解y=x²y=4∴x=2,或x=-2∴y=4与y=x²的交点为(2,4)(-2,4)∴面积=正方形的面积-∫(-2,2)x²dx=4*4-2∫(0,2)x²dx =16-2(1/3x^3)/(0,2) =16-2(8/3-0) =16-16/3...