河海高數(shù)ch5 定積分習(xí)題答案

kFkPkIcos7x1223.JJcos5xsin2xdx=2J2cos6xsinxdx=—2J2cos6xdcosx=—21I=000I7I70龍=朽KK64．J2—xdxx===nJ:cot2tdt=J2(csc21-1)dt=Lcott-1x2血k665．j1xp+3x8d=8Jn+3x8d8=87366．7．f1x2x=tantJKJk1一cos21J1dx===4sin2tdt=4dt=0(1+x2)2002J2<cosxsinxdxKI一2J2cosx一cos3xdx=一22一cos2x3冗2=4

=30sin2f~4兀；2J2、cosxsinxdx0J"f(x)dxf=—J-bf(—t)dt=Jbf(—t)dt一bb一bf(t)dt=J0(1+12)dt+J1e-tdt-100[137121+一e—t0=24一e六、證(1)①(—x)=J-xf(t)dt令=一一0四、五、證解：Jf(x—1)dxA1J12:t3=t+—3色Jxf(u)du=①(x);得證.0Jx-f(-uu)du0:Jx+Tf(t)dt=f(x+T)一f(x)=0;dxxJx+Tf(t)dt因與=無(wú)關(guān)Jif(t)dtf為奇函數(shù)0x令x=-今-Tf(t)dt,得證．AJxf(t)dt=Jx+丁，aaEx30定積分的分部積1耳xe-xdx=一xde-x法1d1[]e一一o+J1e一心=一e+~e一xo22?fxlog2用=2f12log2xdx2=2x2log2x!_2f：詁=2_盤=2_丄4ln2f兀tsinntdt，1=一專申td3．n是正整數(shù)。cosnt=_—tcosnt如

n竺+丄Isinnt如=_還nn2°n4．5．f1|lnxIdx=一f：lnxdx+『Inxdx=ee112=_-+(1一一)+e_(e_1)=2_-eee1f1x(arctanx)2dx=f1(arctanx)2dx2=1xr°2—_f1arctanxdx+f132+—留cosntdt=一_Lm旳+J：dx+Lm乂-腫ee2(arctan肌_f1蘭竺沁1+x2arctanxdx1+x2arctan2x27?f^nQl+x2+arctan2x27?f^nQl+x2+x)dx=xln(J1+x2+x)^一f1xdx1+x2=ln(1+J2)-&11+x2=ln(1+、?2)_\2+1」°、x8?設(shè)f(3x+1)=xe2,解：令t=3x+1，求f1f(t)dt?°則x=上13故f；f(t)dt=ff—3一e6dt=3f°1ue6du=2f°1uude6=u一2f°1e6du-——Larctanx^+f1—xdx+32°f°1+x2兀2兀11兀2兀2兀ln2=_+ln(1+x2)+=_+324203216426?fesin(lnx)dx=txsin(lnx)1_fecos(lnx)dx111=esin1一txcos(lnx)1_fesin(lnx)dx11原式=—Csin1-ecos1+1)_114e_114e"6_121．2．Ex31反常積分廣811．2．Ex31反常積分廣81dx=2【“仙伍!8=兀0v'x(1+x)dx3．4．J；00x(i+x)=JTI￡-x+1卜f+8xdx=1[n(1+x2)】801+x220f￡一J0=2xl^_x+1=+8+8=ln21???J+8Xdx發(fā)散。-81+x2■1■12-■1■_2x-1_0_2x-1_^-dx=-=+8，125!1(2x-1)2dx=f02諾1產(chǎn)+￡(2^-dx=-=+8，12因此發(fā)散。課余練習(xí)一、1.1.-3e-2+丄，2.兀,3.空+丄，4.—亙兀+冬+1叭3—In遼5.TOC\o"1-5"\h\z44444442二、設(shè)F(x)=Jxf(x)dx一3fbf(x)dxa3a則F(a)=-3Jbf(x)dx<0,F(b)=3Jbf(x)dx>03a3a???F(a)?F(b)<0<r當(dāng)F(a)?F(???F(a)?F(b)<0<、當(dāng)F(a)?F(b)=0時(shí),g=a或b第六章定積分的應(yīng)用Ex32幾何應(yīng)用、解：A=J2C2-iG)dr=—03

