求曲线X^(2/3)在-1到8之间的弧长,

弧长s=∫√(1+y'²)*dx
y=X^(2/3) y'=2/3*X^(-1/3) y'²=4/9*X^(-2/3)
因导数在0处不存在,分两段
∴弧长s=∫[-1,8]√(1+y'²)*dx=∫[-1,0)√(1+y'²)*dx+∫(0,8]√(1+y'²)*dx=s1+s2
令x=t^3,则t=x^(1/3),dx=3t^2dt,y'²=4/9*X^(-2/3)=4/9*t^(-2)
对于每一段,均有
s=∫√(1+y'²)*dx
=∫√(1+4/9*t^(-2))*3t^2dt
=∫1/|3t|*√(9t^2+4)*3t^2dt
=±∫√(9t^2+4)*tdt (当t>0时取+,t

应用定积分 求弧长的公式 自己做就是了 简单得很