Question

помогите срочно (((((((((((((

Answer 1

$S= intlimits^2_{-4} {(-x^2-2x+8)} , dx = (- frac{x^3}{3}-x^2+8x )|_{-4}^2 = (- frac{2^3}{3}-2^2+8*2 ) - \ -(- frac{(-4)^3}{3}-(-4)^2+8*(-4) )=(- frac{8}{3} -4+16)-( frac{64}{3}-16-32 ) = \ =- 24+12 +48=36$

$S= intlimits^3_2 {x^2} , dx =( frac{x^3}{3} )|^3_2=( frac{3^3}{3} )-( frac{2^3}{3} )=9- frac{8}{3} =6 frac{1}{3}$