练习6.11.若)(xF=)(xf,则)(xF是)(xf的原函数,)(xf的原函数全体称为)(xf的不定积分。区别是:)(xf的不定积分描述了所有满足导数是)(xf的函数,而原函数只是任一个满足导数是)(xf的函数。2.(1)3xe(2)cxcos(3)a1(4)2(5)-1(6)21(7)213.(1)(10)10xccxdx1010(2)xcxsin)cos2(cxxdxcos2sin2(3)dxxcxd455)(cxdxx5454.解:由题意cxxf2)(,又由1)1(f,知1c,因此12)(xxf。5.解:由题意xxxf1)(ln)(,所以21)(xxf练习6.21.(1)cxxx12ln2(2)cxxx3423cos3arcsin(3)cxeexeexe111(4)=dxxxx)9264(=cxxx99ln166ln244ln1(5)=cxdxxdxxxx81587814121158)((6)=cxdxx454154(7)=cxxxdxxxarctan3)111(322(8)=cxxdxxdxxdxxdxxxdxxarctan11112)1(12222222(9)cxxdxxdxx2sin22cos12cos2(10)=cxxdxxxcottan)cos1sin1(22(11)=cxxdxxcot)1(csc2(12)=cxxdxxx2tan21cos2cos1222.解:由题意知c(x)=7x+cx50由固定成本为1000知c=1000因此c(x)=7x+501000x练习6.38932222232232211(53)5(53)545112(2)21212(21)4431(21)61ln2(3)1lnln(1ln)3(4)csccsccotln(csc(5)csccsccotxxxdxxdxxcxxdxxdxxcxcxdxxdxxcxedxdeecxxxdxxdxdxxx8(1)(5x-3)222222222222cot)ln|csccot|csccot(6)111(7)(1)ln|1|1111111(8)ln(1)ln11212111(9)21(1)111(10)23(3)(1)xxxcxxxxdxdxxdxdxxxxcxxxdxdxxcxcxxdxdxcxxxxdxdxxxxx2222111111ln(3)ln(1)4341441111(11)arctan25(1)22211(12)()()()()2.122(1)222l1112xtdxdxxxcxxxdxdxcxxxfaxbdxaxbdaxbFaxbcaatdxdtdtdttttx342233324431433441n(1)22ln(2)33333111(ln|1|)(1)1ln|11|(2)(1)144444411111111(2)(1)2(3)22212111nxtxttcxxctxdttttcxxxcdxdxtxxxtttdxtdtdtttx211(2)444ln|1|1414ln(11)tdtdttttctxxxc232222222222221(4)(1)sincos1sectancosx=1-x1x=3tandx=3sec3(5)3secseccos13tan3sin3sin3tan33tandxxxxdxtdtdttdttctcdxtdtxxtdtttdtdtcttttt令原式令原式222222222222223+x=-1111(6).ln3911939(7)sindx=acostdtxasincosdx==-acost+ccosx=-a1-+c=-aaa1(8).14cxdxdxxxcxxxattatdtataxxcx令22221ln2(2)3(2)(3)dxdxxxcxx练习6.41.(1)cxxxxdxxxxdxxxxdxxdxx4ln2121ln21ln21ln21ln21ln222222(2)cxxxxxxdxxxxxxxdxxxdxxxxxdxxdxxcos2sin2cossin2sin2cossin2coscos2coscossin222222(3)cxxxdxxxxxddxxxxcot21csc2csc21csc2csc21sincos22223(4)cxxxdxxxxxxdxdxxdxxxarctan22112111211121)1()1(21)1(2222222222(5)xdxxdxxxxdxxxxxxdxxxxdxdxsecsectansectansectansecsectantansectansecsec323因此cxxxxxdx2tanseclntansecsec3(6)tatadxxaxtaxsectan22tan222设tdta2seccxaxxaxcaxaxaaxaxaactsetttadttta2ln22ln2)tanln(tansec)sec(sec222222222232(7)cxxxdxxxxxxdx221arcsin1arcsinarcsin(8)xdxexexexdexexdxexexdexexdxexxxxxxxxxxcossincossincossincoscoscoscos因此cxxexdxexx2)sin(coscos(9)cxxxcttttdtttttdtdtttdtdxxxtcos2sin2cos2sin2sin2sin2sin2cos2coscos2设(10)ceeeecuuuuduuuuuuduuuuuuduuuduuudxeexxxxuexxx)1ln(21lnarctan1)1ln(21lnarctan111arctan1111arctan1arctan1arctan1arctanarctan22222设2.解:由已知得)sin()(sin)(xxxfcxxdxxf或因此cxxxxfdxxfxxfxxdfdxxfxsin习题六1.单题(1)C(2)A(3)C4)D(5)B(6)C(7)B(8)B(9)A(10)D(11)C(12)A(13)C(14)D(15)B(16)C(17)B(18)C(19)C(20)D2.解:已知xxxef')(因)(xfxxxdexeuxceexceueudexxuuu)(由此曲线过(0,0)知.0)0(f即C=1于是1)(zxxexef3;解:由边记函数为xfx220)(知边际函数cxxdxxRx2)(20)220(又0)(xR知C=0于是2)(20xxRx4.根据提议:求函数为pQx5100)(成本为2)(205.01215QQcx既利润关于P的函数为:2)()4100(205.0)5100(15).5100(PPPPLx-12令0)('PL得P=7120即当P为120/7时利润最大5.解:由已知得)()(')(1xxxfffx即cxfx||ln)(又0)1(fC=0即)(xf=ln|x|6.(1)233xdxcxxxd3231)23(21)23()23(31(2)cxdxxdxxx2222cos21sin21sin(3)cxxxddxxxcos2coscoscossin(4)cedeedxeexxxxxsincoscos(5)cxxxddxxx2222252125)25(4125(6)cxxxxxddxxxxx|54|ln21)54(21542254222(7)cxxdxxxdx126sin226cos13sin2(8)cxxxdxxxddxxxx)]ln[ln(lnlnlnlnlnlnlnlnln)ln(ln.ln.1(9)cxxxcuuuduxxddxxx)321ln()2ln(22)1()1(32122222(10)cexdexdxexxxcoscoscoscossin(11)cxxxxdxxxdxxdx5cos_cos32coscos)coscos21(cos)cos1(sin5342225(12)cxxxddxxx3lnarcsinln3lnln3122(13)cxxxdxxdxarcsin11)arcsin1()arcsin1(1)arcsin1(222(14)cxxdxxxdxxdxx4sin27181)4cos1(812sin41cossin222(15)cxarcsunxdxxdx4331)34(31916222(16)cxxxxxddxxxarcsin1arcsin1)1(21112222(17)ceedeeedxxxxxxarctan12(18)cxxxdxxxddxxxxdxtan3tantantantan)1(sectantan322224(19)cxxxdxxxdxdxdxx3cotcotcot)cot1(cotcsccscsin132244(20)cxxxxdxdxx685253cos61cos81coscos)cos1(cossin3335333232385132333383511(21)1(1)(1)331(1)(1)385(1)(1)85xuutxxdxxxdxuuttttdtcxxc122112(22)ln(1)1111111()lnln11111xetxxxdxdtdtttetedtccttte638522375367/65/63/61/666(23)6111661263ln||751661263ln||751xtxttdxtdtdtttxtttttctxxxxxcx3sec222213sectan(24)9sec3tan91119cossin999xtttdxdtttxxxtdttccx22tan3322212sec1(25)cos8sec4(4)1sin444xttdxtdttxxtccx2tan222221sec(26)tansec111cscln|csccot|1111l