Thread: curve length with ln function

1. curve length with ln function

Find the length of the curve:

$y = \frac{x^2}{4} - \ln \sqrt x$ for $1\leq x\leq 2

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If I understand correctly, Length = integral from 1 to 2 of $\sqrt {1+(\frac{dy}{dx})^2} dx$

So in this case, I need to find y' and substitute it in the formula and then find the integral.

y = $\frac{x}{2} - \frac{1}{2}\frac{1}{x} ==> \frac{x^2-1}{2x}$

Then I plug in y' into the formula to get (integral from 1 to 2)
L = $\int \sqrt{1 + (\frac{x^2-1}{2x})^2} dx

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I'm not sure where to go from here to make the integration work. I've been stuck on this for a while, so if anyone could help, I'd appreciate it!

2. After you get your derivative, square it and all that.

$\sqrt{1+\left(\frac{d}{dx}\left[\frac{x^{2}}{4}-ln(\sqrt{x})\right]\right)^{2}}=\frac{x^{2}+1}{2x}$

which is not too bad to integrate...no radical.

Most of these things are made to simplify down with some algebra.

3. am i right that dy/dx = $\frac{x^2-1}{2x}$?

4. Continuing on.. for the integration of x^2+1 / 2x I got:

$\frac{1}{2}\ln\mid\frac{x^2+1}{2x}\mid + C$

I did this by making u = 2x du = 2 and dx = 1/2

Is this even close?

If so, then since the curve length is from 1 to 2, would I input 2 and 1 and subtract to find the difference? Is the "C" ignored in this sense?

length = .1115717757 ??