考研数学高数习题集及其答案9021

1函数、极限、连续解.11x21,0x22,|x|21x2．设limatetdt,则a=________.axxxa解.可得eaatetdt=(tetet)aeaea,所以a=2.12nnnn1nn2n3.limn=________.n2221nnn2nnnnnn解.n22212nnnn12nn1n1nn2nn1nn1n<<n22222212n12nnnn12n<n1nn2nn1所以<n2nnn2222n(1n)12n12n2nn2,(n)nnn2n(1n)12n12nn1n2n12,(n)2121nnnn1nn2nlimn所以=2n2221|x|1|x|1,则f[f(x)]_______.4.已知函数f(x)0解.f[f(x)]=1.5.lim(n3nnn)=_______.n解.lim(n3nnn)lim(n3nnn)(n3nnn)3nnnnnnn3nnnn3nnn=limn2x0时,f(x)e1ax1bx为x的3阶无穷小,则a_____,b______.6.设当xex1ax1bxlimexbxex1axlimexbxex1ax解.klimx03x3x0x0limexbexbxexa(1)3x2x0limex2bexbxex(2)6x2x0由(1):lim(exbexbxexa)1ba0x0由(2):lim(ex2bexbxex)12b0x0b1,a122117.limcotxx0=______.sinxxcosxxsinxlimxsinxlim1cosxlimsinx1limx0解.sinxxsinxx33x26x6x0x0x0n1990A(0),则A=______,k=_______.limn8.已知nk(n1)kn1990n1990Alim解.limnnk(n1)knknk11A，A11所以k－1=1990,k=1991;kk1991二.选择题1.设f(x)和(x)在(－,+)内有定义,f(x)为连续函数(a)[f(x)]必有间断点(b)[(x)]2必有间断点(c)f[(x)]必有间断点(d)1|x|1,且f(x)0,(x)有间断点,则(x)f(x)必有间断点(x)解.(a)反例0|x|1,f(x)=1,则[f(x)]=111|x|1|x|1,[(x)]2=1(x)(b)反例1|x|1(x)(c)反例0|x|1,f(x)=1,则f[(x)]=1(x)(d)反设g(x)=f(x)在(－,+)内连续,则,+)内连续,矛盾.所以(d)是答案.(x)=g(x)f(x)在(－f(x)xtanxesinx,则f(x)是2.设函数(a)偶函数(b)无界函数(c)周期函数(d)单调函数解.(b)是答案|x|sin(x2)解.limf(x),limf(x),f(0)sin2,f(0)sin244x1x0所以在(－1,0)中有界,(a)为答案.x2114.当x1时,函数x1ex1的极限(a)等于2(b)等于0(c)为(d)不存在,但不为x10x10.(d)为答案x21x111limx1ex1lim(x1)ex1x1解..0352n1的值是1222232n2(n1)5.极限lim22n(a)0(b)1(c)2(d)不存在352n11222232n2(n1)limn解.22=lim11111223111lim1n1,所以(b)为答案.n(n1)(n1)2222222n(x1)95(ax1)(x21)5058,则a的值为6.设limx(a)1(b)2(c)58(d)均不对(x1)95(ax1)5=lim(x1)95/x95(ax1)5/x5解.8=limx(x21)50(x1)/x250100x(11/x)95(a1/x)(11/x2)505a5,a58,所以(c)为答案.=limx(x1)(x2)(x3)(x4)(x5)7.设limx,则,的数值为(3x2)13131(a)=1,=(b)=5,=(c)=5,=(d)均不对35解.(c)为答案.8.设f(x)2x3x2,则当x0时(a)f(x)是x的等价无穷小(b)f(x)是x的同阶但非等价无穷小(c)f(x)比x较低价无穷小(d)f(x)比x较高价无穷小lim2x3x22xln23xln3ln2ln3解.=limx0,所以(b)为答案.x1x0lim(1x)(12x)(13x)a6,则a的值为9.设x0x(a)－1(b)1(c)2(d)3解.lim(1x)(12x)(13x)a0,1+a=0,a=－1,所以(a)为答案.x0atanxb(1cosx)limx02，其中a2c20,则必有10.