北大版高数第十一章习题解答

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5pŒ˘61SK)SSSKKK11.11.Oe2´¨'æ5;e´æ,.(1)R+∞0xe−xdx=(−xe−x−e−x)|+∞0=1.(2)R+∞0dx(x+1)(x+2)=lnx+1x+2|+∞0=ln2.(3)1σ√2πR+∞−∞e−(x−a)22σ2dxx−a=√2σt=1√πR+∞−∞e−t2dt=1(σ0).(4)R+∞01+x21+x4dxu=x−x−1=R+∞−∞duu2+2=1√2arctanu|+∞−∞=π√2.(5)R+∞0xsinxdx=(−xcosx+sinx)|+∞0,u.(6)R+∞4dxx√x−1t=√x−1=R+∞√32dtt2+1=2arctant|+∞√3=π3.(8)R+∞−∞dxx2+2x+2dx=arctan(x+1)|+∞−∞=π.(9)R+∞0e−xcosxdx=12e−x(sinx−cosx)|+∞0=12.(10)R1−1dx√1−x2x=sint=Rπ2−π2dt=π.(11)R120dxxlnx=lnlnx|120,u.(12)R120dxxln2x=−1lnx|120=1ln2.3.e¨'æ5.(1)R+∞1dxx2+3√x4+3.ˇǑ1x2+3√x4+31x2,¨'R+∞1dxx2´æ,¨'´æ.(2)R+∞2dx2x+3√x2+1+6.ˇǑlimx→+∞12x+3√x2+1+6/1x=12,¨'R+∞2dxxu,¨'u.(3)R20dx3√x+34√x+x3.ˇǑlimx→013√x+34√x+x3/14√x=13,¨'R20dx4√x´æ,¨'´æ.(4)R10dx3√1−x4.ˇǑlimx→113√1−x4/13√1−x=limx→113√1+x+x2+x3=13√4,¨'R10dx3√1−x´æ,¨'´æ.(5)R10sinxx3/2dx.ˇǑlimx→0sinxx3/2/1x1/2=1,¨'R10dxx1/2´æ,¨'´æ.(6)R+∞0sinxxdx.limx→0sinxx=1,¨' :.ˇǑx→+∞,…Œ1xN“u0,|RA0sinxdx|≤2,)|XO{,¨'´æ.(7)R+∞1xαe−x2dx(α0).ˇǑlimx→+∞xαe−x2e−x22=0,¨'R+∞1e−x22dx´æ,¨'´æ.(8)R10lnx1−xdx.limx→1lnx1−x=limx→11x−1=−1,1·:.ˇǑlimx→0lnx1−x/lnx=1,¨'R10lnxdx=(xlnx−x)|10=−1´æ,¨'´æ.(9)Rπ20dxsin2xcos2x.0π2·:.ˇǑlimx→01sin2xcos2x/1x2=1,¨'Rπ20dxx2u,¨'u.14.e¨'·Ø´æ·^´æ.(1)R+∞0√xcosxx+3dx.).nǑKŒ,R(n+1)πnπ√x|cosx|x+3dx≥R(n+1)πnπ√nπ|cosx|(n+1)π+3dx=2√nπ(n+1)π+3.Rnπ0√x|cosx|x+3dx≥n−1Pk=02√kπ(k+1)π+3.⁄–¨'RA0√x|cosx|x+3dx .,¨'Ø´æ.ˇǑ(√xx+3)′=3−x2√x(x+3)2,⁄–x¿',…Œ√xx+3N4~.dux→+∞,√xx+3→0,qˇǑ|RA0cosxdx|≤2,d)|XO{,¨'´æ.n⁄ª,¨'^´æ.(2)R+∞1cos(3x+2)√x3+13√x2+1dx.).ˇǑ|cos(3x+2)√x3+13√x2+1|≤1x32x23,¨'R+∞1dxx32x23´æ,¨'Ø´æ.5.Qª’u¨')|XO{9CO{.