IrishGemini98
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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
help please. I can't do it
find arc length
- Thread starter IrishGemini98
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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.
\(\displaystyle \large\int_{1}^{3}\sqrt{1+(x^{4}-\frac{x^{2}}{2})^{2}}dx\)
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
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- Feb 9, 2006
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The first reply from stapel is the correct version. Thanks
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!
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
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: \(\displaystyle \,f(x)\:=\:\cosh(x)\)
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!
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
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: \(\displaystyle \,f(x)\:=\:\cosh(x)\)