cosxdx

（2）2(x21)dx22(x2 （3）1x3dx （3）2xdx4 （1）該定積分的幾何意義如右圖所示陰影部分面積的代數(shù)和，且在(22范圍內(nèi)對(duì)稱，所1原式21x12在(1,1范圍內(nèi)關(guān)于軸對(duì)稱，所以等式兩邊相等．（1）1x2dx與1 （2）3x2dx與3 （2）

(lnx)2

（4）0

sinxdx與0

（1）由定積分的比較性可知在(13x2x3由定積分的比較性可知在(34范圍內(nèi)lnx(lnx)2a由定積分的比較性可知在(0范圍sinxx211151（1）1

e-x

5（2）4

(1sin2

（3）

（4）0

sinxdx（1）因?yàn)閑x2在[011，最小值為1e1dx1ex2dx1 e11ex2dx0（2）因?yàn)?sin2x在[,5

22，1 4

ldx4

(1sin2x)dx4

41f1

(1sin2x)dx4fx)fx)

1121

x4(1021x(1x4(109x)21x(1x0x9f(x在(0f(x在[010，最大值是2 100dx11111

dx02dx2ABABsinxADtanxACsinxxtan亦得到：1

sin

0x，從中cosxsinx 2cosxdx

sinxdx

1 1

sinxdx y122（1） xx2 （2）(122

2xx22BC 2（1）2所以原式12

該定積分的幾何意義是以(1,11x方．所以原式1212（1）f

tet

e（2）f(x)e

x3xf(x)

2tet

xf(x)x

(t3x3)sin （1）設(shè)tet2dtf(x)g(e)x2g(x2g(1)令x2f'(x)g'(m)m'x2ex42x（2）設(shè)ln(1t2)dtf(x)g(exg(x)令exf'(x)g'(m)m'g'(x)exln(xe2x)ln(xx2設(shè)et2dtx則f(x)g(x3)g(x)令x3m x12f'(x)g'(m)m'g'(n)n'ex63x212設(shè)(t3x3)sintf(x)g(xf'(x)g'(x)g'(0)

lim

xsin

lim

x1tx0x30 x0x1t

1x

11x21tx1t

0xlntlimx0x

11(xxt

1

t2x

dt

x

(tt x

xx et2 x0xsinx（1） x0x30lim11x0 3

lim x0x20lim

1x2x0x2 21tan2lim1tan2

1sin2tx0x30lim x(

x0x30lim xtanxsinxx0x30lim xx0x30

11t3x0x3 3（4）

x-0

11x21x2 x-0011x2

x0(ln

1

(1sin2t)txx0xx

lim(12x)1

xlnt11x1(xlim1x12xlim1ln

1limxx12x2

2x1 xt

1xt

lim

dt exln(

xlim1lnxet2

e1ex

001

x

x(arctant)2x(x

arctan)2 1x 1x11 0x 1(tt2)et2x2limx

x(tt2)et0xex

(xx2)ex 法則

ex

(12x2 F(x)在[abf(x)F(x)xf(t)dtx1（1）

bfFx)在[ab 設(shè)f(t)dtg(t)1dt ()()()()(bh) 又因?yàn)閒(t0Fx

)')'因?yàn)镕(x)在[ab]F(a)a1deb1debf afbF(b)af(t)dtb又因?yàn)镕(x)在區(qū)間[ab]上連續(xù)．所以在區(qū)間[ab內(nèi)緊有一個(gè)實(shí)根．f(x為連續(xù)函數(shù)，且存在常數(shù)aex-1xafx求f(x)及常數(shù)a．解：設(shè)f(e)deg(t)ex1xg(aex1xg'(a)g'(x)ff(x)1所以af(t)dtaex1dxxex1

aea1ex1 所以a

xx設(shè)x

(xtf(t)dt1cosx，說明

f(x)dx1 x x

t)(f'Qx0x0x

t)((00x

xt)dt0 即P(x)-P(0)1-Q(x)f(x)20202f(x)dxsin 0 8

e2(ln

113（1） （2）3

(exex

cosx

2x

4

44x1x

22ex2ln

04

（10）x6

tan

（11）6

tan2

（12） 1212（13）0

sin

（15）4

xcosxsinxdx(xsinx)2

333（1）3

8

32 x31 （2）1(exex)dx(exex)1 （3）

e2(ln

2edxe

(lnx)2d(lnx)

