定積分、重積分及其應(yīng)01、設(shè)Fxxtcos2tdt02、設(shè)x

et2dt1x3

fx11

e3

1t2dt4dex2dxex2dex2dxex2cdxet2dtex2d3ex2dx5

x9cosxdx0

dx

dx1x 1x2dx bb6、下列式子中正確的 Bbbbabb

fxdxbbb

f

bafxdx

fcafxdx

f

d無(wú)法確007I14xdx00aI1I2

I24tanxdx0b0

I34sinxdx， （ b8、設(shè)fx在a,b上滿足fx0,fx0b

afxdx，

fbba

I1fafbba，則（ 23aI1I2239、1e2xdx1

11e21 13x210dx1 4sin4sin2xdx

44 10

1x2ex3dx1ex311e1 2x21 dx112x2 411

4

xdx

42sin2xx13xxx1

29

x 1sin

dx31 d2 12、 xdx12sin3d2 3 eln3

13

dx1

xdlnx 4

x1exln2xdxe11

xdlnx14、設(shè)0fxdx2 x1t1fx1dx0ft15lnxdx dxp

xln xln

dxp1

xln2

dxp21，收斂ln dxp11ln xx

dxp11

1dxp1，發(fā)散1

dxp21，收斂

1

x

dx

13）1

exdx

elimexe11

exdx

e1lim

4）

2

1 16、設(shè)Dxy1x2y24，則3d341D17、設(shè)Dxyx2y29，則D

9x2y2d142

18Dx0y0xy1xy12I1lnxydD

I2xydD

I3sinxydDI1I2I3yxyx1

11）1

fx,ydy

0 fx,y 2 fx,y 2

dx

fx,

2120、二次積分 fx2y2dy1 21、設(shè)x2y2z29z0則2dv2dv2 xetsint1、

xtan2xx

xetsint2dtxtan2

0

exsin

exx21 2、

x2sine3x1

tlnx2sine3x x2e3x x2

1

1

x0cosxlncosxcos

x0cos

x0sinxlncos

x0lncos

lim x0lncos

x0sin3、2sin4xsinxcosx2

2sin4xsinxcosxdx

2sin5xdx

2sin4xcosxdx0

2sin4xdsin0 2sin5x 1e201e4xdx1e2

1

2

2x

01e4xdx201e2

2

2

02arctan

e2 1lnxe2 exln

1arctane2 e21ln

2e

xlnx2dxexln

2dxlnxxlnx

e2lne2elne

3、4 ；22 x122

xt

dt2

dt

dt2arctantx1x1

1t1t2

1t1t2

11t2 2arctan2 4 11x11x211

xsint

1sin2

cos2

dx

sin4

dsint

sin4

dt

2cot2tcsc2 4

xx dxx2x2

2cot2tdcott6

1cot3t33633

0xcos2dx

2x22

00sin2dx22

2x20x20

xlnx1dx

e1

xlnx1dx

lnx1

lnx1

dlnx2e2

e1x20

e1 dx

e1x2110 x xe

1

e1 xcos2xdx0

2

x1dx

x1【解xcos2xdxxcos2x1dx1xcos2xdx 11xdsin2x 44

x4xt2 【解exdx4xt2 250x1dx2【解：1）x11 x12x1dx11xdxx x 33x4x4dx233

x24x4dxsin3xsin5

sin3xsin5xdx 0200 0x

x26fx

x

求

fx 0fxdx0fxdxx210x 7、設(shè)fx

e2

1x

，求

fx255

fx2

3 3

3

1edt1x3811x2dxx3

11

dx11

x7arctan21

1x7arctan2 1

3

x19、設(shè)fx在02上連續(xù)可導(dǎo)f012fxdx3，計(jì)算2x2f 0【解2x2fxdx022f02

fxdx23010fx2sinxe2x3fxdxfx03：1）設(shè)3

fxdxkfx2sinx 2）3fxdx32 k2cos313）fx2sinxe2x14cos3411fxlnxefxdx，證明efxdx e10.令1fxdxke112xydDy1yxx2D1y

：1）Dx

2）2xyd22x1ydy1x

ydxdyDxy21x0y0y1D0xy2

1y2 ：1）Dy

2）yd0y dx 0y sinxdxdyDy0yxx1 0y：1）Dx0xxsinxd1sindydx1sinxxdxxsinxd1sindydx1sinxxdx1sinxxdx1sin 1x2y2dxdyDy0yxx2y21D0 ：1）D 0 2）1x2y21cos2sin212；dxdyd13）1x2y2dxdy12dd412d1 5）D

x2y2dxdy，其中Dxyx2y22yx0

0：1）x2y22y22sin02sinD

02x2cos2sinx2cos2sin3）

x2y2dxdydd2

2d 1 32 30

2sin3382sin2sind3

21cos2 0xdv，其中x2yz1110z1x2

【解：1） 0y1

；2）xdvx

1x2 22 0x

D

dzdyx2zdv，其中yx2z0zyy1D0z【解：1）x2y

x2zdv

1x21

yzdzdy1x

D

1fxafxdx

fabx ab

afabx

bftdabtbftdtaftdt

fx2、設(shè)fx為連續(xù)函數(shù)，證明

n11ft1dx1

t2 tn

tt

n11ft1dx

nft1dt1 nfudu1

t2 t

t

n 131

12dx1

12x1 11

1 t 1 1 1【證明：x1 12dt1122dt 2dt 1 1tx1t 41xm1xndx1xn1xm 【證明1xm1xndx1xt01tmtnd1t01tmtndt1xn1 5、證明：2sinnxdx 2cosn

2sinnxdx 2cosnxdx2 0cosntdt cosntdt2cosntdt2cosn 2a6、設(shè)fx為連續(xù)函數(shù)，證明a

x3f

1

xfxdx，并求

2x3sinx2

2 ax3fx2dx

x2xfx2dx

1

x2fx2

x2

tftdt1

xfxdx1a2a2 2 2 21a2a2 1 102x3sinx2dx0

2xsinxdx0 0

2xdcosx013、 efxdxeex2fx1y yy

y2xDx10y 0x112）01

yeyfxdx

0

1eyfxdx

1f

1eydy0x1fxey1dx1eex2f0x002001yx22x3yx30A3x3x22x02y2x4x2y40A

24xy2 3y24x及其在點(diǎn)12y：1）y24x2yyy2x1yx02）A1x32x0繞X

1x32dx004y1x2及其在點(diǎn)10y00：1）y1x2y2xy02x1