из системы уравнений 2x-3y+6=0, y=0, x=3 получаем x1 = -3 ; x2 = 3
2x-3y+6=0 <=> y = 2/3 x+2
S = [-3;3] ∫ (2/3 x +2) dx = ( x^2/3 +2x) | [-3;3] =
= ( 3^2/3 +2*3) - ( (-3)^2/3 +2*(-3)) = 9 + 3 = 12
из системы уравнений 2x-3y+6=0, y=0, x=3 получаем x1 = -3 ; x2 = 3
2x-3y+6=0 <=> y = 2/3 x+2
S = [-3;3] ∫ (2/3 x +2) dx = ( x^2/3 +2x) | [-3;3] =
= ( 3^2/3 +2*3) - ( (-3)^2/3 +2*(-3)) = 9 + 3 = 12