Question

求由曲线y=x∧3,该曲线在点(1,1)处的切线,以及x轴所围图形的面积

Answer 1

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