设cln(12x)d(1ex2)(a)b=4d(b)b=－4d(c)a=4c(d)a=－4cabsinxatanxb(1cosx)0cln(12x)d(1ex2)=cos2xa,所以a=－4c,所以(d)为答案.2c2xdex2c解.2=limxlimx0212x三.计算题1.求下列极限lim(xex)1(1)xxln(xex)exlimln(xex)elim1exe1lim(解.xex)xlimeexxex1xxxx(2)lim(sin2cos1)xxxx解.令y1xlim(sin2cos1)xlim(sin2ycosy)y=ey01limln(sin2ycosy)elim2cos2ysinyy0e2ysin2ycosyxxxy01tanx1(3)limx0x31sinx1tanx13lim1xtanxsinx1x3limx01sinx解.1sinxx0tanxsinx1sinxtanxsinxtanxsinx(1sin)xx3xtansinxlim1=elimx0x31sinxx0sinx2sin2xlimsinx(1cosx)x012lim=exe2.=ex3x302.求下列极限ln(13x1)(1)limx1arcsin23x21ln(13x1)~3x1arcsin23x21~23x21.按照等价无穷小代换解.当x1时,,ln(13x1)3x111limlimlimx123x1232x1arcsin23x21x123x211cotx(2)limx02x2解.方法1：sin2xx11cos2x2cos2xlimlimx0cot2xlimx0==xsin2xx2sin2xx22x01(x21)cos2x2xcos2x2(x21)cosxsinxlimlimx0==44x3xx0=lim2xcos2xsin2xlim2x2cosxsinx4x34x3x0x0=lim2cos2x4xcosxsinx2cos2x112x22x0=lim2cos2x2cos2x11lim4cosxsinx4sin2x1124x3212x232x0x0=lim2sin2x111112632324x32x0方法2：sin2xx2cos2x11cos2xlim=limx0cot2xlimx0=xsin2xx2sin2xx0x2211(x21)(cos2x1)1(x1)cos2x22limlim==x4x4x0x01(2x)2(2x)40(x4)1(x21)(1122!4!=limx0x41=2x4=lim3x423x03.求下列极限n(1)limn(nn1)lnnnnn1x(nn1)lim令n1xlimx01ln(1x)解.limnnlnnnlnnn(2)lim1enxn1enx1x0x0x0解.lim1enx0n1enx1nannnbblimn(3)解.,其中a>0,b>0211cxnanlimln(1x0cx)ln2x1/n,cb/aalimxaelimnx22x0limln(1cx)ln2limcxlnc=aex0aex01cxacababxa2(1cosx)x0x0x0x24.设f(x)11cost2dtxx0试讨论f(x)在x0处的连续性与可导性.1cost2dtxxlimcost2dt1xf'(0)limf(x)f(0)limx00解.xxx2x0x0x0limcosx21lim12x202x2xx0x02(1cosx)1lim2(1cosx)x2f'(0)limf(x)f(0)limx2xxx3x0x0x0lim2sinx2xlim2(cosx1)03x26xx0x0f'(0)0,f(x)在x0处连续可导.所以1(1)111f(0)lim2x11,f(02x1)lim1解.11x021x021xx所以x=0为第一类间断点.x(2x)x0x02cosxf(x)sin(2)1x21解.f(+0)=－sin1,f(－0)=0.所以x=0为第一类跳跃间断点;1limf(x)limsinx21不存在.所以x=1为第二类间断点;x1x1x(2x)f()不存在lim,而2,所以x=0为第一类可去间断点;22cosxx2x(2x),(k=1,2,…)所以x=klim为第二类无穷间断点.2cosx2xk26.讨论函数f(x)xx0在x=0处的连续性.ex0时lim(xsin1)不存在,所以x=0为第二类间断点;解.当xx00,lim(xsin1)0,所以当xx01时,在x=0连续,1时,x=0为第一类跳跃间断点.7.