微积分习题答案第七章定积分

1．（1）

n(n1)n2i2212(x2)dxlim(2)lim4(n)60ni1nnnnn

（2）

2．

10exdxlimenn1nin11e(1e)lime11nnn1en

1nsinxdxss0(1)

（2）cosxdx2(s1s2)2cosxdx0

 （3）

（4）3. s=

2024xdx3xdxs22r24

012662

nni1ba(x1)dxlim[(a2(ba)i2ba)1]nn

n(n1)(ba)2n(n1)(2n1)balim[an2(ba)an]2n6n n2=n13(ba3)ba=3

练习7.2

1xdxx3dxx(0,1),xx01. (1) 0

2xx2x(0,1)xxee(2) 

2321e01xdxedx001x2

(3)设x= -t 则

sinxdx220(sint)dt2sinxdx0x(0,)2 -sinx

sinxdx2sinxdx200

x(4)设f(x)=e1x x(0,1)

xxf(x)f(0)0f(x)e10即ex1 推得

所以 （5）2e0221xdx(x1)dx01

ex2cosxdx tx

20e(t)cos2(t)d(t)2

=0

22x(0,)x(x)xx ，

22x(x) x(x) ee

222excos2xdxexcos2xdx22e(t)cos2tdte(x)cos2xdx1nnx1x113.(1)在[0,]上连续，由积分中值定理知：lim2dxlim(0,)当n时，由n01xn11x222n

0

所以lim12n0xndx01x

2. (1) 0(2)

ex2cosxdxexcos2xdxx2x222

111(2x1)0.x.Qf(0)1,f()e4,f(2)e2.2.2设f(x)e2,x[0,2].f(x)e1414x2xMe.me.2e2ee220x2xex01422xdx2e2dx2e. (3)

设f(x)xarctanx,x[M3arctan33x3,3].Qf(x)arctanx0.f(x)在[,3]上单调递增。331x23333，marctan3331833333312(3)3xarctanxdx(3).即，3xarctanxdx.183339333(4)

设，f(x)122sin2x,x(0,).f(x)0.2222121cosx(1cosx)1cosx1sinx2f(x)在[0,]单调递减Mf(0)2.mf()12212dx02121sin2x21xn1xnn1123.(1)在[0,]上连续，由积分中值定理知：limdxlim(0,)n01xn11x2221当n时，由(0,)得:n021nx所以lim2dx0n01x

(2)limsinxdxlimsin4n0nnn4(0,).limsin0lim4sinnxdx04nn0n

4.

f(x)在[a,b]上连续，则f(x)在[a,b]上能取得最大值M，最小值m,而mf(x)MQg(x)0mg(x)f(x)g(x)Mg(x)mg(x)dxg(x)f(x)dxMg(x)dxaaabbbmbag(x)f(x)dxbM由最值定理知(a,b)使f()ag(x)dx即bag(x)f(x)dxf()g(x)dxab

练习7.3

3F(x)sinx1． (1) （2） （3） （4） （5）

F(x)sinx2dtab

11x123x211x8xaF(x)11x8xa2xx2x11x123x2

F(x)xdtf(t)dtaF(x)xF(x)xdtxf(x)2xaf(x)xaF(x)x2xaf(t)dtF(x)2xf(t)dtx2f(x)

（6） F(x)0

2.(1)lim0cost2dt0cosx2x0x0limx0112

lim11t2dtx4lim1x42xx4x3lim1x41 (2)

xx2x22

[(et2lim1)dt]20(et21)dt(ex201)x002 (3)

0limx0ex00etdt

2(x313.(1)

1x3)dx(14x412x2)21148 (2)

2121(xx)2dx1(x22112125x2)dx(3x2xx)146 1112dx2 (3)

21x2arcsin123

02x42x21 (4)

1x21dx01(2x21x21)dx(23x3arctanx)01234 111d(1x2)1 (5) 2xdx01x22201x21x202132

1 (6) e2exlnxdxe21elnxdlnxlnlnxe2eln2

4.

Q11111110f(x)dx0(x20f(t)dt)00xdx20f(t)dt0dxx2120210f(x)dx1.

