Find the length of the arc f(x)=x^3/3+1/4x, x E [1,2]

is it:

a. 5/24
b. 3/8
c. 35/24
d. 59/24

I chose a. Is that correct? If not, please explain

I am assuming that:
y = x^3/3 + 1/(4x).

By re-writing y as x^3/3 + (1/4)x^(-1) and differentiating gives:
dy/dx = x^2 + (1/4)(-1)x^(-2)
= x^2 - 1/(4x^2)
= (4x^4 - 1)/(4x^2), by combining fractions.

Then:
1 + (dy/dx)^2 = 1 + [(4x^4 - 1)/(4x^2)]^2
= 1 + (4x^4 - 1)^2/(16x^4)
= 1 + (16x^8 - 8x^4 + 1)/(16x^4)
= (16x^8 + 8x^4 + 1)/(16x^4)
= [(4x^4 + 1)/(4x^2)]^2
==> √[1 + (dy/dx)^2] = (4x^4 + 1)/(4x^2).
(Note that (4x^4 + 1)/(4x^2) > 0 for all x in the domain.)

Therefore, the arc-length of y = x^3/3 + 1/(4x) on [1, 2] is:
L = ∫ √[1 + (dy/dx)^2] dx (from x=a to b)
= ∫ (4x^4 + 1)/(4x^2) dx (from x=1 to 2)
= ∫ [x^2 + 1/(4x^2)] dx (from x=1 to 2)
= [x^3/3 - 1/(4x)] (evaluated from x=1 to 2)
= (8/3 - 1/8) - (1/3 - 1/4)
= 59/24.

Therefore, the answer is (D) 59/24.

I hope this helps!