解方程组y=2x²和x=y²,得两曲线的交点(0,0)(2^(-2/3),2^(-1/3))
于是,围城的面积=∫[x^(1/2)-2x²]dx
=[(2/3)x^(3/2)-(2/3)x³]│
=(2/3)(1/2)-(2/3)(1/4)
=1/6
y=2x^2 与y=x^2围城的面积 用定积分做.
解方程组y=2x²和x=y²,得两曲线的交点(0,0)(2^(-2/3),2^(-1/3))
于是,围城的面积=∫[x^(1/2)-2x²]dx
=[(2/3)x^(3/2)-(2/3)x³]│
=(2/3)(1/2)-(2/3)(1/4)
=1/6