2022年长沙理工大学高等数学练习册第五章定积分答案.docx
精选学习资料 - - - - - - - - - 习题 5.1 略习题 5.25.3 （A）一 运算以下定积分12sinx3 cosxdx10解：原式23 cosxdcosx14 cosx204042ax2a2x2dxdt0asin ,就dxacostdt解：令x当x0时t0,当xa时t2原式2a2sin2tacostacostdt0a422 s i n2 tdta421c o s t313x24080a424 a1sin4 t216a48840dx1x2解：令xtg,就dx2 secd当x1,3 时分别为4,3原式32 secd4tg2sec3s i n2ds i n4名师归纳总结 223第 1 页,共 20 页3- - - - - - -精选学习资料 - - - - - - - - - 41xdxx154名师归纳总结 解：令54xu,就x51u2,dx1 2udux第 2 页,共 20 页44当x1,1 时,u1,3原式115u2 du138654dx11x解：令xt,dx2tdt当x1时,t1；当x4时,t2原式22tdt22dt2dtt11t111ln22t2ln1t222ln2113611dx13x4解：令1xu,就x1u2,dx2udu当x31,时u10,42原式0u2 udu21uu111du12211027e 2xdxx11ln1ln解：原式e211lnxdlnxe211xd11ln21lnxe 2232180x2dx222x解：原式01dx12arctgx1022x- - - - - - -精选学习资料 - - - - - - - - - a r c t ga r c t g1442901cos 2x dx20cosxdx解：原式022 cosxdx222c o s xdx2c o sdx02si n x2 0s i n x222104 x sinx dx解：x sinx为奇函数x4sinxdx011244 cosx dx2名师归纳总结 解：原式4202cos4xdx2222cos2x2dx22xdx第 3 页,共 20 页0221cos2x2dx212cos2xcos002x2 022cos2xdx21cos4xdx002sin2x2 0212cos4xd4x4031sin4x232402125xx3sin2x1dx542x2解：xx3sin2x1为奇函数42x25xx3sin2x1dx0542x2- - - - - - -精选学习资料 - - - - - - - - - 133xxdx4sin2解：原式3xdctgx4名师归纳总结 x c t g x3c t g x d x2第 4 页,共 20 页4413lns i n x349413ln3ln4922131ln34922dx144lnxdx1x4xdlnx解：原式24 1lnxdx2xlnx411x1dx24ln241x8ln224x12dx18ln24151 0xarctgxdx2解：原式11arctgxdx021x21x2arctgx12001x2811dx11dx20201x2- - - - - - -精选学习资料 - - - - - - - - - 81x111a r c t g x20201名师归纳总结 424第 5 页,共 20 页1602e 2 cosxdx解：原式2e2xdsin x0e2xs i n x22si n x2 e2xdx00e202e2xdc o s xe22 exco s x2202c o s x22 exdx0e2402e2 c o s x d x故22 excosxdx1e205170xsinx2dx解：原式0xsinx2dx0x21cos2xdx210x2dx10x2c o s 2x d x2213 x0102 xdsi n 2x6431x2s i n 2x00s i n 2x2x d x64310xdc o sx6431xc o sx00c o sx d x364618esinlnxdx1- - - - - - -精选学习资料 - - - - - - - - - 解：原式xsinlnxeexcoslnx1dx11x19故ee sin1ecoslnxdxxsinlnx1dx1esin1xcoslnxee11xe sin1e cos 11e 1sinlnxdxsinlnxdxesin1cos11122cosx3 cosxdx4解：原式2cosx12 cosxdx4名师归纳总结 20410c o ss i n xdx2c o ss i n x d x第 6 页,共 20 页042c o s3042c o s322233044233sinxxdx0sin4sinx1sinxdx解：原式01sin2xx1dx4s i n xtg2xdx2 c o s x04dc o s042 se c2 c o sx0210114t g xx4 0242c o s0xsinxxdxcos2- - - - - - -精选学习资料 - - - - - - - - - 解：令x2t,就原式22tsin2tdt2 012x1dx212 cos2t2212c o s1tc o stdt2 s i nt2 s i n21c o stdta r c t g s i n102 si n41xdx222xln1x01ln1xdx2解：原式201x2x1x2ln1x112 x1x12221x0021xx21l n 31l nxx21dx28021ln31dx1xdx122280011ln311lnx12822x1013ln328 B 一 解答1求由yt edtxcos tdt0所打算的隐函数y 对 x的导数dy ；dx00解：将两边对 x 求导得名师归纳总结 eydyc o sx0第 7 页,共 20 页dxdycosxdxey- - - - - - -精选学习资料 - - - - - - - - - 2当 x为何值时,函数Ixxtet dt有极值？