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April 20, 2014

Homework Help: CALCULUS 2!!! PLEASE HELP!!

Posted by Krystal on Thursday, October 19, 2006 at 3:56am.

I'm having trouble with this question on arc length:

y=lnx, (squareroot)3/3 greater than or equal to x less than or equal to 1


It sounds as if you want the length of the y = ln x curve from
x = sqrt(3)/3 (0.57735..) to 1.

The formula for the arc length of a line y(x) is

Length = (INTEGRAL OF) sqrt [1 + (dy/dx)^2] dx

In your case dy/dx = 1/x

so Length = (INTEGRAL OF) sqrt [1 + 1/x^2] dx

Perform the integration between your indicated limits for the answer. I find the integration easier to perform by making the substitution u = 1/x. It's a bit messy.


I have sqrt [1 + 1/x^2] dx , I just don't know what to do next...


Make the substitutions 1/x = u and
dx = u^-2 du

Then the integral becomes
INTEGRAL -[sqrt(1 + u^2)]/u^2 du

According to my table of integrals, the indefinite integral is
[sqrt (1 + u^2)]/u - ln[u + sqrt (u^2 = 1)]
The second term can also be expressed as an arcsinh function.

Integrate that between the two values of u that correspond to your x limits.


Here's a useful site I use to check integrals:
integrals.wolfram. c o m
(delete the spaces)
That's the Wolfram integrator, which is actually Mathematica, a fairly good piece of software for finding anti-derivatives.
When I check sqrt(1 + 1/x^2) I got a slightly different anti-derivative, just thought I'd mention this.

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