文登考研数学--高等数学--习题集及其答案

1 第一章第一章 函数极限连续函数极限连续 一. 填空题 1. 已知,_)(,1)(,sin)( 2 xxxfxxf则 定义域为_. 解. 2 1)(sin)(xxxf, )1arcsin()( 2 xx 111 2 x, 2|, 20 2 xx 2设 a t ax x dtte x x1 lim, 则 a = _. 解. 可得 a ta dttee= aatt eae a ete )(, 所以 a = 2. 3. nnn n nnnn n 222 2 2 1 1 lim=_. 解. nnn n nnnnnn 222 21 nnn n nnnn 222 2 2 1 1 11 2 1 1 222 nn n nnnn 所以 nnn n 2 21 0 解. n nn n ba 2 lim abcnx/,/1 x c x x x x x ae c a 2ln)1ln( lim 1 0 0 2 1 lim =ab a b acaaeae x x x x xc cc x c 1 ln lim 2ln)1ln( lim 00 4. 设 0cos 1 01 0)cos1 ( 2 )( 0 2 2 xdtt x x xx x xf x 试讨论)(xf在0x处的连续性与可导性. 解. 2 0 2 0 0 2 00 cos lim 1cos 1 lim )0()( lim)0( x xdtt x dtt x x fxf f x x x xx 0 2 2 1 lim 2 1cos lim 2 0 2 0 x x x x xx 3 2 0 2 00 )cos1 (2 lim 1)cos1 ( 2 lim )0()( lim)0( x xx x x x x fxf f xxx 0 6 ) 1(cos2 lim 3 2sin2 lim 0 2 0 x x x xx xx 所以 0)0( f, )(xf在0x处连续可导. 5. 求下列函数的间断点并判别类型 7 (1) 12 12 )( 1 1 x x xf 解. 1 12 12 lim)0( 1 1 0 x x x f, 1 12 12 lim)0( 1 1 0 x x x f 所以 x = 0 为第一类间断点. (2) 1 1 sin cos2 )2( )( 2 x x xx xf 0 0 x x 解. f(+0) =sin1, f(0) = 0. 所以 x = 0 为第一类跳跃间断点; 1 1 s inlim)(lim 2 11 x xf xx 不存在. 所以 x = 1 为第二类间断点; ) 2 ( f不存在, 而 2cos2 )2( lim 2 x xx x ,所以 x = 0 为第一类可去间断点; x xx kxcos2 )2( lim 2 , (k = 1, 2, ) 所以 x = 2 k为第二类无穷间断点. 6. 讨论函数 x e x x xf 1 sin )( 0 0 x x 在 x = 0 处的连续性. 解. 当0时) 1 sin(lim 0 x x x 不存在, 所以 x = 0 为第二类间断点; 当0, 0) 1 sin(lim 0 x x x , 所以 1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 x1 x2 xn b, ci (I = 1, 2, 3, , n)为任意正数, 则在(a, b)内至少存在一个, 使 n n ccc cxfcxfc f 21 2211 )()( )(. 证明: 令 M =)(max 1 i ni xf , m =)(min 1 i ni xf 所以 m n n ccc cxfcxfc 21 2211 )()( M 所以存在( a x1 xn 0) 解. 令taxsin ctatadt t a ta tdtata xa dxx 2sin 4 1 2 1 2 2cos1 cos cossin 222 22 22 2 =cxa a x a xa 22 2 2 arcsin 2 5. dxx 32) 1 ( 解. 令txsin dt tt dt t tdtdxx 4 2cos2cos21 4 )2cos1 ( cos)1 ( 22 432 = ctttdtttt4sin 32 1 2sin 4 1 8 3 )4cos1 ( 8 1 2sin 4 1 4 1 =cttx)2cos 4 1 1 (2sin 4 1 arcsin 8 3 =c t ttx ) 4 sin214 (cossin2 4 1 arcsin 8 3 2 =cxxxx)25(1 8 1 arcsin 8 3 22 6. dx x x 4 2 1 解. 