Question

Приложения определенного интеграла. Площадь фигуры

Answer 1

$x^2-3x+2=3x-1, \ x^2-6x+3=0, \ D_1=6, x=3pmsqrt{6}, \ S= intlimits^{3+sqrt{6}}_{3-sqrt{6}} {3x-1-(x^2-3x+2)} , dx = intlimits^{3+sqrt{6}}_{3-sqrt{6}} {-x^2+6x-3} , dx = \ = (-frac{1}{3}x^3+3x^2-3x)|^{3+sqrt{6}}_{3-sqrt{6}} = -frac{1}{3}(3+sqrt{6})^3+3(3+sqrt{6})^2-3(3+sqrt{6}) + \ +frac{1}{3}(3-sqrt{6})^3-3(3-sqrt{6})^2+3(3-sqrt{6}) =$
$frac{1}{3}(3-sqrt{6}-(3+sqrt{6}))((3-sqrt{6})^2+(3-sqrt{6})(3+sqrt{6})+(3+sqrt{6})^2) + \ + 3(3+sqrt{6}-(3-sqrt{6}))(3+sqrt{6}+(3-sqrt{6}))+3(3-sqrt{6}-(3+sqrt{6})) = \ = -frac{1}{3}cdot2sqrt{6}cdot(9-6sqrt{6}+6+9-6+9+6sqrt{6}+6) + \ + 3cdot2sqrt{6}cdot6-3cdot2sqrt{6} = -frac{2}{3}sqrt{6}cdot33 + 36sqrt{6}-6sqrt{6} = -22sqrt{6}+ 30sqrt{6} = \ = 8sqrt{6}$