Find the length of the curve? calculus

f)x) = (3/4)x^(4/3) - (3/8)x^(2/3) + 6 --- interval (1,27)

L=square root 1 + f'(x)^2

f(x) = (x^3)/3 + 1/4x -- interval (1,2)

1) y = (3/4)x^(4/3) - (3/8)x^(2/3) + 6
==> dy/dx = x^(1/3) - (1/4)x^(-1/3).

So, √(1 + (dy/dx)^2)
= √(1 + (x^(2/3) - 1/2 + (1/16)x^(-2/3)))
= √(x^(2/3) + 1/2 + (1/16)x^(-2/3))
= √(x^(1/3) + (1/4)x^(-1/3))^2
= x^(1/3) + (1/4)x^(-1/3).

Hence, the arc length equals
∫(x = 1 to 27) (x^(1/3) + (1/4)x^(-1/3)) dx
= [(3/4)x^(4/3) + (3/8)x^(2/3)] {for x = 1 to 27}
= [(3/4) * 81 + (3/8) * 9] - [3/4 + 3/8]
= 63.

(Double checked on wolfram alpha; it does not seem that the answer is 57.)
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b) This is done similarly.

y = (1/3)x^3 + 1/(4x)
==> dy/dx = x^2 - 1/(4x^2)

So, √(1 + (dy/dx)^2)
= √(1 + (x^4 - 1/2 + 1/(16x^4)))
= √(x^4 + 1/2 + 1/(16x^4))
= √(x^2 + 1/(4x^2))^2
= x^2 + 1/(4x^2).

Hence, the arc length equals
∫(x = 1 to 2) (x^2 + 1/(4x^2)) dx
= [(1/3)x^3 - 1/(4x)] {for x = 1 to 2}
= 59/24.
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I hope this helps!

Only 1 question at a time please.

Note that the length of some curve y = f(x) on the interval [a,b] is defined as:

∫√(1 + (dy/dx)^2) dx from a to b

In this case:

dy/dx = x^(1/3) - 1/4x^(-1/3) => (4x^(2/3) - 1)/(4x^(1/3))

Length = ∫√(1 + (16x^(4/3) - 8x^(2/3) + 1))/(16x^(2/3)) dx from 1 to 27

= ∫√(16x^(4/3) + 8x^(2/3) + 1)/(4x^(1/3)) dx from 1 to 27

= ∫(4x^(2/3) + 1)/4x^(1/3) dx from 1 to 27

= 3/4*x^(4/3) + 3/8 * x^(2/3) eval. from 1 to 27

= 3/4*(81) + 27/8 - 3/4 - 3/8

= 60 + 3 = 63

I made the mistake.. sorry about that. Pretty tired at the moment.