(i))|XO{:…Œf(x),g(x)3(a,b]k‰´,a·§:.3~ŒM0,Ø0εb−a,|Rba+εf(x)dx|≤M.q…Œg(x)3x→a+0N“u0,K¨'Rbaf(x)g(x)dx´æ.(ii)CO{:…Œf(x),g(x)3(a,b]k‰´,a·§:.e¨'Rbaf(x)dx´æ,…Œg(x)3(a,b]Nk.,K¨'Rbaf(x)g(x)dx´æ.SSSKKK11.21.e…Œ4.(1)limk→0Rπ20dϕ√1−k2sin2ϕ.–ϕ,kǑC…Œ1√1−k2sin2ϕ3[0,π2]×[−12,12]ºY,ˇd“=Rπ20limk→01√1−k2sin2ϕdϕ=Rπ20dϕ=π2.(2)limk→1−0Rπ20p1−k2sin2ϕdϕ.–ϕ,kǑC…Œp1−k2sin2ϕ3[0,π2]×[0,1]ºY,ˇd“=Rπ20limk→1−0p1−k2sin2ϕdϕ=Rπ20cosϕdϕ=1.(4)limα→0R1+α0dx1+x2+α2.…Œ11+x2+α23†¡ºY,⁄–limα→0R10dx1+x2+α2=R10dx1+x2=π4.qˇǑ|R1+α1dx1+x2+α2|≤|R1+α1dx|=|α|,⁄–limα→0R1+α1dx1+x2+α2=0.n⁄ª,“=π4+0=π4.(5)limy→0R10exsinxyy+1dx.–x,yǑC…Œexsinxyy+13[0,1]×[−12,12]ºY,ˇd“=R10limy→0exsinxyy+1dx=R100dx=0.2.e…Œ…Œ.(1)g(y)=Ra+kya−kyf(x)dx,¥f(x)3(−∞,+∞)ºY.).g′(y)=kf(a+ky)+kf(a−ky).(2)g(y)=Rcosysinyey√1−x2dx,0≤y≤π2.2).g′(y)=−sinyeysinx−cosyeycosy+Rcosysiny√1−x2ey√1−x2dx.(3)g(y)=Ry0ln(1+xy)xdx,0y+∞.).g′(y)=ln(1+y2)y+Ry011+xydx=ln(1+y2)y+ln(1+xy)y|y0=2ln(1+y2)y.(4)g(y)=Ry20sin(x2+y2)dx,−∞y+∞.).g′(y)=2ysin(y4+y2)+Ry202ycos(x2+y2)dx.3.|^¨'eŒ{e¨'.(1)g(a)=Rπ20arctan(atanx)tanxdx,−∞a+∞.).(x,a)→(0,a0),arctan(atanx)tanx=a·arctan(atanx)atanx→a0.(x,a)→(π2,a0),dY%‰n,arctan(atanx)tanx→0.ˇd¿‰´,–x,aǑC…Œarctan(atanx)tanx3[0,π2]×RºY.⁄–g′(a)=Rπ20∂∂aarctan(atanx)tanxdx=Rπ20dx1+a2tan2x.a0a6=1,g′(a)t=tanx=R+∞0dt(1+a2t2)(1+t2)=11−a2(arctant−aarctanat)|+∞0=π2(a+1).5¿g′(a)·??ºY…Œ,g′(a)=π2(|a|+1).qˇǑg(0)=0,¨'g(a)=π2sgna·ln(|a|+1).(2)g(a)=Rπ20ln1+acosx1−acosx·dxcosx,−1a1.).(x,a)→(π2,a0),acosx→0,ln1+acosx1−acosx·1acosx→2.¿‰´,–x,aǑC…Œln1+acosx1−acosx·1cosx3[0,π2]×(−1,1)ºY.⁄–g′(a)=Rπ202dx1−a2cos2xt=tanx=R+∞02dt1−a2+t2=π√1−a2.ˇǑg(0)=0,⁄–g(a)=πarcsina.(3)I(a)=Rπ20ln(a2sin2x+cos2x)dx,a6=0.).