1(lnx)3

arcsinxdx

2arcsinxd11 1arcsin22

1221s22221s2

)520 200cosxdx2cosxdxcosxdxsin sinxx 2xdx0xdx2xdx1x221

0

0

41 4

12

2 1arctgx204

2 ex2ln

e21ln

2

dx1xdx

dx

1(e21)2

2e1(e26

36tanxdxln36

1ln26

tan2xdx6

1cos2

ldx

3(sec2x6(333(333 x（12）4(x1)2dx4x2x1dx4121x (x

4lnx11212（13）0

sinxdx

6(sinx)dx06

(sinx606026

21(xcosx)1

(cosx231 0

1sin2xdx

sinxcosxdx0

(cosxsinx)dx4

(sinx(sinx

(cosxsinx)x4024024242124xsin（15）4

xcosxsinxdx2(xsinx)2

(xsin

d(xsinx)

3 311（16）1max{1,x}dxldxxdxx1311

f(x)xt(lt)e2tdtxf(x0解：設(shè)

t(tl)e-2tdef(x)g(xfx)gx)x(x1)e2xx(xfx)在(,0(11)0，在(01f(x在(01,)上單調(diào)遞增，在(01所以當(dāng)x0f(x)x1f(x)設(shè)I1I2

2222

sinxcos21(sinx比較I1，I2，I3的大小

I3

2(sin5x2

cosx I1

21

cos2I2

2(sinxcos2 I 2(sin5x2

cosx cos2 cosxdx2 cosxdx2I3I1

ln 1x2a22a1x2a22a2

(1x2(1x21

33exe x1x1

（9）

3lnxx(13lnxx(1lnx)ln

（11）

sin9 x2

1

x 1x22xx（1）x

0

11cos2

darctan(cosx)021（2）1xln(t22xln3，時(shí)t2x0，時(shí)t2222

1t2xxx12[ln32ln(2e2ex1x1ln xsinxdx xsinxdx21ln 23（4）

1 x331x331x2x3(1x2x3(1x2)2

3(1x2) 03(1x2)12222

x2xx21x21 x338

3

a2

aa2a2

arcsin0a4 xtant3311 4

(sect)2tant dt

dcost1cos21

3d(cost)1

d24

34134

24

341341ln(1cost)

1ln(1cost)1ln(22ln(2

1ln3(2221ln2（８）令exexexe dx

t1te dtlnxx21et2t21ee2ee21令tlnxx3lnx3lnxe 2

dx636

(3t2) 令lnxt1與2x(1lnxx(1lnx)lne1 2

令tsin2x4 2sinxdsin 11sin2xsin

2dsin11sin2sin2xsin2x121221224212sin902(1cos2x)4d02(cos8x4cos66cos44cos2x1)d0x2sec3adxsecatan22x2x24

secatanasec2tan4

cosadasin

3334x11 1 1

arctan2212

2(41ln21 1x2e

1(xx)ex

x

0e（4）e

(lnx)3

e（5）ee

lnx

（6）0

xsin1（7）1

x

2ee

2

1e2x(4x e(x ln（10）ln

x3e

0

xsinx

e2sin(ln 11

1x2arcsin

xe （1）0(x2ex11ex0(e112xdxx0(e12e12ex10(3e12e12（2）0(xx)exdx1(xx)e 12xde0(2xex11ex (2e12ex10(2e12e12（3）=1e1ln(x2x1(x2ln(x1)e1e121x xx(lnx)3eex3lnx1 ee13lnee3(xlnxeex1 e

e(xlnx11xxdx)xlnx11xx [1(11)]e(e 0

(xcosx

0sinxx01121(x2arctanx11

01[(x1

1x22 011[1arctanx12 [ln

(x1(xe2lnx11(x

xe11（9）1(4x21[(4x3)e3x11e2x21(7e232

412

2x01[7e232(e22（10）1

1(5e22ln2x2de21[x2e2

ln2

ln22xe1[ln2eln2

ln2dex221[21[2

00ln

]1[2

21(ln24

xsin(xcosxcosxdx)(xcosx22 e2sinln2

dx

sinlnxd 1sinlnxe221coslnx1 1sinlnxe22coslnxd 1sinlnxe21coslnxe2

1sinlnx1x1(2

sinlnxx

coslnx)e1)1(1)e11

1x2arcsin11

(1)

