关于变限积分问题(1)

刘丽英,郭镜明(西交利物浦大学数理教学中心,江苏苏州,215123):2009-12-18;:2010-09-13.:(1979-),,,,,,Email:[e-mail];(1942-),,,,,Email:[e-mail].,,,.;;O1721F(x)=xaf(t)dt.x,t[1].1f(t)[a,b],F(x)[a,b].2f(t)[a,b],F(x)[a,b],ddxxaf(t)dt=F(x)=f(x).1f(t)F(x),f(t):;.,f(t),f(t).22.:.,;,.2,.,f(x),(x)(x),.1ddxbxf(t)dt=-f(x).2ddx(x)af(t)dt=f[(x)](x).3ddx(x)(x)f(t)dt=f[(x)](x)-f[(x)](x).2(1)x,x.F(x)=x0(x-t)f(t)dt=xx0f(t)dt-x0tf(t)dt.(2)x,.F(x)=x0tf(t-x)dt,u=t-x,F(x)=0-x(x+u)f(u)du=x0-xf(u)du+0-xuf(u)du..(3)x,,.F(x)=10f(xt)dt.u=xt,F(x)=1xx0f(u)du.(4),x.F(x)=10|x-t|f(t)dt,(0x1).F(x)=x0(x-t)f(t)dt+1x(t-x)f(t)dt.33.11limx!0x20sin32tdtx0t(t-sint)dt.,31Vol.13,No.6Nov.,2010STUDIESINCOLLEGEMATHEMATICS=limx!02xsin32x2x(x-sinx)=limx!02x3x-sinx=limx!06x21-cosx=limx!06x212x2=12.2limx!01ex-bx+ax0sintt+cdt=1,a,b,c.limx!0x0sintt+cdt=0,limx!0(ex-bx+a)=0,a=-1,limx!0x0sintt+cdtex-bx+a=limx!0sinxx+c(ex-b)=1climx!0x(ex-b)=1,b=1,c=1.3.23dydx,x=t0(1-cosu)du,y=t0sinudu.dydx=sint2t(1-cost).4dydx,y0dtdt+xy0costdt=0.x,ey∀dydx+cos(xy)y+xdydx=0,dydx=-ycos(xy)ey+xcos(xy).5ddxx0sin(x-t)2dt.t-x=u,,ddxx0sin(x-t)2dt=ddx0-xsinu2du=sinx2.3.36[1,e],1.1例6用图:A(x)=x1lntdt+ex(1-lnt)dt,A(x)=0,A(x)=e.,.7x#0,n.f(x)=x0(t-t2)sin2ntdt1(2n+2)(2n+3).,.f(x)=(x-x2)sin2nx=0,(x#0)x=1,k!(k=1,2,∃),x%(0,1),f(x)0,x%(1,+&),f(x)∋0(x=k!),f(1)f(x)[0,+&).f(1)=10(t-t2)sin2ntdt∋10(t-t2)t2ndt=1(2n+2)(2n+3).3.4810xf(x)dx,f(x)=x21sinttdt.f(x)=2sin2xx,32201011f(1)=0,10xf(x)dx=10f(x)dx22=x22f(x)10-10x22f(x)dx=-10xsinx2dx=12(cos1-1).3,,u(x).9∀(x)=x0f(t)dt(-&,+&).f(x)=x,0∋x∋1,2-x,1x∋2,0,x0,x2..,x,t[0,x].,1x∋2,∀(x)=x0f(t)dt=10tdt+x1(2-t)dt.,∀(x)=0,x0,12x2,0∋x∋1,1-12(x-2)2,1x∋2,1,x2.3.5.,.10(x):(x)=ex+x0t(t)dt-xx0(t)dt.,x=0,((x)+(x)=ex,(0)=1.(x)=12(cosx+sinx+ex).4,,..11f(x),f(0)=1,x0,y=f(x)[0,x],.x01+[f(t)]2dt=x0f(t)dt,,f2(x)=1+[f(x)]2,f((x)-f(x)=0,f(0)=1f(x)=coshx(x0).3.612f(x),g(x)[a,b],CauchySchwartz:baf(x)g(x)dx2∋baf2(x)dxbag2(x)dx.ba[f(x)-tg(x)]dx#0,t.,.F(x)=baf(x)g(x)dx2-baf2(x)dxbag2(x)dx,,F(a)=0.F(x)F(x)∋0.F(x),F(b)∋F(a)=0..,.13f(x),g(x)[a,b].:%(a,b),f()bg(x)dx=g()af(x)dx.F(x)=xaf(t)dt∀bxg(t)dt,F(x)[a,b]Rolle.33136,:3.714[2]a0,[0,a]x04a2-t2dt+xa14a2-t2dt=0[].(A)0;(B)1;(C)2;(D)3.5,.F(x)=x04a2-t2dt+xa14a2-t2dt,,F(0)=0a14a2-t2dt0,F(a)=a04a2-t2dt0,F(x)=4a2-x2+14a2-x20,F(x)[0,a],F(x)[0,a].B.15x#0,f(x),g(x),f(x)1g(t)dt=x2-1,f(x)=[].(A)2x+1;(B)2x-1;(C)x2+1;(D)x2.6,.,f(x)g[f(x)]=2x,f(x)∀x=2x,f(x)=2x+C,f(1)1g(t)dt=0,g(x)0,f(1)=1,C=-1,f(x)=2x-1.B.43f(x),(x)=x0f(t)dt.f(x)(),(x)();f(x)T,T0f(x)dx=0,(x).f(x),t=-u,(-x)=-x0f(t)dt=x0f(u)du=(x),(x).f(x),t=-u,(-x)=-x0f(t)dt=-x0f(u)du=-(x),(x).T0f(x)dx=0,,(x+T)=x+T0f(t)dt=x0f(t)dt+x+Txf(t)dt=(x)+T0f(t)dt=(x).(x)T.[1].()[M].3.:,2009:206-212.[2],.GCT[M].:,2009:67-69.OnIntegralswithVariablelimitsLIULiYing,GUOJingMIng(MPTCofXi'anJiaotongLiverpoolUniversity,Suzhou,Jiangsu,215123,PRC)Abstract:Thispaperprovidesasummaryoftherelevantmaterialsandproblempatternsconcerningintegralswithvariablelimits.Someextensionsandtypicalexercisesarealsogathered.Keywords:integralswithvariablelimits;FundamentalTheoremofCalculus;integralequation.34201011