Question

计算曲线y=lnx从x=根号3到x=2的弧长？

如图

Answer 1

题目有问题？详细过程如图rt……希望能帮到你解决问题

Answer 2

y = lnx, y' = 1/x,
s = ∫<√3, 2>√(1+1/x^2)dx = ∫<√3, 2>[√(x^2+1)/x]dx,
x = tanu
s = ∫<π/3, arctan2>[(secu)^3/tanu]du
= ∫<π/3, arctan2>du/[sinu(cosu)^2]
= -∫<π/3, arctan2>dcosu/[(sinu)^2(cosu)^2]
= -∫<π/3, arctan2>dcosu/{[1-(cosu)^2](cosu)^2}
= -∫<π/3, arctan2>{1/(cosu)^2+1/[1-(cosu)^2]}dcosu
= ∫<π/3, arctan2>{-1/(cosu)^2-(1/2)[1/(1+cosu)+1/(1-cosu)]}dcosu
= [1/cosu-(1/2)ln(1+cosu)/(1-cosu)]<π/3, arctan2>
= √5-ln(√5+1)+ln2-2+(1/2)ln3
你可自己算一下。怀疑给定答案选项有误。