Find the length of the arc: y= 1/8[4x^2-2ln(x)] from x=4 to x=5?

Question

a re-asked question because no one answer this question I posted 7 hours ago. My answer of 4.745 was wrong...I need help badly on this one thank guys!!

Answer 1

dy/dx = x - 1/(4x)

(dy/dx)^2 = x^2 - 1/2 + 1/(16 x^2)

(dy/dx)^2 + 1  = x^2 + 1/(16 x^2) + 1/2

= (x + 1/(4x))^2

We must then evalute the integral of (x + 1/(4x)) from 4 to 5.  Forgive my notation.

1/2 x^2 + 1/4 lnx from 4 to 5

1/2(25) + 1/4 ln5 - 1/2(16) - 1/4ln4

12.5 + 1/4(ln5-ln4) - 8

4.5 + 1/4 ln (5/4)

4.5 + .0557 = 4.555

Answer 2

integrate from 4 to 5 (sqrt( (y')^2 + 1) dx

y' = x - 1/(4x)

(y' )^2 + 1 = (4x^2 + 1)^2/(16x^2)

Then integrate the square root of that...

Integrate from 4 to 5 ( (4x^2 + 1)/(4x)) dx

= LN(5/4)/4 + 9/2

Hope that helps.