全国计算机二级vb公共基础知识

第五章定积分一、填空题1．4222sin(31)cosd1xxxxxx2．2040sindlimxxttx3．设1sind,sincos,tuxuuyttt，那么dydx4．()fx在[,ab]上连续，则()dbaxfxx=５．2d(1)pxx当p时收敛，当p时发散６．若()d()()bafxxsfxgx，那么()d()()bagxxfxgx７．设()x在[,ab]上连续，且()ba、()ab，则()()dbaxxx二、计算题１．2120arctand1xxxx２．3e1d1lnxxx３．2d22xxx４．102d1xxx５．10d1exx６．e1/elndxx７．20sind1cosxxxx8．1122d(32)xxx９．13202d1xxx10．设2,01(),12xxfxxx，求0()()dxFxftt在0,2上的表达式，并讨论它的连续性11．设21()edxtfxt，求10()dfxx12．若函数()fx=0dsin()ddxtxtx，求()fx13．设21,0()e,0xxxfxx，求31(2)dfxx第五章定积分复习题答案一、填空题1．02．213．2t4．)()()()(afbfafabfb５．1;1６．sab７．222ba二、计算题１．解：xdxxxarctan11022=102)arctan11(arctandxxxx=101010arctanarctanarctanarctanxxdxxdxx=1021022)(arctan)1ln(214xx=3222ln42２．解：31ln1exxdx=31)ln1(ln11exdx=31ln12ex=2３．解：4324)1arctan(lim)1(1)1(1lim2200202aaaaxxdxxxdx442)1arctan(lim)1(1)1(1lim2200202bbbbxxdxxxdx02022222222xxdxxxdxxxdx４．解：1021xxdx=1)1(lim)1()1(21lim1021201022120xxdx５．解：101xedx=10)11(dxeexx=eeeeedxxxx12ln)1ln(11)1(101010６．解：dxxee/1ln1/1111/11/11lnlnlnlnlnlneeeeeexxdxxxxdxxxdxxdx=e22７．解：20cos1sindxxxx=2020220202tan2cos2cos1sincos1dxxdxxxdxxxdxxx=202tanxxd+20202tan2tanxxdxx=28．解：1122d(32)xxx12231()44xtdttt令3-2=22131()44dttt=213ln()44tt3ln284９．解：31/202d1xxxsin326600sin(cos1)cosxttdttdt令360cos233(cos)338tt10．解：当10x时，xdttfxF0)()(=xxdtt0323当21x时，xdttfxF0)()(=61221102xdttdttx31)01()01(FF，则连续，在]20[)(xF11．解：0)1(f,10)(dxxf=eeedxxexxdfxxfxx2121)()(101010102212．解：令uxt，)(xf=0)(sinxxuuddxd=0sin)1()sin(sinxxxududxd13．解：令ux211)(duuf=eeuudueduuuu137)3()1(1001310012