高数题！！！积分题！！！这题我不懂，老师学霸帮帮忙啊！！！急急急！！！
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y = (√x / 3)(3-x)，dy/dx = (1-x)/(2√x)，(dy/dx)² = (1-x)²/(4x)
The arc length = ∫[1，3] √[1 + (dy/dx)²] dx
= ∫[1，3] √[1 + (1-x)²/(4x)] dx
= ∫[1，3] √[(4x+1-2x+x²)/(4x)] dx
= ∫[1，3] √[(x+1)²/(4x)] dx
= ∫[1，3] (x+1)/(2√x) dx
= (1/2)∫[1，3] (√x + 1/√x) dx
= (1/2)[2/3 * x^(3/2) + 2√x] [1，3]
= (1/2)[(2/3 * 3^(3/2) + 2√3) - (2/3 + 2)]
= 2√3 - 4/3
The arc length = ∫[1，3] √[1 + (dy/dx)²] dx
= ∫[1，3] √[1 + (1-x)²/(4x)] dx
= ∫[1，3] √[(4x+1-2x+x²)/(4x)] dx
= ∫[1，3] √[(x+1)²/(4x)] dx
= ∫[1，3] (x+1)/(2√x) dx
= (1/2)∫[1，3] (√x + 1/√x) dx
= (1/2)[2/3 * x^(3/2) + 2√x] [1，3]
= (1/2)[(2/3 * 3^(3/2) + 2√3) - (2/3 + 2)]
= 2√3 - 4/3
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