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1. y = x^2, y' = 2x, I = ∫<0, 1> x√[1+(y')^2]dx
= ∫<0, 1>x√(1+4x^2)dx = (1/8)∫<0, 1>√(1+4x^2)d(1+4x^2)
= (1/12)[(1+4x^2)^(3/2)]<0, 1> = (5√5-1)/12.
2. I = ∫<0, 1> (x+2x)√[1+(y')^2]dx = 3√5∫<0, 1> xdx = 3√5/2
= ∫<0, 1>x√(1+4x^2)dx = (1/8)∫<0, 1>√(1+4x^2)d(1+4x^2)
= (1/12)[(1+4x^2)^(3/2)]<0, 1> = (5√5-1)/12.
2. I = ∫<0, 1> (x+2x)√[1+(y')^2]dx = 3√5∫<0, 1> xdx = 3√5/2
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