如题所述
第1个回答 2012-08-19
∫(3π/4,5π/4) |sinx| dx
=∫(3π/4,π) sinx dx +(-1) ∫(π,5π/4) sinx dx
=-[cosπ-cos(3π/4)]+[cos(5π/4)-cosπ]
=-{(-1)-[-(√2)/2]}+[-(√2)/2-(-1)]
=2-√2
欢迎追问！本回答被提问者采纳
=∫(3π/4,π) sinx dx +(-1) ∫(π,5π/4) sinx dx
=-[cosπ-cos(3π/4)]+[cos(5π/4)-cosπ]
=-{(-1)-[-(√2)/2]}+[-(√2)/2-(-1)]
=2-√2
欢迎追问！本回答被提问者采纳
