Arc Length of Polynomials

[utah412]

I need to find the arc length of ((x^2)/2 - ln(x)/4) from 2 to 4.


Using the arc length formula, I get that L = ∫ds = ∫ sqrt(1+(f'(x))^2) dx =


∫ sqrt (1 + ((x^3)/6 -1/(4x))^2) dx from x=2 to x=4,


but I cannot solve it any further.


I cannot use u-substitution and I have tried using trig-substitution but it does not seem to work either.


The answer given by WolframAlpha is (1/4)(24 +ln(2)) which is correct, but I want to know how it computed that. I understand that the ln(2) part came from ln(4) - ln(2) = ln(4/2) but beyond that I am lost.

 

[lookagain]

I need to find the arc length of ((x^2)/2 - ln(x)/4) from 2 to 4.

Using the arc length formula, I get that L = ∫ds = ∫ sqrt(1+(f'(x))^2) dx =

∫ sqrt (1 + ((x^3)/6 -1/(4x))^2) dx from x=2 to x=4,....

No, your integrand should be this instead:

∫ sqrt (1 + (x - 1/(4x))^2) dx from x=2 to x=4

The derivative of (x^2)/2 is x.