设f(x)在[a,b]上连续,且a<x<x<…<x<b,c(I=1,2,3,…,n)为任意正数,则在(a,b)内至少存在一个,使n12icf(x)cf(x)cf()1122n.cccn12证明:令M=max{f(x)},m=min{f(x)}ii1in1incf(x)cf(x)c1122nM所以mcccn12cf(x)cf(x)cnx<b),使得f()1122所以存在(a<x1cccnn128.设f(x)在[a,b]上连续,且f(a)<a,f(b)>b,试证在(a,b)内至少存在一个,使f()=..所以(x)f(x)x恒大于0或恒小于0.不妨设x[0,1],(x)f(x)x0.令mmin(x),则m0.0x1证明:令F(x)=x5－3x－2,则F(1)=－4<0,F(2)=24>0所以在(1,2)内至少有一个,满足F()=0.0,求f(0),f'(0),f''(0)及limf(x)3sin3xf(x)12.设f(x)在x=0的某领域内二阶可导,且limx0.x3x2x2x0解..所以sin3xlimf(x)0.f(x)在x=0的某领域内二阶可导,所以f(x),f'(x)在x=0连续.所以f(0)=－3.因为xx0sin3xf(x)sin3x3f(x)3x0,所以limx0x0,所以limx0x2x23sin3xlimf(x)3limxlim3xsin3xlim33cos3x3x2x0x2x2x3x0x0x0=lim3sin3x92x2x0f'(0)limf(x)f(0)limf(x)3limxf(x)3090x0xx22x0x0x0limf(x)39由x02,将f(x)台劳展开,得x2f(0)f'(0)x21!f''(0)x20(x2)3limx0912f''(0)92,于是2,所以x2f''(0)9.2导数与微分一.填空题f(xkx)f(x)13f'(x),则k=________.1.设limx000x0解.klimf(xkx)f(x)1f'(x),所以kf'(x)13f'(x)00kx3000x0所以k13dydx2.设函数y=y(x)由方程exycos(xy)0确定,则解.exy(1y')(yxy')sinxy0,所以______.y'ysinxyexyexyxsinxy3.已知f(－x)=－f(x),且f'(x)k,则f'(x)______.00解.由f(－x)=－f(x)得f'(x)f'(x),所以f'(x)f'(x)f'(x)f'(x)k所以00f(xmx)f(xnx)x_______.4.设f(x)可导,则limx000f(xmx)f(x)f(x)f(xnx)lim0000解.xx0f(xmx)f(x)+nlimf(xnx)f(x)=(mn)f'(x)=mlimx00000mxnx0x0f(x)1x,则f(n)(x)=_______.5.1xf'(x)1x1x(1)121!(1)k2k!(k)(1x)k1f解.(1x)11,假设,则(1x)2(1)k12(k1)!(1x)k11(1)n2n!(1x)n1f(k1),所以f(n)d11,则f'1_______.2fx6.已知dxx2121,所以f'1.令x=2,所以f'x11x22解.f'22x2x3xx2dy解.dxf'(x)cosf(x)f'[sinf(x)]cos{f[sinf(x)]}e8.设y=f(x)由方程2xycos(xy)e1所确定,则曲线y=f(x)在点(0,1)处的法线方程为_______.解.上式二边求导e2xy(2y')(yxy')sin(xy)0.所以切线斜率1.法线斜率为,法线方程为ky'(0)22y112x,即x－2y+2=0.二.选择题f'(x)[f(x)]2,则当n为大于2的正整数时,f(x)的n阶导数是1.已知函数f(x)具有任意阶导数,且(a)n![f(x)]n1(b)n[f(x)]n1解.f''(x)2f(x)f'(x)2![f(x)]3[f(x)]n(d)n![f(x)]n22(c)f1,所以,假设=k()(x)k![f(x)]kf(k1)(x)=(k1)k![f(x)]kf'(x)(k1)![f(x)]k2,按数学归纳法f(n)(x)=n![f(x)]n1对一切正整数成立.(a)是答案.f'(0)2.设函数对任意x均满足f(1+x)=af(x),且b,其中a,b为非零常数,则(a)f(x)在x=1处不可导(b)f(x)在x=1处可导,且f'(1)a(c)f(x)在x=1处可导,且f'(1)b(d)f(x)在x=1处可导,且f'(1)ab11f(1x)f(1)f(x)f(0)x0=limaa1f'(1)af'(1),所以ab.(d)是答案解.b=f'(0)limx0xx0f(x)没有假设可导,不能对于f(1x)af(x)二边求导注:因为.3.