数学分析数学分析PPT (13).pdf
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1、?1.?1?F?102dxxx?2?F?0cos3xdx?3?F?101nnxdx?4?0?nF?011dxxxnm?0?mn?5?dxxxF?024)1(?6?F20217cossin?xdxx?7?8?F?F?0dxexnxm0,?nm?1011)1(dxxxqnp0,?nqp?1?F?102dxxx8)21(81)3()23()23,23()1(22102121?R?RR?S?Fdxxx?2?2cos1xt?142arcsinxt?314221dtdxtt?123cos2(1)xt?F?0cos3xdx)21,41(221)1(221102143S?F?dttt?3?nxt?F?101n
2、nxdx?S?F?)11,1(1)1(110111nnndtttnnnnn?sin?4?8tan2?nx?F?011dxxxnmFF?2021122012cossin2tan2?88888dndnnmnmnm?82sin?tF?011dxxxnm?S?F?)1,(1)1(1101nmnmndtttnnmnmnmn?sin?5?xxt?1?dttdxttx2)1(1,1?1dxxxF?024)1(224sin4)43()41(41)43,45()1(104141?RR?S?F?dttt?6?xt2sin?F20217cossin?xdxx115525643474114152!3)434()43(
3、)4(21)1(2110413?2222?RRR?F?dttt?7?nxt?F?0dxexnxm)1(11011nmndtetntnm?R?F?8?nxt?F?1011)1(dxxxqnp?F?1011)1(1dtttnqnp()$%&qnpBn,1?2?()$%&R?F?nndxenx110?n1lim0?F?dxenxn?nxt?()$%&R?FF?nndttendxentxn1110110?)()1(sssR?RR1)1(11limlim0?R?()$%&?R?Fndxenxnn?3?)(sR0?sF?R01lne)(xdxxsxs?F?R01)(lne)(dxxxsnxsn1.n?10
4、(e)sxxdxs?I?IF10elnsxxxdx?F?000sS?0ss.1,0(?xxxxexsxslnln110?F?101ln0dxxxs?F?101lnexdxxxs0ss.?0sS?),1?x01lnSsxxxexx e?01eSxxdx?F?F?11lnexdxxxs0sS?F?01lnexdxxxss(0,)?)(sR0?sF?R01lne)(xdxxsxs?F?R011)1(lne)(dxxxsnxsn?110elnnsxxxdxs?I-#?,IF?F?01lnedxxxnxs?(0,)?2?)()1(sn?R0?s?F?R01)(lne)(dxxxsnxsn4?R?)(li
5、mss?1)2()1(?R?R?)(sR?Rolle?0?s)2,1(0?0 x0)(0?R x?0ln)(021?R F?xdxexsxs),(0?x?0)(?R s?)(sR),(0?x)1()()(?R?R?Rsss?)(0 xs?R!)1(nn?R?)(limss?5?FR10)(lndxx?tx?1?FR10)(lndxx?R?F10)1(lndttF?R10)1(lndxx?FR10)(ln2dxx*FF?RR?1010)sinln(ln)1()(lndxxdxxx?2lnsinln1sinln010?FF?uduxdx?FR10)(lndxx=?2ln?6?1|),(222?Jz
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