微积分求弧长求y=x^2在0,1间的弧长,哪位好心人给个详解过程,谢了!1楼的谢谢了！但是我正好做到你说的最后一步卡住了，关键是根号下（1+4x^2）dx如何积出来？换元积还是如何？望给个

Question

微积分求弧长
求y=x^2在【0,1】间的弧长,哪位好心人给个详解过程,谢了!
1楼的谢谢了！但是我正好做到你说的最后一步卡住了，关键是根号下（1+4x^2）dx如何积出来？换元积还是如何？望给个详细解答，谢谢！

Answer 1

∵y=x^2,则y'=2x,√(1+y'²)=√(1+4x²)
∴根据弧长公式,得
所求弧长s=∫(0,1)√(1+y'²)dx
=∫(0,1)√(1+4x²)dx
∵设x=1/2tanθ,则dx=1/2sec²θdθ
当x=1时,θ=arctan2 ==>sinθ=2/√5
当x=0时,θ=0
∴所求弧长s=∫(0,arctan2)secθ*1/2sec²θdθ
=1/2∫(0,arctan2)sec³θdθ
=1/2∫(0,arctan2)cosθ/[(cosθ)^4]dθ
=1/2∫(0,arctan2)d(sinθ)/(1-sin²θ)²
=1/8∫(0,arctan2)[1/(1+sinθ)+1/(1-sinθ)+1/(1+sinθ)²+1/(1-sinθ)²]d(sinθ)
=1/8[ln(1+sinθ)-ln(1-sinθ)-1/(1+sinθ)+1/(1-sinθ)]|(0,arctan2)
=1/8{ln[(1+sinθ)/(1-sinθ)]+2sinθ/(1-sin²θ)}|(0,arctan2)
=1/8{ln[(√5+2)/(√5-2)]+(4/√5)/(1-4/5)}
=1/8[ln(9+4√5)+4√5].