y=2+x² y=x+4
2+x²=x+4
x²-x-2=0 D=9 √D=3
x₁=-1 x₂=2
S=₋₁∫²(x+4-(2+x²))dx=₋₁∫²(2+x-x²)dx=(2x+(x²/2)-(x³/3)) ₋₁|²=
=((2*2+(2²/2)-2³/3)-(2*(-1)-((-1)*2+(-1)²/2-(-1)³/3))=(4+2-8/3)-(-2+1/2+1/3)=
=6-(8/3)+1,5-1/3= 6+1,5-(9/3)=7,5-3=4.5.
Ответ: S= 4,5 кв.ед.