0f(x)dx1x212012110f(t)dtf(x)x220f(t)dtx1.练习7.4 1．（1）

2520cosxsinxdx20(1sin2)4sin2xdsinx (1sin7x2sin5x1sin328753x)0105

（2）



 （3）

40tand4tan(sec21)d0314tandtan4tandtan200240lncos401(1ln2)`2

1cos2x11322cosxdxdx(xsin2x)6622268622（4）

e01arctanx11arctanxarctanxdxedarctanxe01x210e41（5）

1|1x|dx(1x)dx(x1)dx(1x)2112 131311

311(x1)224（6） （7）



20|sinx|dx2sinxdx2cosx040

40sinxcosxd(sinxcosx)dx402(sinxcosx)23sin2x1arctansinxcosx4022 11dxdx111109x26x10(3x1)233x104(8) 11dxdx1111

09x26x10(3x1)233x104(9)

22(arctan)242

20x2312233321xdx1xdx(1x)30932052920x21221xdx1x3dx3(1x3)2309332052 9(10)

dx221cosx202dxx2cos222sec20xdx2

300(11)

22cosxcosxdx22cosxsinxdx2234cosxdcosxcos2x32043

(12) 2. (1)

42121dxx11822()dx[lnxln(1x)]ln1xx31xx21225

1212dx122tx2tdt2(1)dt2[tln(t1)]12(1ln)11t11t3 1x41212dx12 2tx2tdt2(1)dt2[tln(t1)]12(1ln)11t11t3(2) 1x (3)

设tex1则xln(t21)1xdx54x1t54x (4)

ln5022(t1)t2exex12t4t2dxdt2(1)dt2(t2arctan)040t24t210t442ex3设x5sint,则dx5costdt4dxx25x23453arcsin5arcsinarcsin445costdt1arcsin511arcsin53dt3csctdtarcsin5sint5cost55arcsin55sint1ln|csctcott|5453arcsin513ln52 (5)

设xsect,则dxsecttantdt

 (6)

21x21tantdx3secttantdt3tan2tdt3(sec2t1)dt(tantt)0330sect00x3设xtant,则dxsec2tdt(7)

3dxx212sect13dt22tantsectsint1x24342233

设xtant,则dxsec2tdtx2tan2t1cos2t1112444sin2tdt4dxsectdtdt(tsint)(2)00(1x2)20sec4t0022281 (8)

3.(1) (2)

x3sin2x3x41dx0

310xexdxxdexxex011011exdxex0e1012e

e1(3)

e1e11212e212exlnxdxlnxdxxlnx1xdxx1122224e12(e1)4

34xdx33xcsc2xdx3xdcotxxcotx4sin2x441313()ln492234133cotxdx()lnsinx494 (4)

1212xsin2xdxxdcos2xxcos2x002220112cos2xdxsin2x2044204(5)

(6)

4lnx11dxlnxd2x2xlnx2xdx8ln24x11xx4414121212418ln24

sin(lnx)dxxsin(lnx)1e1ee1ee1e1xcos(lnx)dxesin1cos(lnx)dx1xeesin1[xcos(lnx)]sin(lnx)dx(esin1ecos11)sin(lnx)dx11(7)

e1sin(lnx)dx1(esin1ecos11)2

10(arcsinx)dxx(arcsinx)21210xd(arcsinx)201(arcsin1)22xarcsinx1x1020dx(arcsin1)22arcsinxd1x2(arcsin1)22(1x2arcsinx10dx01(arcsin1)22242

(8) 4.(1)

e1e0lnxdx1lnxdxlnxdx(xlnxx)11(xlnxx)12(e1)e1e1e

2a0f(x)dxf(x)dx02aaa2aa0f(x)dxa0Qf(x)dxt2axf(2at)dtf(2ax)dxa2a0f(x)dx[f(2ax)f(x)]dx0a

(2)

0xf(sinx)dxtx00(t)f(sin(t))dxxf(sinx)dx002xf(sinx)dxf(sinx)dx即2xf(sinx)dx020f(sinx)dx5.

Qf(x)sinxdxsinxdf(x)f(x)sinx00

0cosxf(x)dx0cosxdf(x)00f(x)cosx0sinxf(x)dxf()f(0)sinxf(x)dx00 6.

10sinxf(x)dxsinxf(x)dxf(x)f(0)5f()5f(0)2011212111111xdf(2x)xf(2x)f(2x)2xdxf(2)xdf(2x)000002222211111(xf(2x)1f(2x)dx)(f(2)f(2x)d2x)000222112111(f(2)f(t)dt)()0220222x2f(2x)dx

练习7.5

1ax1axedxe00aa 1. (1)

dx131x14133 x(2) dx1(n1)(x2)0(x2)nn1(3)

01(n1)2n1

dxdxx22x21x12arctan(x1)(4).

