,

2,x(t

t

2

)sin2n

tdt,(n

N

)p70.三.1.證明:

f

(

x)

f

(

x)

(1

2

x)sin2n

x

2n(

x

x2

)sin2n1

x

cos

xf

(1)

sin2n

1

0,

f

(1)取極大值.當x

k時,可證明f

(x)不取極值.1(t

t

2

)sin2n

tdt01(t

t

2

)t

2ndt

011

x20011

x2

)

ln(1

2

).p70三.(接上頁)1.

f

(0)

0,

0

M

f

(1)

1

1

2n

2 2n

31.(2n

2)(2n

3)11

x212.證明提示

11

xn1dx

ln(

x

習題課(課外作業(yè))(71-72)24.

2e2;03.

2

;

10

p71.

一. 1.

0

;

2.

1

;5.

sin

x

;6.2

2

f

(

x)

1

2

xe2

x

,

f

(

x)

xe2

x

1

.xx[2

f

(

x)

1]dx

e

1,2

ln

232

ln

22ex

1

1

u2

du7.

2arctan

uetu

1xxexet

dt

63ex

1

2(

arctan

1)3

,

x

ln

2.

1dtet

1p71.一.7.2

002

022

001611611a2x

x22cos

t212

d

(cos

t

sin

t

)

[

.sin

t

cos

t

4p71.2.arctanx

1dx(利用第62頁第3題結果)13

[

x

arctanx

1

(

x

2)

x

1]3.

16

23a二.1.dxdtdt1

(sin

t

cos

t

)

(cos

t

sin

t

)sin

t

cos

tdt

x

a

sin

t

12

sin

t

cos

t202200220020cos

t1

sin

tp72.3.

f

(

x)

dt,

f

(0)

0,d

sin

xf

(22)

1

sin

t4f

(

x)df

(

x).4xdx

1

f

(

x) 1

f

(

x)

arctansin

x

2

.

arctan

f

(

x)

2

arctan02n2nx0x01

f

(

xn

)nxn12nx2n1x0x0p72.4.

F

(

x)

f

(u)du,1f

(u)duF

(

x)lim

lim

lim

n2nxn

f

(0)

1

.2n

2nxnu

xn

t

n

1n

0xnn

0xnn1t f

(

x

t

n

)dtn

lim

f

(

x

)

f

(0)xx201

cos2

x202202200t

sin

tdt,1

cos

tx

sin

x

0,得x

.2當

x

時,

f

(

x)

0,2x

sin

xdx當

x

3

時,f

(x)

0,

f

(

)為極大值.sin

xdx1

cos

x1

cos

xd

cos

x1

cos

x4f

(

x)

p72.

5.令f

(x)xf

(

)

2

arctancos

x

2

.

00001p72.三.1.證法1

:令F

(

)f

(

x)dx,f

(

x)dx

1

[

f

(

)

f

(

)],(0

1),

F

(

)

0,得1f

(

x)dx

f

(

x)dx.

F

(

)

1

f

(

)

12

f

(x)為單調(diào)減函數(shù),

F

(

)

F

(1),1即110010100p72.三.1.證法2:f

(

x)dx

f

(

x)dx,f

(

x)dx.1

f

(

)

1f

(

x)dx

f

(

x)dx.f

(

x)dx

0f

(

x)dx

f

(1

)

(1

)

f

(1

)

f

(1

)

(1

)

f

(2

),0

1

2

1,f

(x)為單調(diào)減函數(shù),1202F

(

)

F

(1),由羅爾定理,

(,1),使

F

(

)

0,即f

(

)

1

f

(

),

(,1)

(0,1).p72.2.證明:令F

(x)

xf

(x)1已知2xf

(x)dx

f

(1),由積分中值定理

(0,

1

),使得

f

(

)

f

(1),F

(x)在[,1]可導,且F

(

x)

f

(

x)

xf

(

x),§3.8定積分的幾何應用舉例(一)(73-74)ln

b3

a232640

2

R2

y2

dx

, 2

xdy

,

2

rdr

,

1

R2d

;p73.一.1.

22.3.a

sint

a3cost(sin

t

)dt

;324.

22

2

2

2

6

6

5.

2

.02

1

1y ln

a

e dy

b

a

;1

3

2sin2

d

2cos

2

d

;

p74.二.1.

y

2

x

4,

y

|x0

4,

y

|x3

2,過點(0,

3)的切線方程為y

4

x

3,過點(3,0)的切線方程為y

2

x

6,32202332323320( ,

3),2兩切線交點為3[(4

x

3)

(

x

4

x

3)]dx

13(

x

3)3

9

9

98

8

4

.3[(2

x

6)

(

x

4

x

3)]dx

A

x33

p2[(

3

p

y)

y

]dy2

2

p32過點(p

,p)的法線方程為x

y

2法線與拋物線的兩個交點為p

0,(2

2p

9

p,

p),

(

,

3

p),3

16

p2

.p( ,

p

)2pyp73.2.

y

p

,

y

1,

A

32033

(2303

.

1r

1

cos

9A

2[2232cos

292

2cos

)d

)d(1

cos

22

5

.4cos2

d

](1

cos

)2

d

p74.二.3.

r

3cos

,2

ap74.二.4.解:面積A

02a2

(1

cos

t

)2

dt

8a40024220

2sin

t2

2

16asin

udu

3

a

.5.(微分方程題)

af

(

x)dx

20

f

(

x)dxdtu

t§3.8幾何應用舉例（二）(75-76)2cos222022202

0000p75.一.1.s

1

cos

xdx

22

2

2cosdx

4 2

sin

x

4.

x2.2tt0

2xdx2.s

2r

2

rx23.s

4

y2

dt

12a2

sin

t

cos

tdt

6a.112020121x

12

2

(1)

dx

5.Vy

0

2

xydx

0

2

x

sin

xdx6

2xd

(cos

x)

2

.2a2xp75.一.4.V

aa

b6.V

a

aa

a

(

y

b)2

dx

(

y

b)2

dx[(

y

b)2

(

y

b)2

]dxa

a4b

x2

dx

4

b

1

a2

2

2a2b.

xdx

.

2p75.二.1.dV

A(

y)dy

3

x2dy

3(

R2

y2

)dy,0dV

y2dx,23sin7

t

cos2

tdt2003sin7

t

cos2

tdt20

6

a3

(

6

4

2

8

6

4

2

)

32

a3

.7

5

3 9

7

5

3

105V

233

R3

.3(

R2

y2

)dy

42.(1).V

2

y dx

6

a

6

aRa2

ap76.二.2.(2)V

(2a)2

2

a

(

y

y

)2

dx1

202023202320232

8

2a3

8

2a3

a

8

2a3

a

8

2a3

a

8

2a3(1

cos

t

cos2

t

cos3

t

)dt0

2a3

7

2a3

.

[2a

a(1

cos

t

)]

da(t

sin

t

)[2

(1

cos

t

)]

(1

cos

t

)dt(2cos

t

1

cos

t

)(1

cos

t

)dt334

x2

dx001

y2

dx

22

cos2

tdt

3300p76.二.3.l

x

2sin

t2(1

cos

2t

)dt

2

3

.baxf

(

x)dx.3

2三.證明:

dV

2

xdy

2

xf

(

x)dx,V

2§3.9-§3.10

定積分的物理應用舉例、平均值（77-78）8

104

dx

8

104

(ln

80

ln

40)040

8

104

8

104

ln

2(千克

厘米)

800

ln

2(千克

米).1.設當高度減少x厘米時,壓強為P(x),體積為V

(x)

100

(80

x),氣體對底面壓力為F

(x)

100

P(x),P(