2arcsinxd(1x2)3 2

(1x2)2arcsin3

(1x2)2(1x2)221x2)dx3 xexe（14）ee2e

lnxdx1 e 12lnxx

e

x2dx1 12lnxx

x2dxe2

4e2e0 dx2

221 2ee11186e111

1

3sin2xln(x1x23

cosx2（1）f(xsin2xlnx12f(x)sin2xln(x1x2f(x)f(x)sin2x[ln(x

1x2)ln(x

x2sin2xln(1x2x2)f(xf(x）在（-3，3）f(x)

(212

12f(x)

212

12f(x)f(x)

212

1

1)f(xf(x在（-1，1） 設(shè)5(5)dx4(4)dx4因?yàn)閒(x)cosxarccosxff(x120cosx1求0(x)dx，其中(xx1解：[0

xdx121

(1x)dx]1201 12120[2

(xx)12求[x]dx，其中[xx5 543213f(xg(x)在(f(x)3x22g(x)dx，g(x)x33x22f f(xg(x1解：令0f(x)dxC（C為常數(shù)）則g(x)x33Cx212g(x)dx(1x4Cx3) 4f(x)3x28C2f(x)dx(x38Cx4x)2 816C8Cf(x)g(x)3x24[f(x)g(x)]'6x-3x2x0xx0時(shí)f(xg(x)]minx2時(shí)f(xg(x)]maxx0arcsinarcsin1 1（1） 4

x

4ln(1tanarcsinarcsin1解：（1） 4arcsin1arcsin1x2 424

xxdx1x11x14（2）2

51x14（3）ln21x148f(x)xln(1t)dt(x0)f(xf1 1ln22

x(1x2

exe2

x2x

3

arctanx0x

0(1

xe

n20(1ex)2

0

e解：（1）1ln2（2）4（3）2ln3（4）221ln (nxa

2已知lim

4xe

dx，求常數(shù)axxa a0a1ln20

1 dx11

1 1-0(2-1-

（4）5 5 51-xx2（5）e （6） （7）arccos1-xx21 1 1（1）（2）（3）23223（1）yx2與y2x2 （2）y1與yx，x2x（3）yx2，4yx2，y1 （4）yx3，y2x（5）yex，yex與x （6）yx2，yx，y2x（7）yxlnx與yx，y0 （8）yx(x1)(x2)，y0ysinx，ycosx，x0，x2 （10）y212(x3)，y212(x3)（1）122(x1x3) 3（2）s2(x1 1 (xln 3ln2y（3）y

ty212t041t3 31 2

20 12022

(xx

x)(12)(2（5）exex0ee11x21(x21x3（6） 6 1x211x2e1x2(lnx1) 0 1 211(e21)[1e211(1 1e2 1(x33x22x)dx2(x33x20(14

x2)0

(14

x21111[(484)(11 44(sinxcos)4(sinx022（10）1(1y336y)3 yx24x3及其在點(diǎn)(03和點(diǎn)(30)gx24x3y1y1023-t(0,3處切線的斜率k12x4x04y4xt(3,0處切線的斜率k22x4x32y2x1S0

4x3(x24x3)dx

2x6(x24x3)104

x26dx

(x26xylnxxylnxy1x二曲線的交點(diǎn)為(eeyy01x從而面積S11xdxe1xln0 111x2111x2eelne e 11e2xlnxe(ee 1e(e)(e1)1e2ylnx上的點(diǎn)(e,1）xylnxyex

1x軸的交點(diǎn)為

e2e

,0)從而面eSe

lnxdx

ee

(exe2 1y4xx2、yyb(b0y4xx2yb所圍圖形面積的一半，求b的值．44解：設(shè)yb，與拋物線的兩個(gè)交點(diǎn)的x軸坐標(biāo)是a12 ，a44