设f(x)3x3x2|x|()(0)存在的f,则使n最高阶导数n为(a)0(b)1(c)2(d)34x3x0f''(x)24xx0解.f(x)x0.12xx02x3f'''(0)limf''(x)f''(0)lim24x024x0xx0x0f'''(0)limf''(x)f''(0)lim12x012x0xx0x0所以n=2,(c)是答案.limydy4.设函数y=f(x)在点x处可导0,当自变量x由x增加到x+x时,记y为f(x)的增量,dy为f(x)的微分,0等于x0x0(a)－1(b)0(c)1(d)limydylimo(x)0xx0解.由微分定义y=dy+o(x),所以.(b)是答案.xx05.设f(x)xx0在x=0处可导,则axb(a)a=1,b=0(b)a=0,b为任意常数(c)a=0,b=0(d)a=1,b为任意常数解.在x=0处可导一定在x=0处连续,所以limx2sin1lim(axb),所以b=0.xx0x0x2sin1xlimaxf'(0)f'(0),lim,所以0=a.(c)是答案.xxx0x0三.计算题yln[cos(103x2)]，求y'1.sin(103x2)6x6xtan(103x2)cos(103x2)y'解.yf[ln(xax2)]，求y'2.已知f(u)可导,12xy'f'[ln(xax)]12解.xax22ax2f'[ln(xax2)]=ax2costdtsiny22xy',求.yedt23.已知t00ey'2xcosx22yy'cosy22解.y2xcosx2y'ey22ycosy2lnx2y2arctanyy'确定的,求.4.设y为x的函数是由方程xy'xyx2x2yy'解.2x2y22x2y2y21x2xy'xyxyy'y'xyy,所以四.已知当x0时,f(x)有定义且二阶可导,问a,b,c为何值时f(x)F(x)ax2bxcx0x0二阶可导.解.F(x)连续,所以limF(x)limF(x),所以c=f(－0)=f(0);x0x0因为F(x)二阶可导F'(x),所以连续,所以b=f'(0)f'(0),且'()2axf'(0)x0fxx0F'(x)F''(0)F''(0)F''(0),所以存在,所以limf'(x)f'(0)lim2axf'(0)f'(0)2a,所以xxx0x0a12f''(0)x2五.已知f(x)，求f(n)(0).1x21111解.f(x)121x21xn!1(1)nf(n)(x)12(1x)n12(1x)n1f(2k1)(0)0,k=0,1,2,…f2k(0)n!,k=0,1,2,…六.设yxlnx,求(n)(1).f解.使用莱布尼兹高阶导数公式f(n)(x)x(lnx)(n)n(lnx)(n1)x(1)n1(n1)!n(1)n2(n2)!xnxn1(n1)n1(1)n2(n2)!=(1)n2(n2)!xn1xn1xn1f(1)(1)n2(n2)!(n)所以3一元函数积分学(不定积分)一.求下列不定积分:1.11x21xln1xdx解.11xdx1ln1x1x11x1x2lncln1x21xdln41x21x2.1arctan1xarctan1xdarctan1x11x2dxarctan1x2c1x21x1x1x3.cosxsinx11sinxdx1cosx(1cosx)2解.cosxsinx11sinx1sinx1sinx11sinxx1cosx1cosx21cos2dxdc(1cosx)1cosx24.dxx(x81)1解.方法一:令x1t,xx(81)dxdtt7dtt11ln(1t8)c8t21118tt8181cln1=8x方法二:1)111)dxdxx7dx1)x(7x(xx(xxx888881d(1x8)111ln|x|ln(1x8)c=ln1cdx=x81x88x8812(1sinxcosx)1(sinxcosx)11sinxcosx5．1sinx22dx1sinxcosxdxcosxsinxdx11dx21sinxcosx1dx121sinxcosx2d(1sinxcosx)111x1dx221sincosxx22sincosx2cos2xx2221x1ln|1sinxcosx|11dtanx2222tanx121x1ln|1sinxcosx|1ln|tanx1|c2222二.求下列不定积分:1.dx(x1)2x22x2dt解.d(x1)令x1tantcos2t(x1)2(x1)21tan2tsectdx(x1)2x22x2=cos1x22x2ctdtcsintsintx122.dxx41x2解.