∫上3下-3[(9-x²)∧1/2 ] dx
问题描述：
∫上3下-3[(9-x²)∧1/2 ] dx
答
∫(-3 -> 3) √(9-x^2) dx
∵√(9-x^2) 为偶函数
∴原式=2∫(0 -> 3) √(9-x^2) dx
令x=3sint
dx=3cost dt √(9-x^2)=3cost
当x=0时t=0
当x=3时t=π/2
则原式=2∫(0->π/2) 3cost *3cost dt
=18∫(0->π/2) cos^2 t dt
=9∫(0->π/2) (cos2t +1)dt
=9∫(0->π/2) cos2t dt +9∫(0->π/2)dt
=9/2sin2t |(0->π/2) +9t |(0->π/2)
=9/2 (0-0) +9 (π/2-0)
=9π/2前面不就是？说的是偶函数