1个回答
展开全部
∫(1->2) (2y/π)[ cos(π/2)- cos(π/2)y ] dy
=-(2/π)∫(1->2) y.cos[(π/2)y] dy
=-∫(1->2) y. dsin[(π/2)y]
= -[y. sin[(π/2)y] ]|(1->2) + ∫(1->2) sin[(π/2)y] dy
=1 -(2/π)[cos[(π/2)y] ]|(1->2)
=1 -(2/π) (-1-0)
= 1 +(2/π)
=-(2/π)∫(1->2) y.cos[(π/2)y] dy
=-∫(1->2) y. dsin[(π/2)y]
= -[y. sin[(π/2)y] ]|(1->2) + ∫(1->2) sin[(π/2)y] dy
=1 -(2/π)[cos[(π/2)y] ]|(1->2)
=1 -(2/π) (-1-0)
= 1 +(2/π)
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
