Find the length of the arc formed by

y = (1/8)(4x^2-2ln(x)) from x=4 to x=8.

I found the derivative of the function and got y'= x-(1/4x)

Where I'm lost now is after plugging it into the arc length equation: integral of sqrt(1+(x-(1/4x))^2). Squaring the derivative yields me sqrt(1+x^2+1/16x-1/2). Help please.

I got sqrt [ 1 + x^2 -(1/2) + 1/(16x^2) ]

which is
sqrt [ x^2 + (1/2) +1/(16x^2) ]
BUT
x^2 + (1/2) + 1/(16x^2) = [x+ 1/(4x)]^2
ok ?

Oh yes, sorry, I merely had a typo. I had the same result as you. What I'm confused about is what to do from that point on.

well, the sqrt of that is just

x + (1/4)(1/x)
integral of that is
(1/2) x^2 + (1/4) ln x

Sure! I'll help you step by step.

1. First, you correctly found the derivative of the function: y' = x - (1/4x).

2. To calculate the length of the arc, we need to integrate the expression sqrt(1 + (y')^2) over the interval [4, 8]. In this case, our expression becomes sqrt(1 + (x - (1/4x))^2).

3. Now, let's simplify the expression inside the square root. Squaring the derivative, we get (x - (1/4x))^2 = x^2 - (1/2) + (1/16x^2).

4. Combine like terms to get (17/16)x^2 - (1/2).

5. The expression inside the square root becomes 1 + (17/16)x^2 - (1/2).

6. Now, let's set up the integral. We want to evaluate the integral of sqrt(1 + (17/16)x^2 - (1/2)) with respect to x, from x = 4 to x = 8.

7. After integrating the expression, you will get the length of the arc.

Let me know if you need further assistance!