22

1xdx1x10lim1x111x11x00dx)dxlim00(1x324121lim((1x)21x)003 (5). 032222exex3x32dxlim[(e1)3](e21)310x1dxlim0022x33(e1)(e1)(6).

(7)

adxa2x2100lim0adxa2x210limarcsin00xaa0lim(arcsin011a0)a2

(lnx)dxlim(lnx)dxlim(xsin(lnx)cos(lnx)dx)0lim[(ln)(xcos(lnx)sin(lnx)dx)110lim[sin(ln)cos(ln)1]sin(lnx)dx0101(8)

2.(1)

11(lnx)dx[sinlncosln1)022bln2

2dlnxdtbdtdx1k1tlnxlimlimtln2tkbln2tkbk1x(ln)k2(lnx)k1,k1{(k1)(ln2)k1发散，k1当k1时，dtlntln2tln2发散。dx1收敛于，kk12x(lnx)(k1)(ln2)dx当k1时，2x(lnx)k发散。bdxbln(bx)a发散a(bx) (2) 当k=1时

所以当k1时，bd(bx)dxlima(bx)k0a(bx)k1k1b发散，k1lim(bx){a1,k10k1(k1)(ba)k1b

当k1时

dxa(bx)k1时所以 当k 发散，

bdx11k收敛于(ba)kak1 当k1时 (bx)

b(8)7!353．（1）2(4)(5)23!4!2

135135111()()()()()()()()22222222275553(4)()(4)•()3!2222 （2） 333()(4)()(4)()3!3322(,4)22311975332(4)()()3152222222 （3） 755311311()()()()7522223(,)2222212225!265()2（4）

（5）0(6)

x4exdx(5)4!24



0xe42x2521214t2115131132dxt2tedt2t2etdt()()4028208228222dx11(3)()62!()18223753115()()22222（7）

011x13216sin5tcostdt1623122xsint2sintcos2tdt3(3,)00cost21

121e(xu)222dxt2xu1220et2dt24． 习题 七 1．（1）A (2)D (3)B (4)C (5)C (6)B (7)A (8)C (9)C (10)A

(11)A (12)B (13)C (14)D (15)D (16)D (17)C (18)B (19)B (20)D

et2dtetdt12xxF'(x)2ee, x(,) 2.

1令F'(x)0得xln22

11ln2ln2122F''(ln2)2ee0xxF''(x)2ee2

1xln2是F(x)的极小值点。2 xx2xx1sinxdx(sincos)dxsincosdx00022223. (1)



20xxxx(cossin)dx(sincos)dx22222

xxxx2(sincos)22(cossin)024(21)22022

 (2)

20111sinxdx6(sinx)dx2(sinx)dx022261123(xcosx)02(cosx)1226212(x2)x2

eex2x2f(x2)221(x2)x2 (3) =x4x5 x2

31f(x2)dx(x24x5)dxex2dx12231712(x32x25x)1ex23233e

33111x(1x4)2dx(1x4)2dx2x2sint20 (4) 0 1215134costdt(,)42232 2011222x2cos2xdx2x2dsin2xxsin2x022xsin2xdx00202 (5)

121222xdcos2x2xcos2x2cos2xdx004044 = 1112x212x22113x2x21xedxxde(xedx)ee020222 (6) 0

(7)

10ln(x1)dx(xlnx1)x01011x12x1dx

=

ln211x11x11111dxln2dxln2dx00021x22x1x1

11x11dxln2(xln(x1))100222 x1 =ln2-11xexx1111ede1221dxdxln01ex0exe2x01x211110(e)2ex24222 (8)

ee1ln21ln2 =lne12sinxsinx(1sinx)sinxsinx4444dxdxdxdx22201sinx000cosxcosxcosx (9)

 = =

1cosx401cos2x4dx214(sec2x1)dx200cosx

21(tanxx)042204

4(10)

20sinxcosxdx4(cosxsinx)dx2(sinxcosx)dx

2 (sinxcosx)04(cosxsinx)2(21) 4xx1444dxdxxdtanx01cos2x02cos2x02 (11)

1111 (xtanx0404tanxdx)lncosx04ln2

282844． 证明： F(x)ftdtxf(x)2xf(x)f(t)dtxf(x)

