Question

Интеграл -3,1=(x^2+4x+4)dx

Answer 1

$intlimits^1_{-3} {(x^2+4x+4)} , dx =( frac{1}{3} x^3+2x^2+4x)|_{-3}^1=$
$=( frac{1}{3} +2+4)-( frac{1}{3} (-3)^3+2(-3)^2+4(-3))= frac{19}{3}+frac{27}{3}-6=frac{28}{3}=9frac{1}{3}$