令x=tant,dttcos3tsin2tdsintdsint11dxcos2dtctttan4secsin4tsin4t3sin3tsintx41x2311x1x22c=3xx3.dx(2x21)1x2解.令xtantsec2t(2tan2t1)secdtcost2sin2tcosdsintdxdt(2x1)1xt2t1sin2t22x=arctansintcarctanc1x24.(a>0)xdx2a2x2解.令xasinta2sin2tacostdt1cos2tdt1a2t1a2sin2tc24x2dxa2acost2a2x2ca2=2xxarcsina2x2aa25.(1x2)3dx解.令xsint(1cos2t)2dt12cos2tcos22t(1x2)3dxcos4tdtdt4411=44tsin2t1(1cos4t)dttsin2t1sin4tc31884323arcsinx1sin2t(11cos2t)c=8443=8arcsinx12sintcost(412sin2t)c443=8arcsinx1x1x2(52x2)c86.21xdxx41解.令xt1t2x21dx1dt令tsinusinucos2udut12dtt1t2x4t2t41cos3uc(x1)3c2=33x37.x1dxx2x21解.令xsect,dxsecttantdtx1dxsect1secttantdt(1cost)dttsintcsectanttx2x212x21carccos1xx三.求下列不定积分:1.e3xexe2x1dx4xe解.e3xex4xe2xexexdx()deexx1dx1arctan(exex)c2x1e2x(ee)xx2ee2.dx2x(14x)dt解.令t2x,dxtln21111arctantln2dxdtt2(1t2)ln2ln221tdttln2c2x(14x)t21(2xarctan2x)c=ln2四.求下列不定积分:1.x5dx(x2)100解.dx1x55xx5x5d(x2)9999(x2)4(x2)99dx(x2)100999999x55x454x3(x2)98dx99(x2)999998(x2)989998=x5=99(x2)995x454x3543x29998(x2)98999897(x2)9799989796(x2)965432x54329998979695(x2)959998979695(x2)94c2.dxx1x41dt解.令x1/t1dxtdtdt2t2x1x41t411t421(t2)2tt4tanu1sec2udu1ln|tanusecu|cln11x令t4cx222secu22五.求下列不定积分:1.xcos2xdxxcos2xdx1x(1cos2x)dx1x1xdsin2x2解.2441x21xsin2x1sin2xdx4441x21xsin2x1cos2xc448sec3xdx2.sec3xdxsecxdtanxsecxtanxtanxsecxtanxdx解.=secxtanx(secx1)secxdxsecxtanxln|secxtanx|sec23xdx223.(lnx)3dxx2解.(lnx)3dx(lnx)d11(lnx)3(lnx)2dxx233x2xx(lnx)33(lnx)26ln(lnx)33(lnx)26lnx6dxxxxxxdxxxx22(lnx)33(lnx)26lnx6cxxxx4.cos(lnx)dx解.cos(lnx)dxxcos(lnx)sin(lnx)dxx[cos(lnx)sin(lnx)]cos(lnx)dxxcos(lnx)dx[cos(lnx)sin(lnx)]c2xx5.xcosxcos44111x1x2sin3x2dxxdsin2xxsin2sin2dx828282xxsin3cos3221xsin2sin2d1xsin21cotxc1xxxx824228242六.