(b4x

s1

4b(b4xx241x32x2bx2 1x32x2bx4 1a32a2

1

23 s112所以1a32a2

1a32a23 3 14[(44b)2(44b)]3b3yx2x[01]間t取何值時(shí)，圖6-29S1S2解vyv2ylnx(1xe)ss1v2

圖6-

t(t2x20(t2x1x3)

s1(x2t2t11 (xt

1t22 s4t3t2 s'4t22t2t(2t所以s

(0

1單調(diào)遞減，在

,

s1s2當(dāng)t0s1t1s s1s e y=-求由拋物線y2)2x1y03xxy24yxf(y，則fy2yy03的切線斜率為k2x2y4s3y24y52y03y26y01y36

y29y092727（1）xya2y0xax2a(a0x

ysinx0x

x2

y0xyyx3xy2xyex2x0x1yy5xy6xxx2y24x24y1y0xycosx,x0,x,y2a（1）va(

2a1aa41x12xvx20

sin22020(sinx

2cos200

(1sin20

ldx0

sin2因?yàn)?0x

sin2xdx2所以v4yvyv2v2y12v1ysinx(0xy2 1v2

1 1v1(arcsinx)210134vyv2v1vv2v2xy2(0x1xv1yx2(0x1xv21xdx1x21 v1x6dx1x7 所以vv2v15vxee(lnx)21[x(lnx)2eexx2lnx1 ee[e2(xlnxeex1 [e2(e(e(evyv2v2是以e1yv1ylnx(1xeyv2

v1(ey)21011e201e2y e2 所以vv2v1(e22 v1ldy

lne1[ylny

1ydy]1e1[e1(1e12e156xx11x2xvv2v2xy6(1x5xv1y5(1x5xx5v5(6x)2 v55 ()5 155(x212x1

25

1dx(13

1212

36x)1

1x1241x所以vv2v1643

v1x24y1)(2x2)x2 2v1 22(1x2x4 2(x

1x3

1x5232

16 v22v242332所以v128(8)ycosxx0xy0yy1π0x-y12V20xcosx 20

xcosxdx22

xcos2(xsinxcos2

2(xsinxcosx)220yx2(x0)Ax1AAxx設(shè)切點(diǎn)A(x0y0直線斜率k2x0直線方程：y2x0(xx0y0x軸交點(diǎn)為

y000

xdx

y

(2x0x2x0y0)dx

且有y0x0解得x0y01從而A（1，1y2x1x軸交點(diǎn)1021V22

xx2dx2

x(x2(2xA0yx22xy0，x1，x3S解ss12(x22x)dx3(x2 (x21x3)2(1x3x2 2 v(v4v3)(v2v43y3v2yx22x(1x2)y軸（x軸的下方）v432

v1v3

同理得：v 31333

1y)2

1710(y1

1 t則yt1y3ty0時(shí)t所以原式1196v(v4v3)(v2v1))x軸旋轉(zhuǎn)一周產(chǎn)生的兩個(gè)旋轉(zhuǎn)體體積之比．解：如右圖所示，即要證S1S2y0y0B23ya(x1)(x310a(x1)(x3)dx3a(x1)(x 左邊1ax24ax01ax32ax2 1a2a3a4 右邊3(ax24ax11ax32ax23ax 27a18a9a1a2a 43（2）31V12xa(x1)(x3)dxV220xa(x1)(x3)31V1 ylnxxxy軸yy101e exx軸旋轉(zhuǎn)所得的體積為VxVx1(1x)2dxe(1xlnx)20 1(223y軸旋轉(zhuǎn)所得的體積為V2Vy21x1xdx2ex(1xln

e ) e 從而Vx(2

e)，Vy yax(0a1yx2所圍圖形的面積為S1x1S2試確定a的值，使S1S2S1a(axx20S21(x2aS1S2a(axx2)dx1(x2 1ax21x3a1x31ax2 0 aS1S2求得可的：a21a2

0(S1S2min22622V

(axx2)2dx

(x2ax)21(axx2)2021yy0a12000萬元建成一條生產(chǎn)線，投產(chǎn)后，在t c(t)52t3（百萬元/年R(t)17t3（百萬元/年R'(t)-c'(t)23t3可解得tR'(t)-C'(t)-

2t33故而t8

(123t3)dt1001840(e06解：總計(jì)利息為(380e06501064.75萬元MC2x（萬元/臺(tái)）x表示產(chǎn)量，固定成本C022（萬元，邊際收益MR204x（萬元/臺(tái)求：XXC2200(2XX 2x2

