Question

Помогите пожалуйста ​

Answer 1

Ответ:

$3)\ \ 6yy'=8-9x^2\ \ ,\ \ \ \ y_0=5\ ,\ x_0=2\\\\\displaystyle 6\int y\, dy=\int (8-9x^2)\, dx\ \ ,\ \ \ 3y^2=8x-3x^3+C\ \ ,\\\\\\y(2)=5:\ \ 3\cdot 5^2=8\cdot 2-3\cdot 2^3+C\ \ ,\ \ \ 75=-8+C\ \ ,\ \ C=83\\\\\\\underline{\ 3y^2=8x-2x^3+83\ }$

$\displaystyle 4)\ \ f(x)=x^2\ \ ,\ \ g(x)=6x\\\\x^2=6x\ \ ,\ \ x(x-6)=0\ \ ,\ \ x_1=0\ ,\ x_2=6\\\\S=\int\limits_0^6\, (6x-x^2)\, dx=\Big(3x^2-\dfrac{x^3}{3}\Big)\Big|_0^6=3\cdot 36-\frac{6^3}{3}=108-72=36$