文档介绍：[分部积分公式]定积分的分部积分公式[分部积分公式]定积分的分部积分公式篇一:定积分的分部积分公式第四节定积分的分部积分公式一、定积分的分部积分公式bbudv?uv??vdu?aaab例7(4(1用分部积分公式求下列定积分:??20xcosxdx;?edx;?021x12?lnxdx.;34解???20x2cosxdx??2x2dsinx0?2??xsinx2?2?2xsinxdx00??24??2?2xdcosx0??2???2xcosx2?2?2cosxdx040??24??2sinx2?2;??241?10exdx??dex0?ex11x??ed00?6e?1?5?exdx011?6e?1?5e0x?e?4;?1???1121?2?1?x2????arcsin?x1?212??121?2arcsinx??6???x612?12?1;?243ln2xdx44?xlnx??xdxln2x334?1??4ln24?3ln23?2??xlnx??dx3x???4ln24?3ln23?2?lnxdx344?4ln24?3ln23?2?xlnx?x?3?4ln24?3ln23?8ln4?6ln3?2.?例7(4(2试求定积分解???20sinnxdx?In??则I0??2sin0xdx??2dx????sinxdx??cosx2?1,I1??20020?而In??20sinnxdx???20sinn?1xsinxdx????20sinn?1xdcosx????sinn?1xcosx2??2cosxdsinn?100x???0?n?1??20sinn?2xcos2xdx???n?1??20sinn?2x?sin2x?1?dx????n?1??2sinn0xdx??n?1??20sinn?2xdx??n?1?In??n?1?In?2即In??n?1?In??n?1?In?2整理,得递推公式In?n?1nIn?2那么I?0?2,I1?1I11?222?2I0?2?2,I3?3I1?3?1I331?4424?4I2?4?2?2,I5?5I3?5?3?1????总之I?1n?nn?n?3n?2??31?4?2?2,In?1n?34nn?2??5?2n??3??例7(4(3求定积分解套上面公式,得??20sin5xdx.?20sin5xdx?I5?428??1?.5315二、分段函数的定积分当定积分的被积函数为分段函数时,需利用积分的区间分割性质?babfdx??cafdx??(4(4设函数f?x???解由积分的区间分割性质得?x?1,x???1,0?1?,求f?x???1x?1,x?0,1?????f?x?dx??f?x?dx??f?x?dx?1?101101显然f?x?在x???1,0?与x??0,1?均连续,通过N-L公式均可计算出其积分即?1?10f?x?dx??f?x?dx??f?x?dx?1100??x?1?dx???x2?1?dx??10?1?0?1?1??x2?x???x3?x??2??1?3?0?,比如出现第一类间断点,这时如何积分,?x2?12,x?1?例7(4(5求?g?x?dx,其中g?x???x??1,x?1?解可见g?x?在?0,1?连续,不满足N-L公式要求在?0,1?上连续的条件,这时可在?0,1?内取一点?,g?x?在?0,??上连续则?g?x?dx??0??0??x2???2x2?1dx???x?1?dx???x????0x?1?2?02由lim??101????2?3g?x?dx?lim???????1?2?2235,同理可得?g?x?dx?122得?0g?x?dx?20那么?g?x?dx?4图7(4(,一个第一类间断点不影响定积分的存在性.????sinx,x???,?????2???例7(4(6设函数f?x???,求?f?x?dx.???x,x????,??????2??解由积分的区间分割性质,有????f?x?dx?????2??f?x?dx????2?f?x?dx?2????sinxdx???xdx?2?1??cosx2?x2?2???23??2?,要先把绝对值符号去掉即化为分段函数再求积分.???例7(4(7求下列定积分:解?2?x??5?502?;?x2.?21?2?x,x????,2??x?2,x?2,??????2502??2???dx??dx02151??0222?2??9213;22??x3,x????,0???xx??3x,x?0,????????x2???x3dx??x3dx?2?2010101411??x4