Ответы и объяснения

2012-04-29T14:46:48+04:00

2x-y²=0

2x=y²

x=y²/2

 

2x+y-6=0

2x=-y+6

x=-y/2+3

 

y²/2=-y/2+3

y²=-y+6

y²+y-6=0

y²-2y+3y-6=0

y(y-2)+3(y-2)=0

(y+3)(y-2)=0

y=-3 ∨ y=2

 

\\\int \limits_{-3}^2-\frac{y}{2}+3-\frac{y^2}{2}\, dx=\\ \Big[-\frac{y^2}{4}+3y-\frac{y^3}{6}\Big]_{-3}^2=\\ -\frac{2^2}{4}+3\cdot2-\frac{2^3}{6}-(-\frac{(-3)^2}{4}+3\cdot(-3)-\frac{(-3)^3}{6})=\\ -1+6-\frac{4}{3}-(-\frac{9}{4}-9+\frac{9}{2})=\\ 5-\frac{4}{3}+\frac{9}{4}+9-\frac{9}{2}=\\ \frac{60}{12}-\frac{16}{12}+\frac{27}{12}+\frac{108}{12}-\frac{54}{12}=\\ \frac{125}{12}