这不是定积分,是多元函数极值问题。z' = 2x+2y-4 = 0, z' = 2x-2y+2 = 0, 联立解得驻点 (1/2, 3/2)A = z'' = 2, B = z'' = 2, C = z'' = -2AC - B^2 = -8 < 0, 函数无极值
1/2- sinx>0 sinx 2 0π/2) | 1/2-sinx| dx =∫(0->π/6) ( 1/2-sinx) dx - ∫(π/6->π/2) ( 1/2-sinx) dx = [x/2 +cosx]|(0->π/6) - [x/2 +cosx]|(π/6->π/2) =π/12 + √3/2 -1 - ( π/4- π/12-√3/2) =√3 -1 -π/12