y=2x+2–x2
x2–2x–4=2x+2–x2
2x2–4x–6=0
x2–2x–3=0
D=16
x1=–1; x2=3
S= ∫ ba(f(x)–g(x)dx
S=∫ 3–1(2x+2–x2–(x2–2x–4))dx=
=∫ 3–1(6+4x–2x2)dx=(6x+2x2–(2/3)x3)|3–1=
=6·(3–(–1))+2·(32–(–1)2)–(2/3)·(33–(–1)3)=
=24+16–(2/3)·28=64/3