2个回答
展开全部
∫(0,π/4)(cosx-sinx)dx+∫(π/4,5π/4)(sinx-cosx)dx+∫(5π/4,2π)(cosx-sinx)dx
=(sinx+cosx)|(0,π/4)+(-cosx-sinx)|(π/4,5π/4)+(sinx+cosx)|(5π/4,2π)
=sinπ/4-sin0+cosπ/4-cos0-cos5π/4+cosπ/4-sin5π/4+sinπ/4+sin2π-sin5π/4+cos2π-cos5π/4
=√2/2-0+√2/2-1-(-√2/2)+√2/2-(-√2/2)+√2/2+0-(-√2/2)+1-(-√2/2)
=4√2
解答完毕。
=(sinx+cosx)|(0,π/4)+(-cosx-sinx)|(π/4,5π/4)+(sinx+cosx)|(5π/4,2π)
=sinπ/4-sin0+cosπ/4-cos0-cos5π/4+cosπ/4-sin5π/4+sinπ/4+sin2π-sin5π/4+cos2π-cos5π/4
=√2/2-0+√2/2-1-(-√2/2)+√2/2-(-√2/2)+√2/2+0-(-√2/2)+1-(-√2/2)
=4√2
解答完毕。
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