xR0(20x20x-2x

2 x2C20x

218x3x22MR2x204xx6'18-"3x6x6此時(shí)324824 （Ａ．Ａ．

B．1

C．

D．1

1xlne

0f(x是[ab上的連續(xù)函數(shù),則下列論斷不正確的是xaf(x)dxf(xxbxf(x)dx是f(xbbaf(x)dxf(xbf(x在[abf(xF(x)f(x的原函數(shù)，則f(xF(x)f(xF(x)f(xF(x)f(xF(x)設(shè)在區(qū)間[abf(x)0fx0f"(x)0則下列不等式成立的是(ba)f(a)bf(x)dx(ba)f(a)f (ba)f(b)bf(x)dx(ba)f(a)f (ba)f(a)f(b)bf(x)dx(ba)f (ba)f(a)f(b)bf(x)dx(ba)f xxf(x在(內(nèi)為連續(xù)可導(dǎo)的奇函數(shù)，則下列函數(shù)中為奇函數(shù)的是xxsinf'

0sinxf

0f(sin

xxxf(x在(x0F(x)x0f(t)dt，則下列結(jié)論正確（CF"(x)不存 B．F"(x)存在且F"(x)C．F"(x)存在且F"(x)2f D．F"(x)存在且F"(x)f(xf(0)0f00x0F(x)x(sin2xsin2t)f0xk為同階無窮的，則k f(x)

xetdx

et2dt1f(x為（Ｂ． A．正常 F(x)x2esinxsinxdx則F(x)為x正常 解：1（1）AA2dxarctanx

arctan211 BlnxCln(lnxDx1baf(x)dxf(xbf(x)0,f'(x)0,f(x)f(x(ba)f(a)bf(x)dx(ba)f(a)f xF(x0sinxf F(x)0sin(x)f(t)dtsinx0f由（3）A知

f(t)dt0fxxxF(x)sinx0f(t)dtFxsin2xf(t)dtxsin2tflimF(x)lim x0 lim

0fxkx

lim

f lim2

f kk

kxx2

k(k2

k(k2)(kf(x)

etdt0

etxxetdt

xet2dt12 2x2x0f設(shè)f(x)具有一階連續(xù)導(dǎo)數(shù)，且f(0)0，f'(0)0，求 x0x2

f0x2

fxf

lim x02x

2xf(x2)f(t)dtx2f0

f'(x2)3f(x)

f

f'(x0 f'(x 4f'(x2

x0 x04f求證方lnxx

1cos2xdx在(0) 解：證明f(xlnxx

1 f'(x)1 x(0e)時(shí)f'(xx(e時(shí)f'(xx0f(x) 2f(e)2得

1cos2xdx

2cosx 0f(xf(0)0F(x)

xtn1f(xntn)dtlimF(x) x0limF(x)

1nxn1f(xnn

f(xnx0

x0 nxn1f'(xn

f'(xn

f'

2n2

x0xu

2xx

2xarctan(1x1cosxxsin

x02sinxx

2arctan(1x2)2

22 f(x連續(xù),且xtf(2xt)dt1arctanx2；已知f(11，求2f(x)dx的值 解：令2xtx

t2x 2x(2xu)f(u)du

f(u)du

f f(u)du2xf(2x)2xf(x)2xf(2x)xfxx1則22f(u)duf(1)

xf(u)duxf(x)

1 2f(x)dx1x(11) f(x在[0f(0)0n1n0F(x) 0

f(t)dt

xx在[0上連續(xù)且單調(diào)不減（其中n0解：證明limF(x)lim1xtnf(t)dtlimxnf(x)

x0x

x0F(x)在[1]F'(x) xtnf(t)dtxn1fx21

1arcsin(1（1）11ex1

（2） 2x2x33(15x2)1(15x2)13

110（1）0

xdx

x111 0111 xedx1

01 01311ex 3xxdx x01 1（2）12

arcsin(1x)dx11

1 2arcsin(1x)12(6)72x (15tan2t)1(15tan2t)1tan20

(15tan2)

1

d2sin 14sin2

2

1(2sinxsin

1arctan(2sint 60 603sin2t3 6

dt6

1cos2t21(3

3) 4 f(x在[02f(xf(2x)0x f (2xx2)dx0f(x)f(22f(x)(2x

f(2I

f(x)f(2x)dx

x(2f(2x)f2I2f(x)f(2x)(2xx20f(x)f(222xx2dx48 I3f(x)xsintdt，計(jì)算f(x)dx0 sintdtdxsintdtcost110 ef(x)axt(2at)dt，求af(x)dxe adtatet aet(2at)(a01ea2ae(ta)2d(t212

