-2x^2+2=-6
2x^2=8
x^2=4
x= ± 2
[m]S= ∫ ^{2}_{-2}(-2x^2+2-(-6))dx=∫ ^{2}_{-2}(-2x^2+8)dx=(-2\frac{x^3}{3}+8x)|^{2}_{-2}=[/m]
[m]=(-2\frac{2^3}{3}+8\cdot 2)-(-2\frac{(-2)^3}{3}+8\cdot (-2))=[/m]
[m]=-\frac{16}{3}+16-(\frac{16}{3}-16)=-\frac{32}{3}+32=\frac{64}{3}[/m]