It's xyz!
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Calcuculus problem
Where did I go wrong?
Okay, it's proving the integral of x.
x.dx = 1/2.x^2
2x.dx = x^2
x = x^2/2.dx
1 = x/2.dx
1/x = 1/2.dx
2.dx = x
2x = x
2 = 1
Oh shi...!
Someone help?
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Last edited by Raz on Jan 1st 2000 at 00:00AM
May 8th, 2010 06:49 PM
Admiral Akbar
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Re: Calcuculus problem
quote: (post ) Originally posted by lord xyz
Where did I go wrong?
Okay, it's proving the integral of x.
x.dx = 1/2.x^2
2x.dx = x^2
x = x^2/2.dx
1 = x/2.dx
1/x = 1/2.dx
2.dx = x
2x = x
2 = 1
Oh shi...!
Someone help?
Take the derivative of 1/2x^2
Umm, what did you do with the dx?
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May 8th, 2010 06:58 PM
It's xyz!
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quote: (post ) Originally posted by Admiral Akbar
Take the derivative of 1/2x^2
Umm, what did you do with the dx? Integrated.
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Last edited by Raz on Jan 1st 2000 at 00:00AM
May 8th, 2010 07:04 PM
It's xyz!
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Re: Calcuculus problem
Ah **** it, 2 = 1.
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Last edited by Raz on Jan 1st 2000 at 00:00AM
May 8th, 2010 07:11 PM
Bicnarok
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here´s my calculus problem..
WTF is Calculus?
May 8th, 2010 07:15 PM
King Kandy
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Re: Calcuculus problem
quote: (post ) Originally posted by lord xyz
Where did I go wrong?
Okay, it's proving the integral of x.
x.dx = 1/2.x^2
2x.dx = x^2
x = x^2/2.dx
1 = x/2.dx
1/x = 1/2.dx
2.dx = x
2x = x
2 = 1
Oh shi...!
Someone help?
Your notation is so horribly confusing, that I can't hope to help you here. Especially in this step, I don't have a clue what exactly you're trying to communicate (taking the derivative of both sides?)
2x.dx = x^2
x = x^2/2.dx
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May 8th, 2010 08:21 PM
Rogue Jedi
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Grammar problems to boot.
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May 9th, 2010 06:38 AM
It's xyz!
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quote: (post ) Originally posted by King Kandy
Your notation is so horribly confusing, that I can't hope to help you here. Especially in this step, I don't have a clue what exactly you're trying to communicate (taking the derivative of both sides?)
2x.dx = x^2
x = x^2/2.dx
2x = d(x^2)/dx
I think that's my first error.
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Last edited by Raz on Jan 1st 2000 at 00:00AM
May 9th, 2010 10:30 AM
tsilamini
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I solved for d and got x = x when I subed...
maybe you need to collect all the x's on one side then sub in d=1/2
also, I am assuming the periods in the first equation are multiplication?
even when I collected the x's, I just keep getting x^2 - x^2 = 0
I haven't done calculus in over 5 years....
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Last edited by tsilamini on May 9th, 2010 at 02:10 PM
May 9th, 2010 02:00 PM
Bardock42
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quote: (post ) Originally posted by Bicnarok
here´s my calculus problem..
WTF is Calculus?
It's a part of the mathematical field of Analysis. In fact it's what the German mathematician Leibniz invented, but that prick Newton took credit for.
For the OP, first of all you forgot the constant every time you integrated, that's not good. Anyways lets try to note that a bit better:
1) Integral x = (x²/2)+C
You: x.dx = 1/2.x^2
2) Integral 2x = x² + C
You: 2x.dx = x^2
3) Integral of x²/2 = (x³/6) + C
Your: x = x^2/2.dx
Here you seem to say that the integral of x²/2 is x. That is however not the case. In step one we saw that the Integral of x is x²/2 however the opposite is not true. So that is the first mistake that breaks your whole process, I believe
4) -
You:1 = x/2.dx
You can't just multiply in and out of integrals that's not how they work. Even if x was the integral of x²/2 dividing by it would give you 1 = (1/x) times Integral x²/2, not Integral of x/2
5) -
You: 1/x = 1/2.dx
Same problem and continuing from a false premise.
