Question

Найти площадь фигуры, ограниченной линиями: y=2/x-2 и y=x^2-10x+27.

Answer 1

Объяснение:

$y=\frac{2x}{x-2}\ \ \ \ \ y=x^2-10x+27\ \ \ \ S=?\\\frac{2x}{x-2} =x^2-10x+27\\(x-2)*(x^2-10x+27)=2x\\x^3-10x^2+27x-2x^2+20x-54-2x=0\\x^3-12x^2+45x-54=0\\x^3-6x^2-6x^2+9x+36x-54=0\\(x^3-6x^2+9x)-(6x^2-36x+54)=0\\x*(x^2-6x+9)-6*(x^2-6x+9)=0\\x*(x-3)^2-6*(x-3)^2=0\\(x-3)^2*(x-6)=0\\(x-3)^2=0\\x-3=0\\x_1=3.\\x-6=0\\x_2=6.\ \ \ \ \Rightarrow\\S=\int\limits^6_3 {(\frac{2x}{x-2}-x^2+10x-27) } \, dx =\int\limits^6_3 {\frac{2x}{x-2} } \, dx-\int\limits^6_3 {(x^2-10x+27)} \, dx =\\$

$=\int\limits^6_3 {\frac{2x-4+4}{x-2} } \, dx -(\frac{x^3}{3} -5x^2+27x)\ |_3^6=\\=\int\limits^6_3 {(2+\frac{4}{x-2}) } \, dx -(\frac{6^3}{3} -5*6^2+27*6-( \frac{3^3}{3}-5*3^2+27*3))=\\=(2x+4*ln(x-2))\ |_3^6-(72-180+162-(9-45+81))=\\=2*6+4*ln4-(2*3+4*ln1)-(54-45)=6+4ln4-9= ln4^4-3=ln256-3.$

ответ: S≈2,54518 кв.ед.