第三章：一元函数积分学.pdf

)(xg),()(xg)(xf0 x)(xf),(【解解】因因为为在在上上连连续续,则则有有原原函函数数.而而在在处处为为第第一一类类间间断断点点,则则在在上上不不存存在在原原函函数数.)(xfI)(xfI几几个个重重要要结结论论：在在区区间间上上连连续续在在上上有有原原函函数数；1 1）)(xfI)(xfI2 2）在在区区间间上上不不连连续续在在上上不不存存在在原原函函数数)(xfI)(xfI3 3）在在区区间间上上有有第第一一类类间间断断点点在在上上不不存存 在在原原函函数数；1111)1(1)1d()1(232310231021nnnnnnnnnnnnxnxxna.1)e1(111limlim23123 nnnnnnna)2,0(,sin xxx,sin)sin(sinxx 【解解】由由 知知 xxcos)cos(sin 2020.1dsind)sin(sinxxxx 2020.1dcosd)cos(sinxxxx 则则 故故选选（A A）222222)3()3d(8136)136d(21d1365xxxxxxxxxx.23arctan4)136ln(212Cxxx xx1darcsin 21darcsinxxxxx【解解1 1】原原式式txsin tttdttsin1dsinsin2 sintdsinttt.|cotcsc|lnsinCtttt 【解解2 2】令令原原式式 xxxxxxxxxde2e21d)e(22.22Cexexexxxx 【解解】原原式式 ,sintx 4cos1coscos)sin2(cossin202202 ttdtttdtt【解解】令令原原式式【解解】原原式式 00422dsin|cos|2dcoscos2 xxxxxx axaxax2022d)(222023232dcos2dcos)sin1(attattat【解解1 1】原原式式 )sin(taax axaxax2022d)(axaxaaax2022d)()(aaxxaxa20322d2【解解2 2】原原式式 0000cos)(coscoscos)(xdxxfxdxxxxdxxf,2coscos)(00 dxxxxdxxf .2)(xxf【解解】1103)(3)(dttfxxdttfx 112)(9)(dttfxxf2)(11 dttf.29)(2 xxf【解解】【解解】原原式式 0dcos11xx 0d2sin2xx.22 1010cos1dedsinenxnxnxxx 1010dcose1cose1xnxnnxnxx 101dcosecose11xnxnnx uxt xxuuftxtf00d)(d)(,1)1()(20 xxxduuf,1)(01 duuf,21)(01 duuf.23211)(11 dxf【解解】令令，则则Cxedxxfx )(1)(lndxxxf 1ln)(lnxdxf【解解】0ln1 xx 342232122dsin132sind121231 tttxxxx令令 36d)2cos1(131 tt【解解】.12132sin21613136 t【解解】xxdydxS1221 21)21(dxxx.212ln)21(21 dxxx【解解】202de21aA).1(e414 aa 402d)(tan1 xxs【解解】40cosd xx).21ln(tansecln40 xx 1010)()(dxxdxxxx 【解解】102102)12()12(dxxxdxxxx2011351211 1010)(2)(xdxfdxxxfxdx)1ln(410 【解解】dxxxxfx 1010)1ln(2)(2dxxxxx 101014)1ln(4 282ln4 23121212.ddxxxxxx 121212122141ddxxxxx 231223124121ddxxxxx).32ln(23212|dxxx原式 【解解】.21arcsin)12arcsin(121 x2312412121ln xx300021dd)1arctan(lim2xuttxux 【解解】原原式式20023)1arctan(lim2xdttxx 63)1arctan(2lim20 xxxx,utx xxuufttxf00d)(d)(xxuufxf04sind)()(【解解】令令则则 xuuf00222143)d)(212 202321d)(xxf)(xf2,0 232)(20 dxxf则则在在上上的的平平均均值值为为 200204dsindd)()(xxxxuufxf xaaxxaxaaxxfaxttfafxfafttfafaxafaxttf)()(d)()()(lim)(1d)()()()()(d)(lim.)