求下列不定积分:1.xln(x1x2)(1x2)2dx解.xln(x1x2)dx1ln(x1x2)d1x221(1x2)211111=2ln(x1x1x221x2)dx21x2ln(x1x)1121tan12令xtantsectdt22(1x)tsect22ln(x1x2)1dt212sin2tcost===2(1x2)ln(x1x2(1x2))1d2sint12sin2t222ln(x1x2)2(1)112sintcln4212sintx2ln(x1x2)11x22xc12ln=2(1)x242x2x2.xarctanxdx1x2解.xarctandxarctanxd1x21x2arctanx1x2dxx1x21x21xarctanx1dx1x2arctanxln(x1x2)c2=1x23.arctanexdxe2x解.arctanexdx1arctanexde1earctanex1eex1e2x2x2x2xdxe2x22212e2xarctanex1dx12e2xarctanex1x12ex(1e2x)dxe21e2x12e2xarctanex1(12e1e2xx)dx12(e2xarctanexexarctanx)cex七.设f(x)xln(1x)3x0,求f(x)dx.2(x22x3)exx0(xln(1x2)3)dx(x22x3)exdx解.f(x)dx1x2ln(1x2)12[x2ln(1x2)]3xcx0x02(x24x1)exc1考虑连续性,所以c=－1+c,c=1+c111x2)12[xln(1x)]3xc22ln(1xx0x0f(x)dx22(x24x1)ex1c八.设f'(ex)asinxbcosx,(a,b为不同时为零的常数),求f(x).解.令tex，xlnt,f'(t)asin(lnt)bcos(lnt),所以f(x)[asin(lnx)bcos(lnx)]dxx[(ab)sin(lnx)(ba)cos(lnx)]c=2九.求下列不定积分:1.3x23x(2x3)dx3x23x3x23x(2x3)dx3x23xd(x23)ln3c解.3(3x22x5)(3x1)dx2.2133(3x22x5)2(3x1)dx(3x22x5)2d(3x22x5)解.21(3x22x5)2c553.ln(x1x2)1x2dx解.ln(x1x2)dxln(xx1)dln(xx1)1ln(xx1)c222221x24.xdx(1x2x21)ln(1x21)解.dln(1x1)ln|ln(1x1)|cxdx22(1x2x21)ln(1x21)ln(1x21)十.求下列不定积分:1.xarctanxdx(1x)2解.xarctan(1x)1arctanx1xdx22d(1x2)arctanxd(1x2)1(1x)222212a1rctaxn2x21x2darctanx12a1rctaxnx1111dx(1x2)222令xtant12a1rctaxnx1costdt1arctanx11cos2tdt2221x22221arctanx11tsin2tcarctanx1sintcostc1aextanx121x24821x2441aextanx1arctanx141xxc21x2422.arcsinx1xdxxarcsin解.令t,则xtant21xarcsin1xdxtdtan2tttan2ttan2tdtttan2ttanttcxxxxxarcsinxarcsinc(1x)arcsinxc1x1x1x3.arcsinx1x2dx1x2x2解.arcsinx1xdx令xsint1sin2costdtt(csc2t1)dt2ttcost1x2x2sin2ttcottdttdttcottcottdt1t2c2tcottln|sint|1t2c21x2arcsinxln|x|1(arcsinx)2cx24.arctanxdx)2x2(1xarctanxdx令xtanttsectdtt(csct1)dt22解.x2(1x2)tan2tsect2tcsc2tdttdttdcott1t2tcottcotdt1t222tcottln|sint|1t2carctanxln||1(arctanx)2cx2x21x2arctanx1x21(arctanx)2clnx21x22十一.求下列不定积分:x34xdx21.解.dx令x2sint8sint2cost2costdt32sinx34x233tcos2tdt323232(1cos2t)cos2tdtdcostcos3tcos5tc35(4x2)21(4x2)2c435352.2ax2x解.x2aattansecasecttantdtaat1cos2t2令xasectdtcostx2aatantatcx2a2aarccoscx3.ex(1)exdx1e2x解.ex(1)dx令extt(1t)dt1tdt令tsinu1sinucosuduex1t1tcosu1et2x22ucosucarcsinex1e2xcxdxx4.