2(ta)20

12

f(xf(x

xf(xt)etdt0解：令xt原式可化為xf(u)exudusin0exxf(u)eudusin0exxf(u)euduexf(x)ex0cosxf(x)exxf0sinxf'(x)exxf(u)euduexf(x)e0sinxf'(x)cosxf'(x)sinxcosxf(x)cosxsin（1）3

1cos2

(xsinx

11

（2）2

1

sin4

xcosxsincosxsin4（1）

1cos2

ln11

cos23

ln1xdx01x cos23

xsin33

13333xsec ln333342ln(23)（2）2

1

sin4xdx2

sin4xdx2

11

sin420

sin4xdx2

11

sin42sin4xdx

sin2x(1cos2

1

1 cos2x)

1

(1cos2x)dx1

(1cos22 412

1sin2x2 0

1 2422

(14

14

1sin4x2 0

3又e 又e2 12

sin4xdx2

11

sin43e e（3） 4

cosxdx4

xe2sinx e2sinxdx2 cosxcosx

xe2x2(e

1

e24

448(e8e8 3原積分48e8e83f(xaxb

x0，求常數(shù)a,b1f(x)dx解：f(xaxb44ab

2xab3f(x)dx3

a

3bx3

43x 1 4a2b43

328ab43

31)

24333

3224a2b時(shí)，即a1,b22

1f(x)dxf(x)xet2tdt，求1(x1)2f(x)dx 1et22tdt1(x1)2 0

061e1e(t1)2(t1)2d(t61

exsin5

esin（1）

2

（2）

sin

cosx01cos （1）xsin3 (u)sin3I01cos2xdx 1cos2u

sin32

du

usin3201cos 01cos

sin32

dx

xsin3201cos 01cossin3

xsin3

x2I01cos2xdx01cos2xdx01cos2xsin3

sin201cos2xdx01cos2d2(1cos2 1cos2

d2arctan(cosx)cosx2 I2I2（2）x2e e ecost

0sintdt0

ecosecosx

sinx1

dx2 fx)arctan(x-1)2f(0)0．求1f(x)dx0Ixf(x)11xf f(L)1xarctan(x1)20f(l)1f'(x)dxf(0)1arctan(x1)2 I1(1x)arctan(x1)2dx1(x1)arcsin(x1)2 11arctan(x1)2d(x2令uxI1121 11 uarctan 02

112

1ln1u21

1ln4

(x1)4x2(x1)4x20

arctanx

1ex

e3(1x)(1x)2e

cosbxdx(a0 x1（1）

2b0(1x)(1xx2

b0(1

23x2x2b(lnxb(lnx1

1bd(x2x

1b

6 (xx2(xx2x2x13

x3

20(x1)22

32b

(6

239令secx

sec 2(x1)4(x1)2(x1)4(x1)2132cos2d3

2(1sin2)d3sin21sin3

23233 23x

dtanb

3(2tan2t)b

arctanbtsec23

dtant

tcostdtan

sec

b

(tsintarctanbcost1

b1eexe3exlim1

x

11(eee

e2 1()e2

climeaxcceaxcosbxdx1eaxcosbxcbceaxsin1

b

0ac

ca cosbx0a2

sinbx0a20

ceaxcosbxdx1(1eaxcosbxcbeaxsinbx 1

0 I

a 2

cosbxc0

b2

sinbxca a2

（1）0

（2）0

ln1x2

（1）令tx則

lnsinxdx0

lnsin(x

)dx

2lnsinxdx

lncosxdx2A0

(lnsin2xlnln21lnsin 2 1 2

2

2

ln212 2lnsinxdxAln2 （2）同2lncosxdxAln2 1 01111

11

limx

x(1

(1x2

1111

1

2

0

1x2limxlimx

1

21

11111x1lim

1

1

1x 1

1x1xx1x 1 1 1 0ln(1x)dx0ln(11ln(1x)dx2lntdttlnt22dt2ln2 1ln(1x)dx1lntdttlnt11dt 1又limtlntlimlntlim t

t0t

t011

1

dx2ln2112(1lnyx23x2V21x(x1)(x2)222xx23x21 8

111 yx21上的一點(diǎn)，使該點(diǎn)處的切線與所給曲線及兩坐標(biāo)軸所圍S01(x21)dx(1x3x)1110設(shè)切點(diǎn)(x01x02

直線l:y2x0(xx01x022x0x1S11(1x02)(1