00x =0 =0f(t)dxf(x)dx[f(t)f(x)]dx00xx

xf'()(tx)dx  f'(x)0 t由定积分的性质知 F'(x)0,x(0,)

）内单调递增。 F(x)在（0，

(t,x)

5. 证明：

设F(x)f(t)dt0xx0dtf(t) x[a,b]

1f(x)

x(a,b)时，f(x)0

F'(x)0,F(x)在（a,b)内单调递增

bdtbdtF(a)0af(t)af(t) 又

F'(x)f(x)F(b)f(t)dt0ab由单调性与介值定理知：

dx0在（a,b)内有且仅有一个实根。af(t)

ab6.证明：由积分中值定理知：存在一点1（,b)2bbaba使a+bf(x)dx=f(1)222即f(1)1F(x)xF(a),F(b)异号。

因为f(x)在[a,1]上连续，在（a,1)内可导，且f(a)f(1)1由罗尔定理得：存在（a,1)(a,b)使f'()0

f(xh)f(xh)f(xh)f(x)f(xh)f(x)7.limlim[]h0h0hhhf'(x)f'(x)2f'(x)2(3x22x1)

1lnx11118.(1)dx(1lnx)d()(1lnx)(222xx)dx2x2xx1ln2112ln22x213x1dx1221ln4(2)dxln22x2x212931323(x)2x24222

dx1(3).2dx2arctanx020001x2xxxarctanx1111(4).dxarctanxd()(arctanx)dx122111xxxx1x1xx1()dxlnln212214x1x4421x

(5)Q1212dxx21x2dx20dxx21x26012120dxx21x2120x1x602xsintcostdt6csc2tdt20sintcostcott1212发散发散dxdxx2+1x2（6）.(x-1)4x22x设x1sect，则33dx(x1)4(x1)21

dx(x1)43secttant32dt2costdt2dt2(1sin2t)dsint4x22x3secttant333231(sintsin3t)3233381yx9.(1).e交点（e,1)ylnx1es=(eyey)dy1024128(2).S[y8(4)]dy82314(3).S2(2yy)dy03(4).y'exy'x1

eyex在点（1，e)处的切线方程yexe1eS(ylny)dy0e20(5).S4(cosxsinx)dx(sinxcosx)dx22412y=x(6)。由得交点（2，2）（2，2）222xy8214S1(8x2x2)dx2223

S2r2S1643

20220210.(1)Vxydxsinxdx(2)Vx0124Vy[0124arcsin2y]dy2ydxe2xdx0422133(3)Vx[xx]dxVy[y4y]dy00101012(4)Vx4sinxdx2cos2xdx()0424(5)Vx(11x)dx01221011332dx2x621211.f(x)ex2x2xex,（f-1)=etdt0100100411f(x)x2dx（f-1)-x3exdx1-1224401212xf(x)dxf(x)dxxf(x)11221x4011e1(1)44e12.Qf(x)在[a,b]上可微f(x)在[a,b]上可导,且f(x)在[x,a]内可导, x(a,b)f(x)f()(xa)M13.(1)1(xa)22ba(a,x) 又Qf(x)dxabbaf()(xa)dxM(xa)dxabf(x)dxM0aa0aaaf(x)g(x)dxf(x)g(x)dxf(x)g(x)dxaaa000Qf(x)g(x)dxxtf(t)g(t)dtf(t)g(t)dtf(x)g(x)dxa0af(x)g(x)dx[f(x)f(x)]g(x)dxAg(x)dx00aa(2)g(x)sinxf(x)arctanexxx设F(x)f(x)f(x)arctanearctaneexexF(x)0F(0)1e2x1e2x220f(x)f(x)22212sinxarctanedx1x[0,]2x22sinxdx214.设f(x)(2xx1)f(x)4x12(2x2x1)3令f(x)0得x1Qf(x)141214f()471f()1212141dx14M,m12720(2x2x1)715.(1)设f(x)1x4x[1,1]14f(x)2x31x14令f(x)0得x04121f(1)22f(0)1816.方程两边对x求导得1当x1时f(2)5f(x2x3)(2x23x2)117.方程两边对x求导得：eyy2dxy2e-ysinx2dy2sinx2x0x2

