设函数 f(x)=$(x+pi/2,x)|sin t| dt
证明：f(x+pi)=f(x)
并求f(x)的最值

证明：
f(x+π)=∫(x+π/2+π,x+π)|sint|dt
=(令t=π+u)∫(x+π/2+π,x+π)|sint|dt
=∫(x+π/2,x)|sin(π＋u)|du
=∫(x+π/2,x )|sinu|du
=f(x)
故 f(x+π)=f(x)
即 f(x)为周期为π的函数.
故在区间[0,π]上分析函数f(x)即可.

f'(x)=|sin(x+π/2)|-|sinx|=|cosx|-|sinx|
令 f'(x)=0,有
x1=π/4,x2=3π/4
当x 增大地经过点x1=π/4 时,f(x)由正变负,故在该点f(x)有极大值
f(π/4)=(π/4,π/4 +π/2)∫|sint|dt
=(π/4,π/4+π/2)∫sintdt
=√2
当x 增大地经过点x2=3π/4 时,由负变正,故在该点f(x)有极小值
f(3π/4)=(3π/4,3π/4+π/2)∫|sint|dt
=(3π/4,π)∫sintdt + (π,3π/4+π/2)∫[-sint]dt
= 2-√2
又 f(π)=f(0)=( 0,0+π/2))∫|sint|dt
=(0,0+π/2)∫sintdt
=1
故f(x) 的极大值为√2,极小值为2-√2