Помогите с заданием! Найдите площадь фигуры по рисунку:

Question 1

Question 2

Ответ:

$y=-0,5x^2+2x\ \,\ \ y=1\\\\-0,5x^2+2x=1\ \ \ \to \ \ \ 0,5x^2-2x+1=0\ \Big|\cdot 2\ ,\\\\x^2-4x+2=0\ \ ,\ \ D/4=2^2-2=4-2=2\ ,\\\\x_1=2-\sqrt{2}\ \ ,\ \ x_2=2+\sqrt2\ \ ,\ \ x_{versh}=-\dfrac{b}{2a}=-\dfrac{-4}{2}=2\\\\\\S=\int\limits^{2+\sqrt2}_{2-\sqrt2}\, (-0,5x^2+2x-1)\, dx =2\int\limits^{2+\sqrt2}_2\, (-\dfrac{1}{2}\,\cdot x^2+2x-1)\, dx=\\\\\\=2\cdot \Big(-\dfrac{x^3}{2\cdot 3}+x^2-x\Big)\Big|_2^{2+\sqrt2}=\Big(-\dfrac{x^3}{3}+2x^2-2x\Big)\Big|_2^{2+\sqrt2}=$

$=-\dfrac{1}{3}\cdot \Big((2+\sqrt2)^3-8\Big)+2\cdot \Big((2+\sqrt2)^2-4\Big)-2\cdot \Big((2+\sqrt2)-2\Big)=\\\\\\=-\dfrac{1}{3}\cdot \Big(8+12\sqrt2+12+2\sqrt2-8\Big)+2\cdot \Big(4+4\sqrt2+2-4\Big)-2\cdot \sqrt2=\\\\\\=-4-\dfrac{14\sqrt2}{3}+4+6\sqrt2=\dfrac{-14\sqrt2+18\sqrt2}{3}=\dfrac{4\sqrt2}{3}$

Question 3

Точки пересечения:

$\displaystyle\\-0,5x^2+2x=1\\\\-x^2+4x-2=0\\\\x^2-4x+2=0\\\\D=\sqrt{8}\\\\x_1=\frac{4-\sqrt{8}}{2}=\frac{4-2\sqrt{2}}{2}=2-\sqrt{2} \\\\x_2=\frac{4+\sqrt{8}}{2}=2+\sqrt{2}$

$\displaystyle\\S=\int\limits^{2+\sqrt{2}}_{2-\sqrt{2}} {-0,5x^2+2x-1} \, dx=\bigg(-\frac{x^3}{6}+x^2-x\bigg)\Large\mid^{2+\sqrt{2}}_{2-\sqrt{2}}=\\\\\\=-\frac{(2+\sqrt{2})^3}{6}+(2+\sqrt{2})^2-(2+\sqrt{2})-\bigg(-\frac{(2-\sqrt{2})^3}{6}+(2-\sqrt{2})^2-\\\\\\-(2-\sqrt{2})\bigg)=-\frac{8+12\sqrt{2}+12+2\sqrt{2}}{6}+4+4\sqrt{2}+2-2-\sqrt{2}-\\\\\\-\bigg(-\frac{8-12\sqrt{2}+12-2\sqrt{2}}{6}+(2-\sqrt{2})^2-2+\sqrt{2}\bigg)=\\\\\\$

$\displaystyle\\=-\frac{2(10+7\sqrt{2})}{6}+4+3\sqrt{2}-\bigg(-\frac{2(10-7\sqrt{2})}{6}+4-4\sqrt{2}+2-2+\sqrt{2}\bigg)=\\\\\\=-\frac{2(10+7\sqrt{2})}{6}+4+3\sqrt{2}-\frac{-16+10\sqrt{2}+18-12\sqrt{2}}{3}=\\\\\\=-\frac{10+7\sqrt{2}}{3}+4+3\sqrt{2}-\frac{2-2\sqrt{2}}{3}=\frac{-(10+7\sqrt{2})+12+9\sqrt{2}-(2-2\sqrt{2})}{3}=\\\\\\=\frac{-10-7\sqrt{2}+12+9\sqrt{2}-2+2\sqrt{2}}{3}=\frac{4\sqrt{2}}{3}$