5.2微积分基本公式-习题

142于是得y'(0)=cos0=1，y'()=cos=。442dx0【解】djx21+t2dt=1+(x2)2d(x2)=2x1+x4。dx0dxdxx2x2xdxxdx1dxdxsinxdxsinxdxsinx0=dj0cos(t2)dt+djcosxcos(t2)dtdxsinxdx0dx0dx0dxdxdxlnxtdxlnxtdxlnxt1tdxlnxtdx1tdx1tdx1t2=1d(lnx)+1d(x2)lnxdxx2dx=.+.=.+.2xlnxxx2=+=(2=+=(2)xlnxxxlnx。0dx000000000dx00dx，得=，dxey，得=，dxeydxdx000000dxey得==得==00dx【解】问题是由参数方程求导【解法一】dycosudu=cost=cott。dxdxdjtsinudusintdtdt0【解法二】dx=dj0tsinudu=sintdt=sint=cott。03=.==.=⑴lim0；x0x0【解】这是“”未定型极限，应用洛必达法则，得0x0xx01⑵lim0；x0x20【解】这是“”未定型极限，应用洛必达法则，得0x0x2x02x1=lim1+x2x02112----应用洛必达法则----再次应用洛必达法则⑶lim0；x0x200+(x2)2(x2)'+(x2)2(x2)'lim0=limx0x0x0=limx0x----应用洛必达法则----完成求导(x2)'----整理0004x)0jxte2t2dtx)0xe2x2x)0jxte2t2dtx)0xe2x20x)0xe2x2----应用洛必达法则----完成求导djxet2dtdx0x)0xex2=lim2ex2----再次应用洛必达法则x)0ex2+2x2ex2----分子分母同消去ex2200x1x41x433x31332384432432212714432635==。==。3336630a2+x2a201+(x)2a01+(x)2aaaaa0aaa33aa1x2+11x2+11x2+11x2+11401+x01+x01+x01+x00002020220244841110101062422420【解】0【解】0冗0冗冗0【解】0【解】000冗冗0022冗22|l2x2,001012206126303x=000303x003033x=126x=1dt0010130213226700101x302132626x=2|l2-6111由于初等函数x3在[0,1)内连续，初等函数x2-在(1,2]内连续，故要讨论326由于limC(x)=lim由于limC(x)=limx3=，limC(x)=lim(x2-)=，x)1-x)1-33x)1+x)1+263且C(1)=且C(1)=(x2-)=，26x=13，求C(x)=jxf(t)dt在(-0002202002"202x0jfxdxj801+x20001+x20000404040440030000000000000000=1x32[j2f(x)dx].1x22+[2j1f(x)dx].x23002000000300移项、整理得j2f(x)dx=8+4j1f(x)dx，将其代入题目已知式，得0930fxxxjfxdx2j1f(x)dx，930030再对上式的等号两端在区间[0,1]上积分，得009000330900303930移项、整理得0930039最后得f(x)=x28x(4x2).1=x24x+2。933331tx1tx1t于是又得f'(1)=ln(1+)=xlnx+1，x1xx从而有[f(x)+f(1)]'=f'(x)+f'(1)1=ln(1+x)+xlnx+1.1xxx2xxx2xxxx+1xxxxx+1xxxx2得到f(x)+f(1)=1ln2x。x200