∫(-2~2) √(4 - x) dx = 2∫(0~2) √(4 - x) dx,偶函数 令x = 2sinz,dx = 2cosz dz = 2∫(0~π/2) (2cosz) dz = 4∫(0~π/2) 2cosz dz = 4∫(0~π/2) (1 + cos2z) dz = 4[z + (1/2)sin2z] |(0~π/2) = 4(π/2) = 2π
定积分∫根号下(4-x2)是多少?
∫(-2~2) √(4 - x) dx = 2∫(0~2) √(4 - x) dx,偶函数 令x = 2sinz,dx = 2cosz dz = 2∫(0~π/2) (2cosz) dz = 4∫(0~π/2) 2cosz dz = 4∫(0~π/2) (1 + cos2z) dz = 4[z + (1/2)sin2z] |(0~π/2) = 4(π/2) = 2π