定积分练习题(含答案)

11.10dexx与10de2xx相比,有关系式().(A)10dexx10de2xx(B)10dexx10de2xx(C)10dexx=10de2xx(D)210]de[xx=10de2xx定积分练习题一、单项选择题答案B由于在区间2ee),1,0(xx22.如果)(xf在]1,1[上连续,且平均值为2,则11d)(xxf=().(A)1(B)1(C)4(D)4答案:C.因为平均值211d)()1(11xxf则11d)(xxf=433.xattxdsindd2=().(A)22sinsinax(B)2cos2xx(C)2sinx(D)2sin2xx答案:C.根据变上限求导公式有:xattxdsindd2=2sinx44.baxxxdarcsindd=().(A)0(B)211x(C)xarcsin(D)abarcsinarcsin答案:A.由于定积分是一个常数,而常数的导数等于零所以baxxxdarcsindd=05答案:A.5.设)(xf是连续函数,且xxttfxFed)()(,则)(xF=().(A))()e(exffxx(B))()e(exffxx(C))()e(exffxx(D))()e(exffxx因为)(xF=)()e()e(xffxx=)()e(exffxx66.设)(xf,)(x在点0x的某邻域内连续,且当0x时,)(xf是)(x的高阶无穷小,则当0x时,xtttf0dsin)(是xttt0d)(的().答案:B.(A)低阶无穷小(B)高阶无穷小(C)同阶但非等价无穷小(D)等价无穷小因为0)(sin)(limd)(dsin)(lim0000xxxxfttttttfxxxx77.设xtttxf0de)1()(,则)(xf有().(A)极小值e2(B)极小值2e(C)极大值e2(D)极大值2e答案:A.解因为xxxfe)1()(所以,当1x时,0)(xf当1x时,0)(xf则)(xf有极小值10de)1()1(ttft=e288.设)(xf是连续函数,0a,xattfaxxxFd)()(2,则)(limxFax=().(A)2a(B))(2afa(C)0(D)不存在答案:B.因为axlim)(1)(limd)(afxfaxttfaxxa所以)(limxFax＝)(2afa.99.xttxxcos1d)1ln(lim2sin00=().(A)1(B)2(C)4(D)8答案:C.因为xxxxttxxxsin2cos2)2sin1ln(cos1d)1ln(00limlim2sin0xxxxxxxsin2sin2sin)2sin1ln(2cos200limlimxxxxxxsincossin22sin2sin20lim41010.设xxttttxF10202d11d11)(,则().(A)0)(xF(B)2)(xF(C)xxFarctan)((D)xxFarctan2)(答案:B.因为2221)1(1111)(xxxxF0111122xxCxF)(则2d11d11)1()(102102ttttFxF1111.若23de02kxx,则k=().(A)1(B)2(C)2ln(D)2ln21答案:C.因为23)1e(21e21de20202kkxkxx则k=2ln1212.积分tsxtxftI0d)(与()有关.(A)s,t,x(B)s,t(C)x,t(D)sD:答案因为sutuftxtxftIts00d1)(d)((令tx=u)suuf0d)(所以，积分tsxtxftI0d)(只与s有关1313.设302d)(xttfx,则20d)sin(cosxxxf=().(A)43(B)43(C)2(D)2答案:C.因为2020)sin(d)sin(d)sin(cosxxfxxxf2)sin(2203x1414.aaxxfxfxd)]()([=().(A)axxxf0d)(4(B)axxfxfx0d)]()([2(C)0(D)以上都不正确答案:C.因为])()([xfxfx是奇函数.所以0d)]()([aaxxfxfx1515.设2242dcos1sinxxxxM,2243d)cos(sinxxxN,)cossin(42232xxxPxd,则有().答案:D.(A)MPN(B)NPM(C)PMN(D)NMP因为根据奇偶函数的性质有:0dcosd)cos(sin2242243xxxxxN,0cos)cossin(42242232xdxdxxxxP0dcos1sin2242xxxxM,16二、填空题答案:0.1.10dlimxxnn=2.abbaxxfxxfd)(d)(=因为011limdlim10nxxnnn答案:0.因为baabxxfxxfd)(d)(二.填空题173.1010d)e(xxx=答案:e.4.]cos12cos1cos1[1limnnnnnn=答案:22.因为e10101010ed)e(xxxxx因为原式10dcos1xx22)2sin22(10x10d2cos2xx185.