一元函数积分学及其应用

一元函数积分学及其应用第三章主讲武忠祥西安交大经典考题（爱情）30sinsintantanlimxxxx30limxxxx30sinsintantanlim)1xxxx030sintanlimxxxx30sinsintantanlim)2xxxx2130)cos1(tanlimxxxx30sinsintansintansintantanlim)3xxxxxxxxxx30tantansintantanlim30cos)sin(tanlimxxxx12121第一节不定积分第二节定积分第三节定积分应用第四节反常积分第五节几类简单的微分方程第一节不定积分1．两个概念:1）原函数：)()(xfxF2）不定积分：CxFxxf)(d)(2．基本积分公式：3．三种主要积分法1）第一类换元法（凑微分法）CxFxxxfCuFuuf))((d)())((,)(d)(则若2）第二类换元法：CxFCtFdtttftxxxf))(()()())(()(d)(1taxaxtaxxatataxxasec,iii)tan,ii))cos(sin,i)2222223）分部积分法vduuvudv“适用两类不同函数相乘”.sincosxxdxIxxxdxxxdxIsincoscossincos2dtttttdttdttx)1111()1)(1(212sin22224令【例1】【解】xxdxxxd22sin1sin2sin)sin1(sin例题选讲dxxxI251,tantxtdtdx2sec)(sectan)sec(tantansecsectan4425ttddttttttdttI)sec()1()(sec)1(sec2222tuduutdt【例2】【解1】令则)1(12124224xdxxdxxIdxxxxx2324141【解2】)1(1]1)1[(2122224xdxxxx)(xF)(xf0x.)1(2)()(2xxexfxFx.0)(,1)0(xFF).(xf【例3】设为的原函数，且当时，已知求)1(2))((21)()(22xxexFxfxFxdxxedxxedxexxdxxxexFxxxx2222)1(1)1(1)1()1()(【解1】由Cxedxxexdexxx1)1(12xdxedxxxexFxx11)()1()(22dxxxexxexx1)1()1(【解2】1.定义：nkkkbaxfxxf10)(limd)(2.可积性：1）必要条件：)(xf有界；2）充分条件：)(xf连续或仅有有限个第一类间断点；3.计算：1))()(d)(aFbFxxfba2）换元法3）分部积分法4）利用奇偶性，周期性5）利用公式第二节定积分002020d)(sin2d)(sin(2),32231,221231dcosdsin(1)xxfxxfxnnnnnnnnnnxxxxπnn奇偶xattfd)(],[)(baxf在xattfd)(],[ba).()d)((xfttfxa4变上限积分：上连续，则在上可导且变上限求导的三个类型：ttxxdttxftf(txtF(x)dttxfxxfxxfdttfxbaxxxxxd)sin(dd)),(((3)d))(:1)),(((2))())(()())(())(((1)020)()()()(:例2例5。性质：1)不等式：),()(xgxf.d)(d)(xxgxxfbaba(1)若则)(xf],[ba).(d)()(abMxxfabmba（2)若在上连续，则（3).d|)(|d)(xxfxxfbaba2)中值定理：)(xf],[babcaabcfxxfba),)((d)(（1）若在上连续，则)(),(xgxf],[ba)(xgbcaxxgcfxxgxfbaba,d)()(d)()(（2）若在上连续，不变号，则2π2π223.______dcos)sin(xxxx]8[【例1】._______2202dxxxx]2[【例2】._________2sin50dttxx【例3】例题选讲一、定积分计算]100[.sin0xdxxIndxxxIn0sindxxn0sin2dxxn20sin的奇数为大于为正偶数1,32231,221231nnnnnnnnnn【例4】【解】,sin)(0dtttxfx.)