总习题五

总习题五1.填空:(1)函数f(x)在[a,b]上(常义)有界是f(x)在[a,b]上可积的______条件,而f(x)在[a,b]上连续是f(x)在[a,b]上可积______的条件;解函数f(x)在[a,b]上(常义)有界是f(x)在[a,b]上可积的___必要___条件,而f(x)在[a,b]上连续是f(x)在[a,b]上可积___充分___的条件;(2)对[a,+)上非负、连续的函数f(x),它的变上限积分xadxxf)(在[a,+)上有界是反常积分adxxf)(收敛的______条件;解对[a,+)上非负、连续的函数f(x),它的变上限积分xadxxf)(在[a,+)上有界是反常积分adxxf)(收敛的___充分___条件;(3)绝对收敛的反常积分adxxf)(一定______;解绝对收敛的反常积分adxxf)(一定___收敛___;(4)函数f(x)在[a,b]上有定义且|f(x)|在[a,b]上可积,此时积分badxxf)(______存在.解函数f(x)在[a,b]上有定义且|f(x)|在[a,b]上可积,此时积分badxxf)(___不一定___存在.2.计算下列极限:(1)ninnin111lim;解)122(32)1(32111lim103101xdxxninnin.(2)121limppppnnn(p0);解11111])()2()1[(lim21lim101101pxpdxxnnnnnnnpppppnppppn.(3)nnnn!lnlim;解]ln1)ln2ln1(ln1[lim!lnlimnnnnnnnnnnnnnnnn1)]ln(ln)ln2(ln)ln1[(lnlim10ln1)ln2ln1(lnlimxdxnnnnnn1)ln()ln(10101010xxxdxxx.(4)xaaxdttfaxx)(lim,其中f(x)连续;解法一)()(lim)(limaafxfdttfaxxaxaax(用的是积分中值定理).解法二)(1)()(lim)(lim)(limaafxxfdttfaxdttfxdttfaxxxaaxxaaxxaax(用的是洛必达法则).(5)1)(arctanlim202xdttxx.解4)(arctan1lim1)(arctanlim1)(arctanlim22222202xxxxxxxdttxxxx.3.下列计算是否正确,试说明理由:(1)111111222)1arctan()1(1)1(1xxxdxdx;解计算不正确,因为x1在[1,1]上不连续.(2)因为111122111ttdttxxxdx,所以11201xxdx.解计算不正确,因为t1在1,1]上不连续.(3)01lim122AAAdxxxdxxx.解不正确,因为AAAbbaadxxxdxxxdxxxdxxx2020221lim1lim1lim1.4.设p0,证明10111pxdxpp.证明pppppppxxxxxxx11111111.因为1010101)1(dxxdxdxxpp,而110dx,pppxxdxxpp1)1()1(10110,所以10111pxdxpp.5.设f(x)、g(x)在区间[a,b]上均连续,证明:(1)bababadxxgdxxfdxxgxf)()(])()([222;证明因为[f(x)g(x)]20,所以2g2(x)2f(x)g(x)f2(x)0,从而0)()()(2)(222bababadxxfdxxgxfdxxg.上式的左端可视为关于的二次三项式,因为此二次三项式大于等于0,所以其判别式小于等于0,即0)()(4])()([4222bababadxxgdxxfdxxgxf,亦即bababadxxgdxxfdxxgxf)()(])()([222.(2)212212212)()()]()([bababadxxgdxxfdxxgxf,证明babababadxxgxfdxxgdxxfdxxgxf)()(2)()()]()([222212222])()([2)()(babababadxxgdxxfdxxgdxxf,又2212212212222])([])([])()([2)()(babababababadxxgdxxfdxxgdxxfdxxgdxxf,所以212212212)()()]()([bababadxxgdxxfdxxgxf.6.设f(x)在区间[a,b]上连续,且f(x)0.证明babaabxfdxdxxf2)()()(.证明已知有不等式bababadxxgdxxfdxxgxf)()(])()([222,在此不等式中,取)(1)(xfxf,)()(xfxg,则有bababadxxfxfdxxfdxxf222])(1)([])(1[])([,即babaabxfdxdxxf2)()()(.7.计算下列积分:(1)20cos1sindxxxx;解20202020220)cos1ln()2(tancos1)cos1(2cos2cos1sinxxxdxxddxxxdxxxx20202022ln2cosln222ln2tan)2tan(xdxxxx.(2)40)tan1ln(dxx;解4040cos)4sin(2ln)tan1ln(dxxxdxx404040cosln)4sin(ln2lnxdxdxxdx.令,4ux则40404040coslncosln)44sin(ln)4sin(lnxdxududuudxx,所以2842ln2ln)tan1ln(4040dxdxx.(3)axaxdx022;解令xasint,则20022cossincostttdtxaxdxa.又令ut2,则2020cossinsincossincosuuudutttdt,所以adtdtttttxaxdx0202022421cossincossin21.(4)202sin1dxx;解2020|sincos|2sin1dxxxdxx2440)sin(cos)sin(cosdxxxdxxx)12(2)cos(sin)cos(sin2440xxxx.(5)202cos1xdx.解2022022202tan2tan)1(seccoscos1xxdxxdxxdx422tanarctan2120x.8.设f(x)为连续函数,证明xtxdtduufdttxtf000])([))((.证明xtxtxtduuftdduuftdtduuf000000])([)(])([xxdtttfduufx00)()(xxxdttxtfdtttfdttfx000))(()()(.9.设f(x)在区间[a,b]上连续,且f(x)0,xaxbtfdtdttfxF)()()(,x[a,b].证明:(1)F(x)2;(2)方程F(x)0在区间(a,b)内有且仅有一个根.证明(1)2)(1)()(xfxfxF.(2)因为f(x)0,ab,所以0)()(abtfdtaF,badttfbF0)()(,由介值定理知F(x)0在(a,b)内有根.又F(x)2,所以在(a,b)内仅有一个根.10.设011011)(xexxxfx,求20)1(dxxf.解200110111111)(1)1(dttdtedttftxdxxft令)1ln()1ln()1ln(11110010110etedttdteettt.11.设f(x)在区间[a,b]上连续,g(x)在区间[a,b]上连续且不变号.证明至少存在一点x[a,b],使下式成立babadxxgfdxxgxf)()()()((积分第值定理).证明若g(x)0,则结论题然成立.若g(x)0,因为g(x)不变号,不妨设g(x)0.因f(x)在[a,b]上连续,所以f(x)在[a,b]上有最大值M和最小值m即mf(x)M,因此有mg(x)f(x)g(x)Mg(x).根据定积分的性质,有bababadxxgMdxxgxfdxxgm)()()()(,或Mdxxgdxxgxfmbaba)()()(.因为f(x)在[a,b]上连续,根据介值定理,至少存在一点x(a,b),使)()()()(fdxxgdxxgxfbaba,即babadxxgfdxxgxf)()()()(.*12.(1)证明:020)1(,2122ndxexndxexxnxn证明010)(2122xnxnedxdxex=)](|)[(21100122nxxnxdeex)1(21022nxdexnxn(2)证明))(1(210122Nnnxdexxn