∵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=...