令 t x 1 dtttdt t t t t dx x x 2 2 4 2 2 4 2 1 1 1 1 1 utsin令 uduu 2 cossin 15 =c x x cu 3 32 3 3 ) 1( cos 3 1 7. dx xx x 1 1 22 解. 令 tdttdxtxtansec,sec cttdtttdtt tt t dx xx x sin)cos1 (tansec tansec 1sec 1 1 2 22 c x x x 11 arccos 2 三. 求下列不定积分: 1. dx ee ee xx xx 1 24 3 解. cee ee eed dx ee ee dx ee ee xx xx xx xx xx xx xx )arctan( 1)( )( 11 22224 3 2. )41 (2 xx dx 解. 令 x t2, 2lnt dt dx c t t dt tttt dtdx xx 2ln arctan 2ln 1 1 11 2ln 1 2ln)1 ()41 (2 2222 =c xx )2arctan2( 2ln 1 四. 求下列不定积分: 1. dx x x 100 5 )2( 解. dxxx x x xdxdx x x 994 99 5 995 100 5 )2( 99 5 )2(99 )2( 99 1 )2( = dxxx x x x x 983 98 4 99 5 )2( 9899 45 )2(9899 5 )2(99 = 96 2 97 3 98 4 99 5 )2(96979899 345 )2(979899 45 )2(9899 5 )2(99 x x x x x x x x c xx x 9495 )2(9596979899 2345 )2(9596979899 2345 16 2. 4 1xx dx 解. 22 2 4 4 4 2 4 )(1 2 1 111 1 /1 1t dt t tdt t t t dt t tx xx dx 令 c x x cuudu u u ut 2 42 2 1 ln 2 1 |sectan|ln 2 1 sec sec 2 1 tan令 五. 求下列不定积分: 1. xdxx 2 cos 解. xxdxdxxxxdxx2sin 4 1 4 1 )2cos1 ( 2 1 cos 22 xdxxxx2sin 4 1 2sin 4 1 4 1 2 cxxxx2cos 8 1 2sin 4 1 4 1 2 2. xdx 3 sec 解. xdxxxxxxxdxdxtansectantansectansecsec3 = xdxxxxxxdxxxx 32 sec|tansec|lntansecsec) 1(sectansec cxxxxxdx |tansec|ln 2 1 tansec 2 1 sec3 3. dx x x 2 3 )(ln 解. dx x x x xx dxdx x x 2 2 33 2 3 )(ln3 )(ln 11 )(ln )(ln dx x x x x x x 2 23 ln6)(ln3)(ln dx xx x x x x x 2 23 6ln6)(ln3)(ln c xx x x x x x 6ln6)(ln3)(ln 23 4. dxx)cos(ln 解. dxxxxxdxxxxdxx)cos(ln)sin(ln)cos(ln)sin(ln)cos(ln)cos(ln cxx x dxx )sin(ln)cos(ln 2 )cos(ln 17 5. dx xx x x xddx xx x x dx x x x 2 sin 8 1 2 sin 8 1 2 sin 8 1 2 cos 2 sin 2 cos 8 1 sin 2 cos 222 33 4 3 4 c xx x x d xx x 2 cot 4 1 2 sin 8 1 22 sin 4 1 2 sin 8 1 222 六. 求下列不定积分: 1. dx x xxx 22 2 )1 ( )1ln( 解. 2 2 22 2 1 1 )1ln( 2 1 )1 ( )1ln( x dxxdx x xxx = dx x xx xx 2 22 2 1 1 1 1 2 1 1 1 )1ln( 2 1 txtan令 tdt ttx xx 2 22 2 sec sec 1 tan1 1 2 1 )1 (2 )1ln( =dt t t x xx 22 2 sin21 cos 2 1 )1 (2 )1ln( = t td x xx 22 2 sin21 sin2 22 1 )1 (2 )1ln( =c t t x xx sin21 sin21 ln 24 1 )1 (2 )1ln( 2 2 =c xx xx x xx 21 21 ln 24 1 )1 (2 )1ln( 2 2 2 2 2. dx x xx 2 1 arctan 解. dx x x xxxxddx x xx 2 2 22 2 1 1 arctan11arctan 1 arctan =cxxxxdx x xx )1ln(arctan1 1 1 arctan1 22 2 2 3. dx e e x x 2 arctan 解. dx e e eeedeedx e e x x xxxxx x x 2 222 2 12 1 arctan 2 1 arctan 2 1arctan 18 dx e e ee x x xx 2 2 12 1 arctan 2 1 dx ee ee xx xx )1 ( 1 2 1 arctan 2 1 2 2 cxeeedx e e e ee xxx x x x xx )arctanarctan( 2 1 ) 1 1 ( 2 1 arctan 2 1 2 2 2 七. 设 x exx xx xf ) 32( 3)1ln( )( 2 2 0 0 x x , 求dxxf)(. 解. dxexx dxxx dxxf x ) 32( ) 3)1ln( )( 2 2 1 2 2222 ) 14( 3)1ln( 2 1 )1ln( 2 1 cexx cxxxxx x 0 0 x x 考虑连续性, 所以 c =1+ c1, c1 = 1 + c dxxf)( cexx cxxxxx x 1) 14( 3)1ln( 2 1 )1ln( 2 1 2 2222 0 0 x x 八. 设xbxaef x cossin)( , (a, b 为不同时为零的常数), 求 f(x). 解. 令txet x ln，, )cos(ln)sin(ln)( tbtatf, 所以 dxxbxaxf)cos(ln)sin(ln)( =cxabxba x )cos(ln)()sin(ln)( 2 九. 求下列不定积分: 1. dxx xx ) 32(3 3 2 解. cxddxx xx xxxx 3ln 3 )3(3)32(3 3 233 2 22 2. dxxxx) 13()523( 2 3 2 解. )523()523( 2 1 ) 13()523( 2 2 3 2 2 3 2 xxdxxdxxxx cxx 2 5 2 )523( 5 1 19 3. dx x xx 2 2 1 )1ln( 解. cxxxxdxxdx x xx )1(ln 2 1 )1ln()1ln( 1 )1ln( 2222 2 2 4. )11ln()11 ( 222 xxx xdx 解. cx x xd xxx xdx | )11ln(|ln )11ln( )11ln( )11ln()11 ( 2 2 2 222 十. 求下列不定积分: 1. dx x xx )1 ( arctan 2 解. 122 2222 )1 (arctan 2 1 )1 ( )1 ( arctan 2 1 )1 ( arctan xxdxd x x dx x xx dx xx x xd xx x 22222 )1 ( 1 2 1 1 arctan 2 1 arctan 1 1 2 1 1 arctan 2 1 dt t x x tdt x x tx 2 2cos1 2 1 1 arctan 2 1 cos 2 1 1 arctan 2 1 tan 2 2 2 令 cttx x xaex ctt x x cossin 4 1 arctan 4 1 1 tan 2 1 2sin 8 1 4 1 1 arctan 2 1 22 c x x x x xaex 22 14 1 arctan 4 1 1 tan 2 1 2. dx x x 1 arcsin 解. 令txt x x 2 tan, 1 arcsin 则 ctttttdttttdtdx x x tantantantantan 1 arcsin 2222 cx x x xc x x x x x x 1 arcsin)1 ( 1 arcsin 1 arcsin 3. dx x x x x 2 2 2 1 1arcsin 解. dttttdt t t t t txdx x x x x ) 1(csccos cos sin1 sin sin 1 1arcsin 2 2 2 2 2 2 令 cttdtttdtttdtt 2 2 1 cotcotcot 20 ctttt 2 2 1 |sin|lncot cxx x x x 2 2 )(arcsin 2 1 |ln 1 arcsin 4. dx xx x )1 ( arctan 22 解. dtttdtt tt t txdx xx x ) 1(cscsec sectan tan )1 ( arctan 22 2222 令 222 2 1 cotcot 2 1 cotcsctdtttttdtdttdttt cx x x x x ctttt 2 2 2 )(arctan 2 1 | 1 |ln arctan 2 1 |sin|lncot cx x x x x 2 2 2 )(arctan 2 1 1 ln 2 1arctan 十一. 