∫x - 1∫ = 2 - ( x - 1)^2
∫ x - 1∫ = 2 - ( x^2 - 2x + 1)
∫ x - 1∫ = 2 - x^2 + 2x - 1
∫ x - 1∫ = - x^2 + 2x + 1
x - 1 = - x^2 + 2x + 1 или x - 1 = - ( - x^2 + 2x + 1)
x^2 - 2x - 1 + x - 1 =0 или x - 1 = x^2 - 2x - 1
x^2 - x - 2 = 0 или - x^2 + x + 2x - 1 + 1 = 0
D = b^2 - 4ac = - x^2 + 3x = 0
= 1 - 4*( -2) = 9 x^2 - 3x = 0
x1 = ( 1 + 3) / 2 = 2 x( x - 3) = 0
x2 = ( 1 - 3) / 2 = - 1 x1 = 0, x2 = 3
∫ - это у меня знак модуля