Posted: 12/23/2011 8:06:53 PM EDT
Okay, it's been 3 years and I'm having a problem integrating a problem. I was supposed to be "helping" someone but now I look like an idiot. 
Integral from a to b [1/2,2] of (1 + [((x^2)/2) - (1/(2x^2))]^2)^.5 dx I remember the chain rule in derivation but going backwards I've forgotten where to start. I'm thinking substitution but with what? Oy. Any help would be appreciated. (And no this isn't "homework" I've been out of college for some time now.) PS - if anyone knows a online thing so I can convert this thing to look like a problem I'll do that. ETA: Testing pic 
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Okay, it's been 3 years and I'm having a problem integrating a problem. I was supposed to be "helping" someone but now I look like an idiot.
Integral from a to b [1/2,2] of (1 + [((x^2)/2) - (1/(2x^2))]^2)^.5 dx I remember the chain rule in derivation but going backwards I've forgotten where to start. I'm thinking substitution but with what? Oy. Any help would be appreciated. (And no this isn't "homework" I've been out of college for some time now.) PS - if anyone knows a online thing so I can convert this thing to look like a problem I'll do that. Use an integral table? That doesn't look familiar. |
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Which college isn't on winter break right now? Or are you just doing this for fun? Because fuck that. I'm not in college nor trying. I'm supposed to be tutoring my student but this problem has me friggin' stumped. Technically the problem is finding the arc length of y = ((x^3)/6) + (1/(2x)) which using pythag's rule I should get Int from a to b of sqrt(1+f'(x)^2)dx So... yeah. Oh and fuck calculus. |
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wolframalpha will get you the answer and clicking on 'show steps' will help.
www.wolframalpha.com ETA: enter this: "integrate <the text version of your integral equation> Also, since x is positive in this case, it seems that (x^4-3)/6x is a solution per Wolfram But it won't show the steps for this equation :-( |
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Unless I made a mistake.. which is possible, it's 1:22 am..
If you expand what is under the radical and multiply by 4, it becomes x^4 + 2 +1/x^4. That can be factored into (x^2+1/x^2)^2. When you do square root of that it's just x^2+1/x^2.. which is easy to integrate. ETA: Forget the multiply by four part.. it doesn't actually make anything clearer. You can just integrate it as sqrt(0.25x^4+0.5+0.25x^-4).. which also factors and it becomes straightforward. |
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wolframalpha will get you the answer and clicking on 'show steps' will help. www.wolframalpha.com ETA: enter this: "integrate <the text version of your integral equation> Also, since x is positive in this case, it seems that (x^4-3)/6x is a solution per Wolfram But it won't show the steps for this equation :-( Neat site, I don't necessarily need the solution so much as the steps which it doesn't show.  The solution doesn't match what I should be getting either so I don't know... perhaps I'm using the wrong formula. Thanks though. 
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Multiply then factor the expression under the radical. The expression simplifies to: 1/2 * integral from 1/2 to 2 of x^2 + 1/x^2 dx so 1/2 *[(1/3x^3 - 1/x) | 1/2 to 2] 33/16 Duh, thanks. :) It's all coming back to me now... including integration by parts, substitution, etc. This one was a rather easy one in retrospect but my brain froze at the integral sign.
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A kcolg30 post in a Razoreye thread.
It's the perfect storm of fail. I feel like I should go to the free clinic and get tested for AIDS, I'm sure I'm positive now. No one said you had to participate and yet here you are throwing your worthless herpes in. Now we all have AIDS and Herps together. 
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33/16 is correct...
Go here (Wolfram Alpha) and enter this.. Integral from 1/2 to 2 of (1 + [((x^2)/2) - (1/(2x^2))]^2)^.5 dx (sorry no work, buy you can check your answer:) |
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Multiply then factor the expression under the radical. The expression simplifies to: 1/2 * integral from 1/2 to 2 of x^2 + 1/x^2 dx so 1/2 *[(1/3x^3 - 1/x) | 1/2 to 2] 33/16 Duh, thanks. :) It's all coming back to me now... including integration by parts, substitution, etc. This one was a rather easy one in retrospect but my brain froze at the integral sign. I always try to multiply through and factor first before i go on to u substittution, by parts, etc., and I usually have to write the exponents out. I can't remember 1/x is x^-1 by looking at it unless I write it that way. Trig functions are my nemesis. Thank God for MATLAB, doing that crap by hand drives me nuts. |
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Quoted: Unless I made a mistake.. which is possible, it's 1:22 am.. If you expand what is under the radical and multiply by 4, it becomes x^4 + 2 +1/x^4. That can be factored into (x^2+1/x^2)^2. When you do square root of that it's just x^2+1/x^2.. which is easy to integrate. ETA: Forget the multiply by four part.. it doesn't actually make anything clearer. You can just integrate it as sqrt(0.25x^4+0.5+0.25x^-4).. which also factors and it becomes straightforward. I ain't much on numbers and such..... but 1:22am don't sound like the karrect answer ta me..... |
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Multiply then factor the expression under the radical. The expression simplifies to: 1/2 * integral from 1/2 to 2 of x^2 + 1/x^2 dx so 1/2 *[(1/3x^3 - 1/x) | 1/2 to 2] 33/16 Duh, thanks. :) It's all coming back to me now... including integration by parts, substitution, etc. This one was a rather easy one in retrospect but my brain froze at the integral sign. I always try to multiply through and factor first before i go on to u substittution, by parts, etc., and I usually have to write the exponents out. I can't remember 1/x is x^-1 by looking at it unless I write it that way. Trig functions are my nemesis. Thank God for MATLAB, doing that crap by hand drives me nuts. Occam's Razor. I should have tried expanding the problem first and simplifying. I just see pythag's theorem and think trig substitution because that was easiest but I was looking for a fractional answer and that wouldn't have helped me. |
