D: y=x^2 + 1; x+y =3
[m]S= ∫∫ _{D}dxdy= ∫_{-2} ^{1}( ∫_{x^2+1}^{3-x}dy )dx=∫_{-2} ^{1}(y)|_{x^2+1}^{3-x}dx=[/m]
[m]=∫_{-2} ^{1}((3-x)-(x^2+1))dx=∫_{-2} ^{1}(-x^2-x+2)dx=(-\frac{x^3}{3}-\frac{x^2}{2}+2x)|_{-2} ^{1}=[/m]
[m]=(-\frac{1^3}{3}-\frac{1^2}{2}+2\cdot 1)-(-\frac{(-2)^3}{3}-\frac{(-2)^2}{2}+2\cdot (-2))=[/m]
[m]=-\frac{1}{3}-\frac{1}{2}+2-\frac{8}{3}+2+4=4\frac{1}{2}[/m]