Question

СРОЧНО СРОЧНО СРОЧНО ​

Answer 1

Ответ:

$y=4-x^2\ ,\ y=4-2x\\\\Tochki\ peresecheniya:\ \ 4-x^2=4-2x\ \ ,\ \ x^2-2x=0\ ,\ x\, (x-2)=0\ ,\\\\x_1=0\ ,\ x_2=2\\\\\displaystyle \int\limits^2_0\, (4-x^2-(4-2x))\, dx=\int\limits^2_0\, (-x^2+2x)\, dx=\Big(-\frac{x^3}{3}+x^2\Big)\Big|_0^2=\\\\\\=-\frac{8}{3}+4=\frac{12-8}{3}=\frac{4}{3}$