find arc length

[IrishGemini98]

s=the interval from 1 to 3 of the square root of (1+(x^4-1/2x^2)^2)dx

help please. I can't do it

 

[stapel]

Do you mean the following?

. . . . .$\displaystyle \large{\int_1^3{\,\sqrt{1\,+\,\left(x^4\,-\,\frac{1}{2x^2}\right)^2\,}\;}dx}$

Thank you.

Eliz.

 

[pka]

Please post the complete statement of the problem.
I cannot understand your question from your post!

 

[galactus]

I believe they mean:

$\displaystyle \large\int_{1}^{3}\sqrt{1+(x^{4}-\frac{x^{2}}{2})^{2}}dx$

Either way, it's a booger of an integral to do the olf-fashioned way. I would find a computer or calculator to tackle it.

Hey, this would be a real dilly of an integral for Soroban and/or pka and/or Unco to tackle.

 

[IrishGemini98]

The first reply from stapel is the correct version. Thanks

 

[galactus]

OK. I ran this through my trusty TI and, after some groaning and grunting, it gave

me 48.2437905509.

Are you expected to do this by hand?.

 

[soroban]

Hello, IrishGemini88!

If Eliz is correct and the intergral is: \(\displaystyle \;\;\;\L \int^{\;\;\;3}_1\sqrt{1\,+\,\left(x^4\,-\,\frac{1}{2x^2}\right)^2\,}\:dx\)

$\displaystyle \;\;$I see no elementary method for integrating it.


Please give us the original statement of the problem.
$\displaystyle \;\;$I suspect that the above integral is not the correct one.

I seriously doubt that $\displaystyle \,x^4\,-\,\frac{1}{2x^2}\,$ is the derivative of $\displaystyle f(x)$.


It is highly likely that: \(\displaystyle \L\,f(x)\:=\:x^4\,+\,\frac{1}{32x^2}\;\) . . . or something similar.

Then: \(\displaystyle \L\:f'(x)\;=\;4x^3\,-\,\frac{1}{16x^3}\;\;\Rightarrow\;\;[f'(x)]^2\:=\:16x^6\,-\,\frac{1}{2}\,+\,\frac{1}{256x^6}\)

Hence: \(\displaystyle \L\:1\,+\,[f'(x)]^2\;=\;1\,+\,\left(16x^6\,-\,\frac{1}{2}\,+\,\frac{1}{256x^6}\right)\)

\(\displaystyle \L\;\;\:=\:16x^6\,+\,\frac{1}{2}\,+\,\frac{1}{256x^6}\:=\:\left(4x^3\,+\,\frac{1}{16x^3}\right)^2\) . . . a perfect square!

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Over the years, I've seen this type of problem many times.

The length-of-arc intergral is truly awful to integrate
$\displaystyle \;\;$unless they deliberately choose an $\displaystyle f(x)$ to be convenient.

A "nice" one is: $\displaystyle \,f(x)\:=\:x^3\,+\,\frac{1}{12x}$ . . . even nicer is: $\displaystyle \,f(x)\:=\:\cosh(x)$