设2210de)(xttxf,则)(xf=答案:22)1(2xxe.6.设xttfxlnd)(102,其中)(xf为连续函数,则)5(f=答案:81.因为,根据变上限求导公式:)())(())(()(xxfdttfxa2222)1(2)1(2)1()(xxxexexf.解从xttfxlnd)(102两边求导:xxxf12)1(22221)1(xxf所以,由连续性有:81421)12(lim)5(22xffx197.设)(xf是连续函数,且0d)(sin)(xxfxxf,则)(xf=答案:12sin)(xxf因此12sin)(xxf解:令axxf0d)(则axxfsin)(两边积分得到:00)(sin)(dxaxdxxf从而有:aa2故12a208.0sin20dcos1limxxttx=答案:1.因为根据洛必塔法则xttttxxxxx0sin200sin20dcoslimdcos1lim11cos)cos(sinlim20xxx219.设.21,1,10)(xxxxf,则20d)(xxf=答案:23分段函数的定积分,一般采用分段积分则211020d1dd)(xxxxxf23121121102x2211.202dcos1sinxxx=答案:4.10.10d)ee(xxx=答案:1ee.因为.101010eed)ee(xxxxx1ee.因为.202202cosdcos11dcos1sinxxxxx4)cosarctan(20x2312.若Cxxxf22d)(,则202d)1(xxfx=答案:24.因为2022202)1(d)1(21d)1(xxfxxfx24125)1(2212022x2413.115de2xxx=答案:0.14.设)(xf是连续奇函数,且1d)(10xxf,则01d)(xxf=答案:1由于被积函数是奇函数.因为)(xf是连续奇函数,则10d)(xxf0d)(01xxf从而01d)(xxf1d)(10xxf2515.若1dln1bxx,则b=答案:0b.或eb因为Cxxxx)1(lndln所以1)1(lndln11bbxxxx从而得到0b.或eb2616.40dexx=答案:)1(22e因为)(d2ede2040txttxtx令20de2tt2020dte2e2ttt202e2e4t)1(22e27三、计算题1.设函数)(xfy在),0(内可导,且xttfxxf1d)(11)(,求)(xf.解由已知可得xttfxxfx1d)()(,两边求导:)(1)()(xfxfxxf,xxf1)(,从而Cxxfln)(,又从xttfxxf1d)(11)(可得，1)1(f故1ln)(xxf三.计算题282.求连续函数)(xf,使它满足xxxfttxfsin)(d)(10.解因为)(d)(1d)(010uxtuufxtxtfx令.所以xxxxfuufxsin)(d)(20.两边求导:xxxxxfxxfxfsin2cos)()()(2.从而)sin2cos()(xxxxf.dxxxxxf)sin2cos()(Cxxxxcos2cossinCxxxsincos293.计算10de11xx.解1010)e1(edede11xxxxxttttxe)1(de1令tttd)111(e12ln)e1ln(11lne1tt304.计算212dxxx.解012212d)(dxxxxxx212102d)()d(xxxxxx212310320123)2131()3121()2131(xxxxxx611)21312438()03121()21310(315.计算03dsinsin.解0203dcossindsinsin220dcossindcossin220sindsinsindsin2232023)(sin32)(sin32340dcossin326.求4321d)1(arcsinxxxx的值.解43214321d)1(arcsin2d)1(arcsinxxxxxxx4321)arcsind(arcsin2xx43212)arcsin(x21447337.设220de)(xttxF,试求:(1))(xF的极值;(2)曲线)(xFy拐点的横坐标;(3)322d)(xxFx之值.解(1)0)0(2e)(4FxxFx0)0()41(e2)(24FxxFx故)(xF在0x点取极小值0)0(F.(2)因为在0)0(,21Fx并且在21x两侧，)(xF异号所以，)(xFy的拐点横坐标为21x34(3)323322d2d)(4xexxxFxx)d(213244xex)(21211681324eeex358.设xttxftxF022d)()(,其中)(xf在0x点的某一邻域内可导,且0)0(f,)0(f=1,求40)(limxxFx.解)(d)(21d)()(2200222utxuufttxftxFxx令用洛必塔法则求40)(limxxFx，(注意运用条件)320404)(221lim)(limxxfxxxFxx0)0()(41lim220xfxfx41)0(41f369.计算102