Приложения определенного интеграла. Площадь фигуры

Решите задачу:

$x^2-3x+2=3x-1, \\ x^2-6x+3=0, \\ D_1=6, x=3\pm\sqrt{6}, \\ S= \int\limits^{3+\sqrt{6}}_{3-\sqrt{6}} {3x-1-(x^2-3x+2)} \, dx = \int\limits^{3+\sqrt{6}}_{3-\sqrt{6}} {-x^2+6x-3} \, dx = \\ = (-\frac{1}{3}x^3+3x^2-3x)|^{3+\sqrt{6}}_{3-\sqrt{6}} = -\frac{1}{3}(3+\sqrt{6})^3+3(3+\sqrt{6})^2-3(3+\sqrt{6}) + \\ +\frac{1}{3}(3-\sqrt{6})^3-3(3-\sqrt{6})^2+3(3-\sqrt{6}) =$
$\frac{1}{3}(3-\sqrt{6}-(3+\sqrt{6}))((3-\sqrt{6})^2+(3-\sqrt{6})(3+\sqrt{6})+(3+\sqrt{6})^2) + \\ + 3(3+\sqrt{6}-(3-\sqrt{6}))(3+\sqrt{6}+(3-\sqrt{6}))+3(3-\sqrt{6}-(3+\sqrt{6})) = \\ = -\frac{1}{3}\cdot2\sqrt{6}\cdot(9-6\sqrt{6}+6+9-6+9+6\sqrt{6}+6) + \\ + 3\cdot2\sqrt{6}\cdot6-3\cdot2\sqrt{6} = -\frac{2}{3}\sqrt{6}\cdot33 + 36\sqrt{6}-6\sqrt{6} = -22\sqrt{6}+ 30\sqrt{6} = \\ = 8\sqrt{6}$