18(1)121411arcsinx15222dx12arcsinxdarcsinx(arcsinx)12144x(1x)44(2)(3)2aax2a21dxxasectx4a21edxy1e32-2x30tan2t1dtsec3ta232133sintdsintsint203a23038a2320ln2-2x003t2111t2dt(1)dt(tln)01t21t221tln(23)(4)设xtant,则2sin(t)1ln(1x)costsint444444dxln(1tant)dtlndtlndtln2dtlnsin(t)01x200000costcost440lnsin(t)dttu4440sin(u)d(u)4lncostdt02原式4ln2dt08ln230(5)arcsin03xxdxxarcsin1x1x303x11xdxdx(xarctanx)0x12x(1x)3043xxe(cosxsinx)sinx4422(6)dxecosxdxedxcosxcosx4444x2Q4e4x2xxsinxdx24e2d原始cosx2e2cosxcosx4x2424e4x2cosxdx原式2e1cosx4448(e8e8)1011ln(1x)1ln(1x)(7)dxln(1x)d00(2x)22x2x1111111dxln2()dxln2(l02x1x302x1x3n100(8)Qcos(n1)xcosnxcosxsinnxsinx原式sinxcosnxcosxdx2sinn1xsinnxsinxdx

1nQsinsinxsinnxndxsinnxsinnxdx原式0000n02101101ex19.f(x1)dxtx1f(t)dtdxdxdxln(1x)10xxx0111e01x1(1e)en11xcosnxdsinxsinnxcosnxn11exx1(ex1ex)deln2ln(1ex)001ln2ln(1e)330(t)sinttsintxsin3xsin3t20.(1)xtdtdtdt01cos2x1cos2t01cos2t01cos2t322xsinx1cost2(1cost)sin3t2dxdtdcostdcost01cos2x01cos2t01cos2t01cos2t2(2arctancostcost)02

(2)Q2010101010sin10xcos10xcostsintcosxsinxdxxt2d(t)2dx04sintcost04sinxcosx4sinxcosx210101010sintcostsintcostcos10tsin10t2222dtdtdt004sintcost04sintcost04sintcost1010costsint2dt004sintcost0f(sinx)f(cost)f(cost)(3)Q2dxxtdt2dt0f(cosx)f(sinx)02f(cost)f(sint)2f(cost)f(sint)220f(sint)f(sint)f(cost)2dt=dt=2dt0f(cost)f(sint)0f(cost)f(sint)220f(sint)dt=f(cost)f(sint)4xexxex1xdxdxxd(0(1ex)201ex1ex)(1ex)200 21.(1)001dx1ex1ex1exx0(1ex)exdx0(ex1ex)deln(1ex)(2)x1,x5为瑕点ln2

105151dxlim(x1)(5x)1001221dxx32limarcsin()2102x3201()25112lim(arcsin

20212arcsin2)22

+dlnxdx22.(1)当=1时,2xlnx2lnx发散。+++dx1(1)t1(1)t当1时,2xlnxtlnxln2tedtln2[(1)t]ed(1)t由函数可知：f1时，上式收敛。+p1时，上式发散。综上：+223.证明：badx当f1时，上式收敛。当1时，上式发散。xlnx

当f(x)dx0时，不等式显然成立.设f(x)dx0,abQf(x)在[a,b]上连续，x[a,b]f(x)在[a,b]上连续，于是存在一个，使得f()Mmaxf(x)由题设有，f(x)f(x)f(a)f(1)(xa)1(a,x)f(x)f(x)f(b)f(2)(xb)2(x,b)推得f(x)M(xa)ba2abf(x)M（bx)ba2abaf(x)dxbaf(x)dxbaf(x)dxM2ba(ba)2(xa)dxMba(bx)dxM42b又Qf(x)dxf(x)dxM1(ba)2xba(ba)2f(x)dxM(ba)2M4baf(x)dx24.证明：设F（x）（f(t)dt)(xa)f2(t)dtaa2xx）QF（2f(t)dtf(x)f2(t)dt(xa)f2(x)aaxx2f(t)f(x)dtf2(t)dtf2(x)dtaaaxxx[f(x)f(t)]2dt0axx[a,b]时F（x） 单调递减F（b）F（a）=0（f(t)dt)2(ba)f2(x)dxaaxb

11125.(1)ss1s2(axx)dx(x2ax)dxa3a0a323a21sa2s2aa12令s0得as(1(舍负）21)202122时，s1s2最小，最小值为：s1s226a10a(2)Vx[(ax)2x4]dx[x4(ax)2]dx2130