将两曲线联立解得x= -1±√3,所以
面积=< -1-√3→-1+√3>∫[(2-x²)-(2x)]dx
=< -1-√3→-1+√3>∫(-x²-2x+2)dx
=[(-1/3)x³-x²+2x)]|< -1-√3→-1+√3>
=[(-1/3)*(-1+√3)³- (-1+√3)²+2*( -1+√3))]- [(-1/3)*(-1-√3)³- (-1-√3)²+2*( -1-√3))]
=4√3