欢迎来到天天文库
浏览记录
ID:31931359
大小:30.00 KB
页数:3页
时间:2019-01-28
《《微积分》上册部分课后习题答案》由会员上传分享,免费在线阅读,更多相关内容在行业资料-天天文库。
1、《微积分》上册部分课后习题答案习题五(A)1.求函数fx,使f′xx23x,且f10.解:f′xx25x615fxx3x26xC321523f106C0C3261523fxx3x26x32612.一曲线yfx过点(0,2),且其上任意点的斜率为x3ex,求fx.21解:fxx3ex212fxx3exC4f023C2C112fxx3ex14∫23.已知fx的一个原函数为ex,求f′xdx.22解:fxex′2xex∫f′xdx2fxC2xexCdx4.一质点作直线运动,如果已知其速度为3t2sint,初始位移为s0
2、2,求s和t的函dt数关系.解:St3t2sintStt3costCS021C2C1Stt3cost15.设lnfx′1,求fx.1x2解:lnfx′1lnfxarctanxC11x2fxearctanxC1CearctanxCgt0116.求函数fx,使f′xe2x5且f00.1x1x2111解:fxex5fxlnx1arcsinxe2x5xC1x1x2211f0000C0C2212x1fxlnx1arcsinxe5x227.求下列函数的不定积分xx2∫∫dt(1)dx(2)xat1x21∫∫xmn(3)xdx
3、(4)dx21x411sin2x(5)∫x21dx(6)∫sinxcosxdx1cos2x∫∫cos2x(7)dx(8)dxsinxcosx1cos2x∫sin(10)cos2sin2xdx∫cos2xx(9)22dxxcosx2cos2x12x1∫sin∫ee(11)dx(12)dx2xcosx2x12×8x3×5x2x15x1(13)∫8xdx(14)∫10xdxexxe-x(15)∫xdx∫(16)ex2x13xdx1x1xx211x25x(17)∫dx1x1x(18)∫x1x2dx1x21cos2x(19
4、)∫1x4dx(20)∫1cos2xsin2xdxx3x1x4x2(21)∫x1x22dx(22)∫1x2dx1335∫22解:(1)x2x2dxx2x2C351dt1∫12(2).1t12Caat12nnm∫xmdxmxmCm≠nm≠0nmn∫(3)xmdxInxCmndxxC∫m02(4)1∫x21dxx2arctanxCx2x21x21x3(5)∫x12dx3x2arctanxCsin2xcos2x2sinxcosxsinxcosx2(6)∫sinxcosxdx∫sinxcosxdx∫sinxcosxdxs
5、inxcosxCcos2xsin2x(7)∫sinxcosxdxcosxsinxdx∫sinxcosxC1cos2x∫2cos∫cos111x(8)2dx21dxtanxCx2x22cos2xsin2x11(9)∫sin2xcos2xdx2∫sinxcos2xdxcotxtanxCcosx11cos2xcosxcos2x(10)∫22dx221dx∫11xsinxsin2xC24cos2xsin2xcos2xsin2x∫∫cos1(11)22dx22dx2tanxCsinxcosxx∫(12)ex1dxexxCx
6、5x5(13)2dx3dx2x38C∫∫85ln8xx(14)2dxdx∫5∫1112x152xC52ln55ln2(15)exdxexlnxC∫1x∫2x3ex6x(16)ex6x2x3exdxexCln2lln3ln61x1x∫∫1(17)dx2dx2arcsinxC1x21x2x21(18)∫dx1x2lnx5arcsinxC5x21x2∫1(19)dxarcsinxC1x21cos2x11∫2cos∫1x(20)dx1dxtanxC2x2cos2x22xx211111∫∫1(21)dx2xdxlnxarc
7、tanxCx21x2x1x2xx41x2122x3(22)∫1x2dxx22∫21xdx32x2arctanxC8.用换元积分法计算下列各题.x4(1)∫x2dx∫(2)3x28dx.
此文档下载收益归作者所有