(2)()(2)()(1)(d)()()(lim)(12afafafafafxfaxttfaxafxfafxaax 【解解1 1】原原式式 【解解2 2】uxt ,)()()()(0 xfxxxfduugdtxtg,)1ln()()(02 xfxxduugxxxxxfxfg 1)1ln(2)()(2【解解】令令xxxxxfx 1)1ln(2)(2,1)1ln(2)(xxxxf ,)1ln()12()(Cxxxxf 0)0(f.0 C由由得得 )1(nxn )(d|cos|0 xSxxn.d|cos|)1(0 nxx,2d|cos|d|cos|00nxxnxxnn ).1(2d|cos|)1(0 nxxn).1(2)(2 nxSn【解解】（1 1）当当 )1(nxn.)1(2)()1(2 nnxxSnn .2)(lim xxSx（2 2）由由（1 1）知知，当当10 ttt )1ln(.d|ln|d)1ln(|ln|1010 ttttttnn【解解】（I I）当当时时，1010d|ln|d)1ln(|ln|0ttttttunnn,)1(1d11dlnd|ln|2101010 nttnttttttnnn（I II I）由由（I I）知知 1d)()(xttfxxF0)1()0(FF).(d)()(1xxfttfxFx 【证证法法一一】（1 1）设设，则则 xcffxfxf)()0()()(),0(xc【证证1 1】由由拉拉格格朗朗日日中中值值定定理理得得 )(xf 1,0M.m又又在在上上连连续续，则则必必有有其其最最大大值值和和最最小小值值则则Mcfm )(Mxxfmx )(101010)(xdxMdxxfxdxm2)(210Mdxxfm Mdxxfm 10)(2 10)(2)()(dxxfxxfxF,0)0(F 10101010)(2)()(dxxfxdxdxxfdxxF【证证2 2】令令则则 0)()(1010 dxxfdxxf),1,0(c0)()(10 cFdxxF 由由积积分分中中值值定定理理得得，使使得得 ),0(c ,0)(F.)(2)(10 dxxff 由由罗罗尔尔定定理理得得，使使得得 1010)1()()(xdxfdxxf 1010)()1()()1(dxxfxxfx 10)()1(dxxfx 10)1()(dxxf【证证3 3】babadxxgfdxxgxf)()()()()(21 f 202020sin)(sin)(cos)(xdxxfxxfxdxxf 20sin)(xdxxf 2020cos)(cos)(xdxxfxxf 20cos)()0()2(xdxxfff 20.0)cos1)(dxxxf【证证明明】nknknknknknkfndxxfnkfndxxf111101)(1)()(1)(nknknknknknkdxnkfdxxf1111)()(dxnkfxfnknknk 11)()(dxnkxfnknknkk 11)(【证证】（拉拉格格朗朗日日中中值值定定理理）nMnMdxxnkMnknknknk221)(1211 0)(xf),1(x)(xf xxxxfxxxffxf1122d)(1d)()1()(xxx12d11【证证】显显然然则则递递增增.124d11xx41)(xf)(xf)(limxfx 41 即即上上有有界界.则则 存存在在且且不不超超过过 2112d2yxV【解解】2112d)1(2yy.49 21123d)2)(1(10ygyyW 22123d)2)(210ygyyy g 827 14)1(2112112 edxxxdxysee)3)(1(161)ln2141(2221 eedxxxxe.127121)ln2141(321 edxxxe 127121)1)(1(161322eeex.)7(4)3)(1(3322 eee【解解】（）（）,d28d)2(33d422221xxxxxV .d28d)4(33d422222xxxxxV 2142104221d)28(2d)28(2)(2xxxxxxVVV .15448d)28(22042 xxx【解解1 1】130222)3(2)3(2xDdyydxdyV 154489)1(2222 dxx【解解2 2】1212bxxbxa)10(114 aab【解解】1 1）法法1 1.12 xbxaab 114,0法法2 2.有有唯唯一一解解，则则2 2）bayxbxaxV02)d(2)10().1(2222 aaaba DxydV 2 yyydxdy1212 38 DyxdV 2 6112121 dxxdyyy DxdxV )2(22)2ln4625()2(2121 dxxdyyy DydyV )2(22)2ln65(4)2(2121 yydxydy22),(yxxyyxr DdyxV 22dyyxdxxx 22210 602【解解】02yDdS dd cos100223 【解解】（1 1）02yDxydV ddsin2cos1020 03sin)cos1(32d38 （2 2）