0名师归纳总结 解：Ixxex2,令Ix0得x0第 8 页,共 20 页当x0时,Ix0当x0时,Ix0当x0时,函数Ix有微小值；3dcosxcost2dt；dxsinx解：原式daxcost2dtacostcost2dtdxsindasinxcost2dtacosxcost2dtdxc o s2 s i nxs i n x2 c o s c o sxc o s xc o s2 s i nc o sc o s2 c o sxs i n xc o s2 s i nxc o ss i n xc o s2 s i nxs i n xc o sc o s2 s i nx4设fxxx,1x1,求2fxdx；12,x102解：2fxdx1x1dx21x2dx00121x2x11x328206135x limxarctgt2dt；01x2解：x limxarctgt2dt型x lim1arctgx22x0x21x21122x limx212 a r c t g xx limx11x2 a r c t g xx2x- - - - - - -精选学习资料 - - - - - - - - - x l i m11 2arctgxx224名师归纳总结 6设fx1sinx ,0x,求xxftdt；01sintdtx0dt1第 9 页,共 20 页200,其它解：当x0时,xxftdtx0 dt000当0x时,xx1sintdt1cosx022当 x时,xxftdt0ftdtxftdt020 ,当0 时时；故x11cosx,当 0x2,1当x时2fx1 dx；7设fx11x,当x0 时,求11当x0 时0x e解：fx11,11,当x1 时x1当x1 时1dxx e2fx1dx11dx121100ex1x1 1ex1xex1dx12dx01e11x1ln1x e11ln20ln1e8lim n1n2nn2；n2解：原式lim n12n1nnnn- - - - - - -精选学习资料 - - - - - - - - - lim nini11xdx21nn03k9求lim nkn1ne n；2knenk解：原式lim nkn11en2k1fx；enn11exxdxx 1a r c t a n 0a r c t a n 402 e10设fx是连续函数,且fxx21ftdt,求0解：令1ftdtA,就fxx2 ,0从而1fxdx1x2Adx12A002即A12A,A122fxx111如2ln2dt16,求 x ；xte解：令t e1u,就tln1u2,dt12 u2duu当t2ln2时,u3当tx时,ux e12ln2dt1312uduu2arctgu3x1xete x1u2e23a r c t g ex16从而xln212证明：2 e11ex2dx2；2212名师归纳总结 - - - - - - -第 10 页,共 20 页精选学习资料 - - - - - - - - - 证：考虑1,1上的函数yex2,就22yx2xex2,令y0得xy01,且yex 2在x1处取最小值e11, 0时,y0当2当x,01时,y012yex 2在x0处取最大值221e1dx1ex 2dx故2 122 1；2 11 dx222即2e11ex2dx22212名师归纳总结 13已知lim xxaxa4 x2e2xdx,求常数 a ；第 11 页,共 20 页xa解：左端lim x12 axe2axa右端a22 xe2xd2xa2x2de2x2x2e2xaa2xe2xdx2 a2e2a2a2 x d ex2a2e2a2xe2xaae2xdx2 a22 a1e2a2 a22 a1e2ae2a解之a0或a1；14设fx1xx2,x0,求3fx2 dx；e,x01解：令x2t,就- - - - - - -精选学习资料 - - - - - - - - - 3fx2dx1ftdt01t2dt1etdt7111103e15设fx有一个原函数为1sin2x,求2xf2xdx；3fxdx最0解：令2xt,且fx1sin2xsin2x2xdx0tft1dt10fttdt2xf16设022410tdft1tft00ftdt44xf1tsi n 2t012 si nt004axblnx,在,13上fx0,求出常数 a,b 使1小；名师归纳总结 y解：当3fxdx最小,即3axblnxdx最小,由fxaxbln x0知,第 12 页,共 20 页11axb在ylnx的上方,其间所夹面积最小, 就yaxb是ylnx的切线,而y1 , 设 切 点 为 xx 0lnx 0, 就 切 线y1xx 0lnx 0, 故a1,x0x 0bln0x1；于是I3axblnxdxax2bx33lnxdx12114 a21lna3 1lnxdx令I a420得a1a2从而x 02,bln21又I a20,此时3fxdx最小；a2117已知fxex2,求1fxfxdx；0解：fx2 xex21fxfxdx1fxd fx1fx210020- - - - - - -精选学习资料 - - - - - - - - - 12xex2212e22018设fxx2x2fxdx21fxdx,求fx；2A00解：设1fxdxA,2fxdxB,就fxx2Bx001B2AA1fxdx1x2Bx2Adx10032c o s x d xB2fxdx2x2Bx2Adx82B4A003解得：A1,B4,于是33fxx24x2dcosx33190fcosxcosxfcosxsin2xdx；解：原式0fcosxcosxdx0sinfxcosx0fc o s0fc o sc o s xdxs i n xfc o s00f20设xx0时,Fxdtxx2t2f2tdt的导数与2 x 是等价无穷小,试求00；x2t22xftfxtdtlim x 0解：lim x 0003 xx3名师归纳总结 故flim x 02xftdtt2lim x 02xf,求,0xd2yt2；第 13 页,共 20 页0xx2 f0110221设xcost21cos ududy ,dxytcost2dx212u- - - - - - -精选学习资料 - - - - - - - - - 解：dyytcost2tsint22 t2212cost22 tttdxx tsint2 td2yt2ttt22 t12t2121dx2xs i n t22习题 5.4 一 略二 11x2dx1x2dx2011dx1x4解：原式0x 121x4x2x220x1dx112x2x2arctgx1 x2220名师归纳总结 22lnsinxdxsinxcosxdx令x2t24ln2lnsintlncos tdt第 14 页,共 20 页0解：原式2ln202202ln224lnsintdt4lnc o s tdt00t2u2ln2204lnsintdt2lnsi n udu42ln222lns i n tdt0- - - - - - -精选学习资料 - - - - - - - - - 故2lnsinxdx2ln203010xdx 2 1x0xtdtt01xdxx解：令x1,就 tdx1dtt2原式01t1dtt01t 221t212t2tdxdx1x21x0121xx21故00112dxa r c t g x 02xdx1x21x4自测题名师归纳总结 1f1 设fx是 任 意 的 二 次 多 项 式 ,gx是 某 个 二 次 多 项 式 , 已 知第 15 页,共 20 页xdx1f04f1f1,求bgxdx；062a解：设xbata,就dtIbgxdx1gbatabaa0ba1gbatadt,f 1gb0令gbataft于是f 0ga,f1gb2a2由已知得Ib6aga4gb2agb- - - - - - -精选学习资料 - - - - - - - - - 2设函数fx在闭区间a,b上具有连续的二阶导数,就在a,b内存在,使得bfxdxbafa2b1ba3f；a24证：由泰勒公式fxx0fxx0a ,fx0xx0f.2xx02其中,b,位于x 与 x 之间；两边积分得：名师归纳总结 bfxdxbfx0dxbfx 0xx 0dxbf.2xx 02dx第 16 页,共 20 页aaaabafx0fx0bx 02ax 02f6bx03ax 032令x0a2b,就ba2b2aa2b2bfxdxbafa2bfa2b1a2f6ba2b3aa2b3a,b；bafa2b1ba3f,243 fx在a,b上 二 次 可 微 , 且fx0,fx0； 试 证bafabfxdxbafb2fa；知fx是严格增及严格凹的,a证明：当xa ,b时,由fx0,fx0从而fxfa及fxfafbfaxaba故bfxdxbfadxbafaaabfxdxbfafbfaxadxaababafafbfa1ba2ba2bafb2fa- - - - - - -精选学习资料 - - - - - - - - - 4设函数fx在a,b上连续,fx在a,b上存在且可积,fafb0,名师归纳总结 试证fx1bfxdxaxb；ftdt第 17 页,共 20 页2a证明：由于在a,b上fx可积,故有bfxdxxftdtbftdtaax而fxxftdt,fxbftdtax于是fx1xftdtbftdt2axfx1xftdtbftdt1b2ax2a5设fx在0 1,上连续,1fxdx0,1xfxdx1,求证存在一点 x ,000x1,使fx4；证：假设fx4,x1,0由已知1fxdx0,1xfxdx1,得00dx11xfxdx11fxdx1x1fx020021x1fxdx41x1dx02024；从而1fxdx4或4 ,411xdx1x1dx1212202故1x1fxdx41x1dx0022从而1x1fx4dx002fx40由于fx在0 1,连续,就fx4或fx0这与1fxdx0冲突；故fx4；0- - - - - - -精选学习资料 - - - - - - - - - 6设fxx可微,f0F0,f02f1,Fxx 0 tfx2t2dt,求lim x 0Fx；x4解：令2t2u,就x1xudu,明显Fxxfx220名师归纳总结 于是lim x 0Fxlim x 0Fxlim x 0fx2lim x 0fx22f0b1 4f01；第 18 页,共 20 页x44 x342 x4x047 设fx在a,b上 连 续 可 微 , 如fafb0, 就b42bfxdxmaxa x bfx；aa,b上均满意拉证：因fx在a,b上连续可微, 就fx在a ,a2b和a2格朗日定理条件,设Mamaxfx,就有bf
