Question

Найти площадь фигуры, ограниченной линиями
Y=-x^2+10x-16, y=2

Answer 1

Ответ:

$y=-x^2+10x-16\ \ ,\ \ y=2\\\\-x^2+10x-16=2\ \ ,\ \ x^2-10x+18=0\ \ ,\ \ D/4=25-18=7\ \ ,\\\\x_1=5-\sqrt7\ ,\ \ x_2=5+\sqrt7\\\\\displaystyle S=\int\limits^{5+\sqrt7}_{5-\sqrt7}\, (-x^2+10x-16-2)\, dx=\int\limits^{5+\sqrt7}_{5-\sqrt7}\, (-x^2+10x-18)\, dx=\\\\\\=\Big(-\frac{x^3}{3}+5x^2-18x\Big)\Big|_{5-\sqrt7}^{5+\sqrt7}=-\frac{(5+\sqrt7)^3}{3}+5(5+\sqrt7)^2-18(5+\sqrt7)+$

$\displaystyle +\frac{(5-\sqrt7)^3}{3}-5(5-\sqrt7)^2+18(5-\sqrt7)=-\frac{1}{3}\, (125+75\sqrt7+105+7\sqrt7)+\\\\\\+5(32+10\sqrt7)-90-18\sqrt7+\frac{1}{3}\, (125-75\sqrt7+105-7\sqrt7)-\\\\\\-5(32-10\sqrt7)+90-18\sqrt7=-\frac{164\sqrt7}{3}+100\sqrt7-36\sqrt7=\frac{28\sqrt7}{3}$