y = x^2+2x; y = x+2
x^2+2x=x+2
x^2+x-2=0
D=1+8=9
x_(1)=(-1-3)/2= - 2 или х_(2)=(-1+3)/2=1
S=∫^(1)_(-2)(x+2-(x^2+2x))dx= ∫^(1)_(-2)(2-x^2- x)dx=
=(2x-(x^3/3)-(x^2/2))|^(1)_(-2)=
=2*(1-(-2)) - ((1/3)-(-8/3))-((1/2)-(4/2))=
=6-3+1,5=4,5
см. рис.