ПОМОГИТЕ РЕШИТЬ
x^2-2=3x+2 x^2-3x-4=0 x_(1)=-1; x_(2)=4 S= ∫ ^(4)_(-1)(3x+2-(x^2-2))dx=∫ ^(4)_(-1)(3x+2-x^2+2))dx= =∫ ^(4)_(-1)(4+3x-x^2)dx=(4x+3*(x^2/2)-(x^3/3))|^(4)_(-1)= =4*(4-(-1))+(3/2)*(4^2-(-1)^2)-(1/3)*(4^3-(-1)^3)= =20+(3/2)*15 -(1/3)*65=