{y = x^2 + 4x + 3
{y = x + 3.
x^2+4x+3=x+3
x^2+3x=0
x*(x+3)=0
x=0; x=-3
S= ∫_(-3)^(0)(x+3-(x^2+4x+3))dx=∫_(-3)^(0)(x+3-x^2-4x-3))dx= ∫_(-3)^(0)(-x^2-3x)dx=((-x^3/3)-3(x^2/2))|_(-3)^(0)=
=((-0^3/3)-3(0^2/2)) - ((-(-3)^3/3)-3((-3)^2/2))=-9+(27/2)=[b]9/2[/b]