国开电大 微积分基础 形成性考核册作业14答案

微积分基础形成性考核作业（一）————函数，极限和连续21(x)f．ln(x2)x2)0x20x31解：{{f(x))，x2ln(x2)1(x)f．5x15x0x5，f(x)(5x1(x)4xx1f2．x2)ln(x2)01x20x2f(x)4x2(2,(x2)4x202x2(xx2x7f(x)f2．(xx2x7x2x16(x6f(x)x6f222220xx2(x)f(0)f(0)0222f．ex0x(xx2xf(x)f2．(xx2xx2x11(x1f(x)x1f222，2x22x3x1y．x22x3x110x1x1xy1sin11xlimxsinlimxsinlim1．．1xxxxxx1sin4xsinkxlim2k．x0sin4xsin4x4lim44xsinkxlim2k2sinkxkkx0x0kxsin3xlim2k．kxsim3x3x0sim3x3232limlimkkxk3xkx0x02eexxy2e()exeeeexxxxxy(x)yyB222xsinxy2(x)(x)sin(x)xsinxyyxsinxAy22222xx(x)xf2xx．y．轴．y轴222222x(x)xxxxf(x)(x)xf(x)f(x)x22222xx(x)xfD2．xsinxC．lnx．ln(x1x)．xx221yln(xx45x4x4x5x0且x5x4D．且．x．．x40，x5Dx5021(x)fx)．()．(0,2)(2,)2)(2,)．．ln(x0x21(x)2)(2,)D，，fx10x1x(xx1f(x)f2（）．x(xx2x(x．．(xxf(x)x(x2)C．(xx1(xx(xx2]f2(x)(x)2g(x)x，()．fx2g(x)xx，．f．f(x)lnxg(x)2lnxf(x)lnx3，g(x)3lnx．2，Dx0.1．xsinxxx)．．．xx2limx)0x0x)Cx00x2x()x0.k（fxk,x021．01．limf(x)lim(x1(0)2，fkx0x00xx2(x)0f(0)limf(x)1kBfxk,x0x0x0ex()在x0.k（fxk,x0201C．．3f(0)limf(x)lim(e2)3kDxx0x0x33x2x3(x)f）x2x2xxx3．x．．3x2xx2Ax7x23x2．x24x2x23x2(xx2)(x2)(xx114x24x2x2x2x2x25x61x2x1x25x6(xx6)(xxx671x12x2x1x1x1x29．x22x3x3x292x3(xx(xxx363x142x2x3x3x3xx226x85x4x4xx226x85x4(x2)(x4)(xx4)x22x13x4x4x4xx226x85x6．x2xx226x85x6(x2)(x4)(x2)(xx42x3x2x2x21x1．xx01x1(1x1xx(1xxlimlimx(1xxx0x0x01121x1x01x14xx041x14x(1x1xsin4x(1xlimx0x0xsin4x(1x1118limlimsin4x4x4x0x0(1x4x．x42x04xsin4x(x4(x42)(x4x0limx42x04x(x4sim4x4x4(x4xx0x05微积分基础形成性考核作业（二）————导数、微分及应用2f(x)x1在．11f(x)kf22xf(x)e(在．xf(x)ekf(0)e1x0f(x)eyx1x在1yx．21112332yxkyx222x1x111yx2y1(xx2y30212ln2x(2)(2)2ln2．x．xx2x2xyy(0)(2)(6y=xx–x–x–=f)x33xff(x)3x3ln3f2727ln3，=2x11f(x)lnxf(x)f(x)f(x)，=xx2f(x)xef(0)x．f(x)exef(x)e(exe)2exef(0)2x，xx，xxxx．y6(x0x1，)f(x)ax21)a．f(x)20x0a026y(x(D）2f(x)0yf(x)C.．f(x)ecosxf(0)=（C.2.1（Bx.-1.-2．yf(x)f(cos2x)（D．．．2f(cos2x)dxf(cos2x)sin2d2x2f(cos2x)sin2dxf(cos2x)sin2d2x．．．．ye1x2C2在xee4．4．2．．2f(x)xxf(x)（C．cosxxsinxf(x)xa3．cosxxsinx2sinxxcosx2sinxxcosx．．af(x)（C．cosx3a2．sinx6asinxD．cosx．Bf(x)xx在x．．.00f(x)xx在x.00.f(x)f(x)0.fx(B0fx．limf(x)AAf(x)00xx0fxDfx00BAxex3-xx2Affx是fxx)=0x是fxf000007ff(x)xfx)=xf00071yxey．