這是曲線積分問題；
y² = 4x ==> x = y²/4
==> dx/dy = y/2
==> dx = (y/2)*dy
在(x,y)點(diǎn)的弧長(zhǎng)微元為：
dL =√[(dx)²+(dy)²
= √[(y/2*dy)²+(dy)²]
= √(y²/4+1) *dy
L = [0,2]∫[√(y²/4+1) *dy]
作變量代換 y =2tanu ==> dy =2sec²u,原式化為：
L = [0,π/4]∫[√(tan²y+1) *2sec²udu]
= [0,π/4]∫ (2sec³udu)
= [0,π/4]∫ (2/(1-sin²u)² *cosdu)
= [0,π/4]∫ [2/(1-sinu)²(1+sinu)² dsinu]
= [0,π/4]∫ 1/2 *1/[(2+sinu)/(1+sinu)² + (2-sinu)/(1-sinu)²]*du
= 1/2* [0,π/4]∫ [1/(1+sinu)² +1/(1+sinu)+1/(1-sinu)² +1/(1-sinu)]*du
= 1/2*[ -1/(1+sinu) +ln(1+sinu) +1/(1-sinu) -ln(1-sinu)] |[0,π/4]
= 1/2 *(2√2 + ln[(2+√2)/(2-√2)]
= √2 + ln(√2+1)