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November 28, 2014

November 28, 2014

Posted by **Arc Length** on Sunday, August 24, 2008 at 9:11pm.

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.

- calculus -
**Damon**, Sunday, August 24, 2008 at 9:47pmI 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 ?

- calculus -
**Arc Length**, Sunday, August 24, 2008 at 10:12pmOh 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.

- calculus -
**Damon**, Monday, August 25, 2008 at 3:50amwell, the sqrt of that is just

x + (1/4)(1/x)

integral of that is

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

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