(a>0)2axx解.dx令ux2du令u2asint8asin4tdt2xu42ax2au2(1cos2t)2dt2a2(12cos2tcos22t)dt=8a24=2a2t2a2sin2t2a21cos4tdt3a2t2a2sin2tsin4tca424=3a2t4a2sintcosta2sintcost(12sin2t)c=3a2t3a2sintcost2a2sin3tcostc=3a2arcsinx3a2x2ax2a2xx2axc2a2a2a2a2a2ax3ax=3a2arcsinx(2ax)c2a2十二.求下列不定积分:1.dxsinx1cosx解.sinxdxsin2x1cosxd(1cosx)2d1cosxdxsinx1cosxsin2x1cosx1cos2x令1cosxu22u2(2u2)dudu1(u21)2(11)du11ln|2u|c222uu22u2uln|21cosx|c121cosx11cosx222.2sinxdx2cosx解.2sinxdx212cosx2dtdxd(2cosx)2cosx2cosx2ln|2cosx|22dtln|2cosx|1t2令tanxt21t23t221t24t41arctanln|2cosx|carctan(tanx)ln|2cosx|c=333323.sinxcosxsinxcosxdx解.sinxcosxdx112sinxcosx1sinxcosxdxsinxcosx21=2(sinxcos)21dx1(sinxcosx)dxsinxcosx212sinxcosx1dxd(x)1(sinxcosx)24=2sin(x)44=12(sinxcosx)42ln|tan()|c28x十三.求下列不定积分:1.xdx1xx解.dx令xt2t21t23d(1t3)4xdt1tc31xx31t3341x2c332.ex1dxex1解.ex1ex1dxex1e2x1sect1tantdt(sect1)dtdx令exsecttantln|secttant|tcln(exe2x1)arccos1cex3.x1arctanx1dxxttant22x1arctanx1tantt2secttantdt2ttantdt2t1cos2tdttdx22sec2tcosx22dt2tdt2tdtantt22ttant2tantdtt2tcos2t2ttant2ln|cost|t2c2x1arctanx1ln|x|(arctanx1)2c4一元函数积分学(定积分)bf(x)(x)dx0,则f(x)0.一．若f(x)在[a，b]上连续,证明:对于任意选定的连续函数(x),均有a证明:假设f()0,a<<b,不妨假设f()>0.因为f(x)在[a，b]上连续,所以存在>0,使得在[－,+]上2(x)2,其它地方(x)=0.所以bf(x)(x)dxf(x)(x)dxm20.2abf(x)(x)dx0矛盾.所以f(x)0.和adx11I二.设为任意实数,证明:dx=4.021(tanx)021(cotx)dx4=f(sinx)f(sinx)f(cosx)f(cosx)f(sinx)f(cosx)dx证明:先证:2020令t=x,所以2f(sinx)f(sinx)f(cosx)f(cost)dxf(cost)f(sint)d(t)0202f(cost)f(cosx)f(cosx)f(sinx)f(cost)f(sint)dtdx=2200于是f(sinx)f(sinx)f(cosx)dxf(sinx)f(cosx)f(cosx)f(sinx)f(sinx)f(cosx)dx02dx22200(sinx)f(cosx)ff(sinx)f(cosx)dx2dx=2200dxf(sinx)f(sinx)f(cosx)f(cosx)f(sinx)f(cosx)dx.所以204=2011dx(cosx)(cos)(sinx)所以Idx021(tanx)22x4sinxcosx001dx1同理I021(cotx)4.三．已知f(x)在[0，1]上连续,对任意x,y都有|f(x)－f(y)|<M|x－y|,证明kMf1f(x)dx1n2nnn0k11kkkkf(x)dxnf(x)dx,n()nn()fdx1证明:0fnnk1nk1nk1k1k1nn1f(x)dx1nf()|f(x)f(k)dx|kknnnn1nk0k1k1nnf(x)f(k)dxnM(xk)dxnkknn1nk1kk1k1nn1MkMkMnxdxnn1n2n22nkk1k1n11Itannxdx,n为大于1的正整数,证明:2(n1)I2(n1).四.设40nn1tn证明:令t=tanxItannxdxdt,则4001t2n1t2t>0,(0<t<1).