(0dxxf【例5】设计算0)(dxxfdxxxxxxf00sin)(dttt0sindxxxx0sin2sin0xdx【解1】0)(dxxf0)()(xdxfdxxxxxfx00sin)()()(2sin0xdx【解2】;)0(1022aadxxax,sintax20cossincostttdt20cossinsintttdtdttttt20cossinsincos2142120dt【例6】计算定积分【解】令则原式)(xfxxxfxttxtf020d)(,cos1d)(求【例7】已知连续，的值.,utxduufuxdttxtfxx)()()(00xxduuufduufx00)()(xxxduufxxfxxfduufdttxtfdxd000)()()()()(【解】令得xduufxsin)(02x2012sin)(duuf从而有令得：;12111limnnnnn【例1】求极限nnnnnn112111111lim2ln)1ln(11010xxdx【解】原式=二、与定积分有关的综合题,)2()2)(1(1nnnnnnynnnnnynnnln)]2ln()2ln()1[ln(1limlnlim)1ln()21ln()11ln(1limnnnnnn【解】令则10)1ln(dxx12ln2ee412ln2则原式=【例2】求极限nnnnnnn)()2()1(1lim2222)(11)2(11)1(111nnnnnn411102dxx原式【例3】求极限;122111lim222222nnnnnnnnn【解】222222222122111)(11)2(11)1(111nnnnnnnnnnnnn222)(11)2(11)1(1111nnnnnnn【例4】求极限.sin)11(d)1ln(lim33sin002xxttxx4sin0031)1ln(lim2xdttxx32034cossin2)sin1ln(limxxxxx23342lim320xxxx【解】原式)(xf,0)0(f.)()()(lim000xxxdttxfxdttftx【例5】设函数连续，且求极限xdttxf0)(xduuf0)()(utxxxxxdttfxdtttfdttfx0000)()()(limxxxxxfdttfxxfxxfdttf000)()()()()(lim【解】原式=xxxxxfdttfdttf000)()()(lim)()()(lim0xxfcxfcxfx21)0()0()0(fff)(tf).()(,0)(tftftfaaaxadttftxxF.)(||)(【例6】设连续,令)(xFy],[aa1)试证曲线在上是凹的.x)(xF2)当为何值时,取得最小值.)(xF.1)(2aaf).(tf3)若的最小值可表示为试求aadttftxxF)()(xaaxdttfxtdttftx)()()()(xaxaaxaxdttfxdtttfdtttfdttfx)()()()(xaaxdttfxxfxxfxxfxxfdttfxF)()()()()()()(xaaxdttfdttf)()(0)(2)()()(xfxfxfxF【解】1)xaaxdttfdttfxF0)()()(,0)0(F,0)(xF)(xF0x2)令得又在取最小值.aaadtttfdttftF0)(2)()0(1)()(202aaafdtttfaafaaf2)()(21)(2aCeaf,1)0(f2C12)(2tetf3)又则从而)(xf]1,0[.0)(10dxxf【例7】设在上连续,且),1,0().()(0fdxxf求证:使0)()(0fdxxfxdttfxxF0)()(0)1()0(FF),1,0(0)(F0)()(0fdxxf【证】只要证明令则由罗尔定理知使即)(xf],[ba),0,0(ba),()()(22)(22bfabdxxfebaabx),,(ba.0)()(2ff【例8】(2012,1,14)设在上可导且满足证明：存在使)()(2xfexFx【证】令),()(2222bfedxxfeabbbaax),()(22bfecfebc)2,(baac)()(bFcF),,(ba0)(F使)(),(xgxf]1,0[)1,0(,)(3)(13210dxxfdxxf)1,0(,)].()()[()(ffgf【例9】设在上连续，在内可导，且试证存在两个不同的点使得)(xf]1,0[,0)0(f],1,0[.)(2)(10dxxff【例10】设函数在上有连续一阶导数，且试证至少存在一点使xcffxfxf)()0()()(),0(xc【证1】由拉格朗日中值定理得)(xf]1,0[M.m又在上连续，则必有其最大值和最小值则Mcfm)(Mxxfmx)(101010)(xdxMdxxfxdxm2)(210MdxxfmMdxxfm10)(210)(2)()(dxxfxxfxF,0)0(F10101010)(2)()(dxxfxdxdxxfdxxF【证2】令则0)()(1010dxxfdxxf),1,0(c0)()(10cFdxxF由积分中值定理得，使得),,0(c,0)(F.)(2)(10dxxff由罗尔定理得，使得],,[)(2baCxf],,[babafabbafabdxxf)()(241)2()()(3【例11】(2009,1,19)设试证存在使【证1】由Taylor定理得2)2(!2)()2)(2()2()(baxfbaxbafbafxfdxbaxfbafabdxxfba