三、解：A=2J4—（4cos20）三、解：A=2J4—（4cos20）J0=4_24解：點(diǎn)（—，0）處的法線方程為：—A=2J—眄”2）_五a=2妙_ax3_(xz—)!34a32a當(dāng)且僅當(dāng)a￡時(shí)取得最小值。四、A=J2lnxdx=2ln2一——x=ey,dV=(兀?22一兀x2)dy,V=J2氣4兀一兀x2)dy二兀Jln2(4-e2y)dy二兀(4ln2-2).002五、（—）解：五、（—）解：星形線參數(shù)方程為x=acos30y=asin30J:sin70d0_J:sin90d0=2Ka3_6428642_100丿_753一9753_=2他332兀a3—05V=2JaKy2dx=_2J0冗(a2sin60)acos20sin0d0=2Ka3J2(in70一sin90^do0K0六、解：V=k(2a)2-2Ka一J2Ka兀(2a_y)2dx0=8k2a3_J如兀￡一a(—_cost)?a(—-cost)dt0=8ka3_a3KJ如(—+cost)sin2tdt=7a3k20Ex33其它應(yīng)用、解：由題意知k=4（kg/cm）設(shè)彈簧被壓縮至xcm,如圖1Jox201Jox20-x三、解：設(shè)水的比重為yoW=J4(20_x)dx=—92(kg.cm)如圖建立—0坐標(biāo)系9(—_25)dx,W=9kyfx(—_x2)dx=225叼25254

依題意:取水面為y軸，垂直向下為x軸，建立坐標(biāo)系。貝ydF=Cx）20drF=20yfxdx=1000y階段練習(xí)（定積分及應(yīng)用）一、填空1、設(shè)f（x）=f5+12dt，則f⑴=-2邁。x2處的切線平行x軸。J和（4一x2）3dx=3兀。2、曲線f處的切線平行x軸。J和（4一x2）3dx=3兀。3、J1-dx=1-ln1+e。4、01+ex25、用定積分表示lim[1+1lim[1+1+v'n2+12n2+22+]=Jdx，其值為lnQ2+1)\n2+n20U1+x2二、計(jì)算下列定積分1．2．3．41．2．3．4．4x1t4卩一dxt=4x4J01+、x=======0匹1-4sec2x"4dtanx1dt<22J/d一=Jdx=J=J=arctan01+cos2x0sec2x+102+tan2x02+t222J11(x2J1-x2+x3cosx)dx=J11x2J1-x2dx=J專sin21cos2tdt=8一2Jj兀|x一兀|sinxdx=J(兀一x)sinxdx+亍(x一兀)sinxdx0n=Jnsinxdx一xsinxdx=2冗一2l一cos一*=0001+cos2x0sec2x+1dt=4J(t2—1+)dt=n——1+120t2+13ndtanx2+tan2x三、1、Jf(x)f"(x)dx=1f\x)df'(x)=2(f'(x))2=1f2e-x2三、1、0L」02L、'2已知f(x)=x(1-x)5+J：f(x)dx，求f(x)解：設(shè)Jf(x)dx=C，則f(x)=一(1-x)5+2C，等式兩邊在區(qū)間[0,1]上20求定積分，得C=fx(1_x)5dx+1C。011因此C=2Jx(1—x)5dx1="2J(1—t)t5dt=42?:f(x)=x(1-x)5+42四、解0當(dāng)x<0時(shí)，原式=0當(dāng)0<xV2時(shí)，原式=0訥=xxJf(tt)dt=*00xV0x2—0<xV24x-1x>2當(dāng)x>2時(shí)，原式=f-tdt+了1d=x-1022五、解：V]=丫兀(10y)dy=500兀cm3，0?:V=嶺一V2=95兀cm3,六、解：設(shè)水的比重為取y為積分變量，ye0,10V2=J兀[10(y-1)]dy=405兀cm31:.G=Vp=741?？藋Oh_2_hdW=(h-yMydyW=J(h-y)y^ydy=—h312七、解:f(x)-f(0)=f?°x(0<g<x);f(x)-