y'=3x^2
在点(1,1)处
y=1,y'=3
切线方程为y-1=3(x-1)
x轴所围图形的面积
=∫[0,1][(y-1)/3+1-y^(1/3)]dy
=(y^2/6+2/3y-3/4y^(4/3))[0,1]
=1/6+2/3-3/4
=1/12