求定积分 上限是π/2 下限是-π/2 （x+1）min｛1/2 ，cosx｝dx

Question

如题所述

Answer 1

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

Answer 2

2-根号3+pi/3

Answer 3

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