定积分求解。
- 你的回答被采纳后将获得:
- 系统奖励15(财富值+成长值)+难题奖励20(财富值+成长值)
3个回答
展开全部
题目中的㏒x应该是lnx吧。 若是这样,则方法如下:
∵∫x(lnx)^2dx
=(1/2)∫(lnx)^2d(x^2)
=(1/2)x^2(lnx)^2-(1/2)∫x^2d[(lnx)^2]
=(1/2)x^2(lnx)^2-(1/2)∫x^2·2lnxd(lnx)
=(1/2)x^2(lnx)^2-∫x^2lnx·(1/x)dx
=(1/2)x^2(lnx)^2-∫xlnxdx
=(1/2)x^2(lnx)^2-(1/2)∫lnxd(x^2)
=(1/2)x^2(lnx)^2-(1/2)x^2lnx+(1/2)∫x^2d(lnx)
=(1/2)x^2(lnx)^2-(1/2)x^2lnx+(1/2)∫xdx
=(1/2)x^2(lnx)^2-(1/2)x^2lnx+(1/4)x^2+C,
∴∫(上限为e,下限为1)x(lnx)^2dx
=[(1/2)x^2(lnx)^2-(1/2)x^2lnx+(1/4)x^2]|(上限为e,下限为1)
=[(1/2)e^2-(1/2)e^2+(1/4)e^2]-[0-0+(1/4)]
=(1/4)e^2-1/4。
∵∫x(lnx)^2dx
=(1/2)∫(lnx)^2d(x^2)
=(1/2)x^2(lnx)^2-(1/2)∫x^2d[(lnx)^2]
=(1/2)x^2(lnx)^2-(1/2)∫x^2·2lnxd(lnx)
=(1/2)x^2(lnx)^2-∫x^2lnx·(1/x)dx
=(1/2)x^2(lnx)^2-∫xlnxdx
=(1/2)x^2(lnx)^2-(1/2)∫lnxd(x^2)
=(1/2)x^2(lnx)^2-(1/2)x^2lnx+(1/2)∫x^2d(lnx)
=(1/2)x^2(lnx)^2-(1/2)x^2lnx+(1/2)∫xdx
=(1/2)x^2(lnx)^2-(1/2)x^2lnx+(1/4)x^2+C,
∴∫(上限为e,下限为1)x(lnx)^2dx
=[(1/2)x^2(lnx)^2-(1/2)x^2lnx+(1/4)x^2]|(上限为e,下限为1)
=[(1/2)e^2-(1/2)e^2+(1/4)e^2]-[0-0+(1/4)]
=(1/4)e^2-1/4。
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询