6) -
2.dx = x
Same
7)-
2x = x
Here you suddenly remember the integral again though you disregarded it before. You also miss the constant again.
8) -
2 = 1
And that's why this is wrong.
I hope that helps give some clarity.
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Last edited by Bardock42 on May 9th, 2010 at 02:18 PM
May 9th, 2010 02:15 PM
Admiral Akbar
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EDIT- Removed
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May 9th, 2010 03:44 PM
It's xyz!
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quote: (post ) Originally posted by Bardock42
It's a part of the mathematical field of Analysis. In fact it's what the German mathematician Leibniz invented, but that prick Newton took credit for.
For the OP, first of all you forgot the constant every time you integrated, that's not good. Anyways lets try to note that a bit better:
1) Integral x = (x²/2)+C
You: x.dx = 1/2.x^2
2) Integral 2x = x² + C
You: 2x.dx = x^2
3) Integral of x²/2 = (x³/6) + C
Your: x = x^2/2.dx
Here you seem to say that the integral of x²/2 is x.
No, that isn't (x^2/2)dx, I divided by dx.
Which I believe should've been written as d(x^2)/dx integration, see.
quote: That is however not the case. In step one we saw that the Integral of x is x²/2 however the opposite is not true. So that is the first mistake that breaks your whole process, I believe
4) -
You:1 = x/2.dx
You can't just multiply in and out of integrals that's not how they work. Even if x was the integral of x²/2 dividing by it would give you 1 = (1/x) times Integral x²/2, not Integral of x/2
5) -
You: 1/x = 1/2.dx
Same problem and continuing from a false premise.
6) -
2.dx = x
Same
7)-
2x = x
Here you suddenly remember the integral again though you disregarded it before. You also miss the constant again.
8) -
2 = 1
And that's why this is wrong.
I hope that helps give some clarity.
Yeah, I did forget the constant. My bad.
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Last edited by Raz on Jan 1st 2000 at 00:00AM
May 9th, 2010 04:57 PM
Bardock42
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quote: (post ) Originally posted by lord xyz
No, that isn't (x^2/2)dx, I divided by dx.
Which I believe should've been written as d(x^2)/dx integration, see.
Yeah, I did forget the constant. My bad. [/B]
Ah okay, I see, well you can't do that. So I guess you found your answer. I'm still not sure about the thing King Kandy posted a while back though.
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May 9th, 2010 05:28 PM
FistOfThe North
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if bardock eats 6 donuts in 60 seconds outta the dozen and x is what's left in the box, what does x=?
hehe..
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May 14th, 2010 02:22 AM
King Kandy
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quote: (post ) Originally posted by Bardock42
Ah okay, I see, well you can't do that. So I guess you found your answer. I'm still not sure about the thing King Kandy posted a while back though.
I think the problem here is that he used the exact same notation for derivatives and integrals...
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May 14th, 2010 04:49 AM
Bardock42
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quote: (post ) Originally posted by FistOfThe North
if bardock eats 6 donuts in 60 seconds outta the dozen and x is what's left in the box, what does x=?
hehe..
That's unsolvable. You gave a speed and an amount. We'll still need time to solve your equation.
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May 14th, 2010 07:21 AM
Symmetric Chaos
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quote: (post ) Originally posted by Bardock42
That's unsolvable. You gave a speed and an amount. We'll still need time to solve your equation.
Well he said "in 60 seconds" not "per 60 seconds" so the answer should be 6.
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May 14th, 2010 04:27 PM
Bardock42
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quote: (post ) Originally posted by Symmetric Chaos
Well he said "in 60 seconds" not "per 60 seconds" so the answer should be 6.
Well that still assumes that I stop after 60 seconds, or that it is counted after 60 seconds, which is not necessarily clear from the phrasing.
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May 14th, 2010 06:24 PM
It's xyz!
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Ermm, bump?
I probably failed that question anyway.
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Last edited by Raz on Jan 1st 2000 at 00:00AM
Aug 31st, 2011 04:50 PM
King Kandy
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You can't just "divide" by dx, that's not an operation that makes any kind of sense.
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Aug 31st, 2011 05:38 PM
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