所以1t2111221't1t2(1t2)2因为1于是111tnt2dt11tn1tdtndt22000112n2(n1).1立即得到2(n1)In五.设f(x)在[0,1]连续,且单调减少,f(x)>0,证明:对于满足0<<<1的任何,,有f(x)dxf(x)dx0xf(t)dtF()),F(x)xf(t)dtf(t)dt0.证明:令(x00F'(x)f(t)dtf(x)[f(t)f(x)]dt0,(这是因为t,x,且f(x)单减).00f(x)dxf(x)dxF()F()0所以,立即得到0f''(x)六.设f(x)在[a,b]上二阶可导,且<0,证明:bf(x)dx(ba)fab2af(x)f(t)f'(t)(xt)f''()(xt)f(t)f'(t)(xt)2证明:x,t[a,b],2!ab,所以f(x)fababx2abt令f'222abxababbf(x)dxfdxf'dxbb二边积分222aaaf(x)dx1f(x)dx00xf(t)dtF(x)f(t)dtx0000F'(x)f(x)f(x)0,所以F(x)单增ftdt1f(t)dt0,即()100f(x)dx1f(x)dx00f(x)dxf();方法二:由积分中值定理,存在[0,],使01f(x)dxf()(1)由积分中值定理,存在[,1],使因为,所以f()f().所以1f(x)dxf(x)dx1f(x)dxff()()(1)200f()f()(1)f()f(x)dx20f'(x)在[a,b]内存在而且可积,f(a)=f(b)=0,试证八.设f(x)在[a,b]上连续,:|f(x)|1b|f'(x)|dx,(a<x<b)2a证明:|f'(x)|f'(x)|f'(x)|,所以x|f'(t)|dtf(x)f(a)x|f'(t)|dt,aax|f'(t)|dtf(x)x|f'(t)|dt;即即aab|f'(t)|dtf(b)f(x)b|f'(t)|dtxxb|f'(t)|dtf(x)b|f'(t)|dtxx所以b|f'(t)|dt2f(x)b|f'(t)|dtaa即|f(x)|1b|f'(x)|dx,(a<x<b)2a九.设f(x)在[0,1]上具有二阶连续导数,且f(0)f(1)0，f(x)0,试证:f''(x)f''(x)dx41f(x)0证明:因为（0，1）上f(x)0,可设f(x)>0因为f(0)=f(1)=0x0(0,1)使f(x)=0max(f(x))0x1f''(x)1f''(x)dx11所以0f(x)dx>fx()（1）00在（0，x）上用拉格朗日定理0f(x）f'()(0,x)00x0在（x1）上用拉格朗日定理0,f(x)f'()1x(x,1)000所以1f''(x)dxf''(x)dxf''(x)dxf'()f'()0f(x)x(1x)4f(x)0000ab)2ab(（因为）211f''(x)dx4所以f(x)00由（1）得f''(x)dx41f(x)0十.设f(x)在[0,1]上有一阶连续导数,且f(1)－f(0)=1,试证:1[f'(x)]2dx10[f'(x)]2dx1[f'(x)]dx111dx(f'(x)1dx)2(f(1)f(0))211证明:22000000[0,2],取使|f()|=max|f(x)|(0x2)使|f()||f(x)|.所以a|(x1)f(x)dx||x1||f(x)|dx|f()||x1|dx|f()|2220005一元函数积分学(广义积分)一.计算下列广义积分:2(2)1(x21)(x24)(3)exdxdxdx(1)0(ex1)13(1x2)320(4)1sin(lnx)dx(5)11dx(6)arctanxdx2xx21(1x2)3200解.22d(ex1)dx3(e21)23exdxlim0(1)20(ex1)3(ex1)3111(x21)(x24)dxlim11b13x1x24dx12(2)(3)20b0dx(1x2)3211,所以dx因为limx3x积分收敛.所以(1x2)2(1x2)2330=2sectdxdx2令xtant2(1x2)32dt22costdt202sec3t0(1x2)3201sin(lnx)dxlim1sin(lnx)dxlim1(xsin(lnx)xcos(ln))x11(4)220001dxsecttant1dt(5)(6)2xx212ttsectan33arctanxtsec2t2dx32dttcostdt12sec3t2(1x2)0006微分中值定理一.设函数f(x)在闭区间[0,1]上可微,对于[0,1]上每一个x,函数f(x)的值都在开区间