求下列不定积分: 1. dxxx 23 4 解. dtttdtttttxdxxx 23323 cossin32cos2cos2sin8sin24令 ctttddttt 5322 cos 5 32 cos 3 32 coscos)cos1 (32 cxx 2 5 2 2 3 2 )4( 5 1 )4( 3 4 2. x ax 22 解. dt t t adttta ta ta tax x ax 2 222 cos cos1 tansec sec tan sec令 c x a aaxcattaarccostan 22 3. dx e ee x xx 2 1 )1 ( 解. udu u u utdt t t t dt t tt tedx e ee x x xx cos cos sin1 sin 1 1 1 )1 ( 1 )1 ( 222 令令 ceecuu xx 2 1arcsincos 4. dx xa x x 2 (a 0) 21 解. dx xa x x 2 xu 令 du ua u 2 4 2 2 tausin2令 tdta 42 sin8 = dtttadt t a)2cos2cos21 (2 4 )2cos1 ( 8 22 2 2 =ct a tatadt t atata 4sin 4 2sin23 2 4cos1 22sin22 4 22222 =ctttattata)sin21 (cossincossin43 2222 =cttattatacossin2cossin33 3222 =c a xa a x a x a a xa a x a a x a 2 2 22 2 2 2 2 3 2 arcsin3 222 =cxax xa a x a )2( 2 3 2 arcsin3 2 十二. 求下列不定积分: 1. xx dx cos1sin 解. x xd xx xd xx dxx xx dx 2 22 cos1 cos1 2 cos1sin )cos1 ( cos1sin sin cos1sin )2( 2 ) 1(1 2cos1 2222 uu du u du ux令 c u u u du uu | 2 2 |ln 22 11 ) 2 11 ( 22 c x x x | cos12 cos12 |ln 22 1 cos1 1 2. dx x x cos2 sin2 解. x xd dx x dx x x cos2 )cos2( cos2 1 2 cos2 sin2 t x 2 tan令 |cos2|ln 3 2 2|cos2|ln 1 1 2 1 2 2 2 2 2 2 x t dt x t t t dt =cx x cx t |cos2|ln) 2 (tan 3 1 arctan 3 4 |cos2|ln 3 arctan 3 4 3. dx xx xx cossin cossin 22 解. dx xx xx dx xx xx cossin 1cossin21 2 1 cossin cossin = dx xx dxxxdx xx x cossin 1 2 1 )cos(sin 2 1 cossin 1cos)(sin 2 1 2 = ) 4 sin( ) 4 ( 4 2 )cos(sin 2 1 x xd xx =c x xx| ) 82 tan(|ln 4 2 )cos(sin 2 1 十三. 求下列不定积分: 1. dx xx x 1 解. ct t td dt t t txdx xx x 3 3 3 3 2 1 3 4 1 )1 ( 3 2 1 2 1 令 cx 2 3 1 3 4 2. dx e e x x 1 1 解. dttdtt t t tedx e e dx e e x x x x x ) 1(sectan tan 1sec sec 1 1 1 1 2 令 c e eecttt x xx 1 arccos)1ln(|tansec|ln 2 3. dx x xx 1arctan1 解. 令ttdxtxxtxttansec2,sec, 1tan, 1arctan 22 dt t t tdtttdttt t tt dx