Arc Length

**veronicak5678** · October 5th 2008, 07:36 AM

Find the length of the curve y= (x^2/2) - (ln x / 4) from x = 2 to x = 4.

My attempt:

y' = x- (1/4x)

length = Int 2 - 4 root ( (1 + (x- (1/4x))^2)) dx

= Int 2 - 4 root (1+ x - 1/2 + 1/16x^2) dx

I cant work this out. Is it because I did this part wrong, or just because I can't figure out the integral?

**skeeter** · October 5th 2008, 07:52 AM

$y = \frac{x^2}{2} - \frac{\ln{x}}{4}$

$y' = x - \frac{1}{4x}$

$(y')^2 = x^2 - \frac{1}{2} + \frac{1}{16x^2}$

$S = \int_2^4 \sqrt{1 + \left(x^2 - \frac{1}{2} + \frac{1}{16x^2}\right)} \, dx$

$S = \int_2^4 \sqrt{x^2 + \frac{1}{2} + \frac{1}{16x^2}} \, dx$

$S = \int_2^4 \sqrt{\left(x + \frac{1}{4x}\right)^2} \, dx$

$S = \int_2^4 \left(x + \frac{1}{4x}\right) \, dx$

can you finish?

**veronicak5678** · October 5th 2008, 07:57 AM

I see. I would never have thought to do that. Thanks again.

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