求定积分 上限是π/2 下限是-π/2 （x+1）min｛1/2 ，cosx｝dx
3个回答
展开全部
min{1/2,cosx}
=1/2 -π/3≤x≤π/3
=cosx -π/2≤x<π/3或π/3<x≤π/2.
它是一个偶函数，从而xmin{1/2,cosx}是奇函数。
故∫[x=-π/2,π/2](x+1)min{1/2,cosx}dx
=∫[x=-π/2,π/2]min{1/2,cosx}dx
=1/2*∫[x=-π/3,π/3]1dx+2∫[x=π/3,π/2]cosxdx
=π/3+2sinx|[x=π/3,π/2]
=π/3+2(sinπ/2-sinπ/3)
=π/3+2-√3
=1/2 -π/3≤x≤π/3
=cosx -π/2≤x<π/3或π/3<x≤π/2.
它是一个偶函数，从而xmin{1/2,cosx}是奇函数。
故∫[x=-π/2,π/2](x+1)min{1/2,cosx}dx
=∫[x=-π/2,π/2]min{1/2,cosx}dx
=1/2*∫[x=-π/3,π/3]1dx+2∫[x=π/3,π/2]cosxdx
=π/3+2sinx|[x=π/3,π/2]
=π/3+2(sinπ/2-sinπ/3)
=π/3+2-√3
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