…Œln(a2sin2x+cos2x)3a6=0ºY,⁄–I′(a)=Rπ202asin2xdxa2sin2x+cos2x.a0a6=1,I′(a)t=tanx=2aR+∞0t2dt(1+a2t2)(1+t2)=2aa2−1(arctant−1aarctanat)|+∞0=πa+1.5¿I′(a)3a6=0ºY,“I′(a)=πa+13a=1?‰⁄Æ.qˇǑI(1)=0,⁄–a0I(a)=πlna+12.qˇǑI(a)·…Œ,⁄–I(a)=πln|a|+12.SSSKKK11.31.?e¨'3‰«m´æ5.(1)R+∞0sintx1+x2dx,(−∞t+∞).).|sintx1+x2|≤11+x2,¨'R+∞0dx1+x2´æ,M-O{,¨'´æ.(2)R+∞0e−t2x2dx,(0t0t+∞).).tt00,|e−t2x2|≤e−t20x2,¨'R+∞0e−t20x2dx´æ,M-O{,¨'3«m(t0,+∞)´æ.(3)R+∞0e−αxsinxdx,(i)(0α0≤α≤+∞),(ii)(0α≤+∞).3).(i)x∈[0,+∞)0α0≤α≤+∞,|e−αxsinx|≤e−α0x,¨'R+∞0e−α0xdx´æ,M-O{,¨'3«m[α0,+∞)´æ.(ii)α0,R+∞Ae−αxsinxdx=−e−αx(αsinx+cosx)1+α2|+∞A=e−αA(αsinA+cosA)1+α2.KMıo,α=1M9A=2kπMA2M,|R+∞Ae−αxsinxdx|e−122.⁄–¨'3«m(0,+∞)´æ.(4)R+∞1e−bxcosx√xdx,(0≤b+∞).).e−bx·xN…Œ,|e−bx|≤1.¨'R+∞1cosx√xdx´æ.dCO{,¨'3«m[0,+∞)´æ.(5)R+∞0te−txdx,(i)(0c≤t≤d),(ii)(0t≤d).).(i)x∈[0,+∞)0c≤t≤d,|te−tx|≤de−cx,¨'R+∞0de−ctdx´æ,M-O{,¨'3«m[c,d]´æ.(ii)t0,R+∞Ate−txdx=−e−tx|+∞A=e−tA.KMıo,t=1M,A=2M,R+∞Ate−txdx=e−12.⁄–¨'3«m(0,d]´æ.(6)R10dxxt,(0t≤b1).).x∈(0,1]0t≤b1,|1xt|≤1xb,¨'R10dxxb´æ,M-O{,¨'3«m(0,b]´æ.2.e¨'.(1)R+∞0e−ax−e−bxxdx,(0ab).).R+∞0e−ax−e−bxxdx=R+∞0dxRbae−txdt.ˇǑ¨'R+∞0e−txdx3«m[a,b]´æ,⁄–“=RbadtR+∞0e−txdx=Rba1tdt=lnba.(2)R10xa−xblnxdx,(a−1,b−1).).5,bab.R10xa−xblnxdx=R10dxRabxtdt.ˇǑ¨'R10xtdx3«m[b,a]´æ,⁄–“=RabdtR10xtdx=Rab11+tdt=ln1+a1+b.(3)R+∞−14e−(2x2+x+1)dx.).t=√2(x+14),“=R+∞0e−t2−78dt√2=√π2√2e−78.(5)R+∞0sinαx·cosβxxdx(α0,β0).).“=R+∞0sin(α+β)x+sin(α−β)x2xdx=π4(1+sgn(α−β)).3.¨'R+∞0e−xsintxxdx…ŒL“.).¨'PI(t).t∈[−a,a]x∈(0,+∞),|e−xsintxx|≤|e−xt|≤ae−x,¨'R+∞0ae−xdx´æ,⁄–¨'I(t)3«m[−a,a]´æ.⁄–t∈(−a,a),I′(

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