2x111111y2xe()2e(2xe2xxxxxx2ysin4xcosxy3.y4cos4x3cosxsinx21yeyx1.x11yex12x1x2yxxlncosxy.32sinx3yxxtanxcosx2xyxy4dy.yy(x)2222(xdy)022y2x2yxyy(x)xy2xy1dy.22xy2xy12x22(y)022xy0(xy)y(xy)y1，，yyy(x)eex4dy.xy2edxedxxedy2xdx0xyyee2xxyxedy(ee2x)dxyxy，ycos(xy)e1dy．y8cos(xy)e1yy)sin(xy)ye0ysin(xy)ysin(xy)ye0y[esin(xy)]ysin(xy)yxy)yexy)yxy)yexy)y9微积分基础形成性考核作业（三）———不定积分，极值应用问题2(x)lnxf(x)xlnx2xcff2。2(x)xef(x)e2xx2。f1xex(x)dxecf(x)．．x(x)dxsin2xcf(x)2cos2xf1(x)dxxlnxcf(x)f．x(x)dxcos2xcf(x)4cos2xf．dexedxx．．x(sinx)xsinxc．．12f(x)dxF(x)cf(2xxF2x3c．1xfx)dx(x)dxF(x)cF1xc22．f22df(x)dxf(x)．f(x)dxf(x)．df(x)dxf(x)df(x)f(x)．．dxA(xxxecf(x)（f22.x2xex)2xe2e2xe2x2x22x(x)2xe2xe2xex)f2x222Axx()dfxx（.(x)xx(x0)f31233xc2xxcxxcxxcx22AA22223）d3xdxdx13dxx)dxlndd()．ln3．12．．xxxx2xx．xx（()d）101x2()()xfxc()xc()fxc(()xfxc2()dxx()()()()()xxfxdxxfxcAda2dx．x=（．a2x22xlnd．a．a2xd2xd．axcCaxx(x)e1dxe1Cf(x)（f）xx1111xx2xx211(x)e1e1()fxfBxxx2x273xxx3x．x3xxx1dxxdxsin3x3xdxxx233lnxxxc23(2xx．101211(2xd(2x(2xx(2xc101010121011(2xc11221sinxdx．x21sin111xdsind()coscxx2xxxsin2x．x11xsin2xxdcos2x(xcos2xcos2xdx)2121xcos2xsin2xc24．xexdxxexdxxde(xeedx)xeecxxxxx111．x60x60x体积V2(60)x，即：Vxx32VxdVdVdxxx02dxx04060x20x40602．216xx216648L2x32xLxxdLdx648dL21620x1812x2dxx18125(x)xefx()1()10fxfxfex（xex(x)xex12微积分基础形成性考核作业（四）———定积分及应用、微分方程21xx2)d.．xx212311121xxx(sincos2)dsincos22xxxdxx2dxx2dx1110(x4xcosx)dx______.．522(x4xcosx)dx(x4x)dxcosxdx2cosxdx2sinx2552222200222f(x)xxy。2233xdxxcyxc223321354cc2332133xy231dx3x3x．11133)12414x3xdxx3x1110dxaxa22=。0dxax2y2a2a2x20121yaada圆x22a2x2x2440dexxln(．2.dx1dexx0ln(d2dx10edx=．2．x1121120000edlimxlim(2)limelim)2e2xdxe2xdxe2x2bx22bbbbbbb,(0)1yyy.13dydxyyy得y，lnyxcy(0)1ln10cc0lnyxyy,y(0)1yexy3y0.dyy3y0，3y0，0lny3xc，dxy1yc3xye，yeec3x1c3x11y3y0yce3x(y)4xyysinx3(4)7．(y)4xyysinx4阶3(4)72xAyx2yx1y=x+3y=x+4．2．2221(2xk)dx=k（A012110．0Aeeee32xxxx1dx1dx(xcosxx(xsinxx．．．．2211af(x)f(x)dx（D）-a00a2f(x)dxf(x)dxf(x)dx．．．D．0-a-a0．sinxx（D2-220．．．2B11exxedxxdx．．x．．xx0011B1411xdx．sxxind．ed2x．．xxx0011Dlnyyyyxyee．yxysineln．yxyxyx．yx2．2xyy0CCxyxCyCy0．y．．．B）ydxdydxdydyxyxyyxysinxx(yx)；；；dxdx7ln2ee)dx．x2x013819ln2ln20ln2ee)dx)))392e2deexxxxx330015lnxedx．x115lnx15eeed5ln)lnxd5ln)5ln)xxdxxx11111112e