-x^2-2x+8=0 x^2+2x-8=0 D=4+32=36 x_(1)=-4; x_(2)=2 [m]S= ∫^{2}_{-4}(-x^2_2x+8)dx=(-\frac{x^3}{3}-x^2+8x)|^{2}_{-4}=(-\frac{2^3}{3}-2^2+8\cdot 2)-(-\frac{(-2)^3}{3}-(-4)^2+8\cdot (-4))=...[/m]