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#1 2005-11-07 00:15:59

szk_kei
Member
Registered: 2005-11-04
Posts: 21

Integral relationship

Relationship between the integral of the function and
the integral of its inverse function
INTGR(y)dx + INTGR(x)dy = xy + C

Please click below.
http://www.freewebs.com/keiichi_suzuki/

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#2 2005-11-07 08:23:47

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,535

Re: Integral relationship

BTW, you can use the "∫" symbol (I have those symbols just under the forum title - just drag you mouse across one, copy then paste, and you can get: ∫y dx + ∫x dy)

So ... why does ∫y dx + ∫x dy = xy + C ??


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#3 2005-11-07 08:57:58

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Integral relationship

I tried to look at szk_kei's proof, but when I went to his site and clicked the link to it I just got an acrobat reader thing with 4 blank pages. Probably my computer doesn't support something that it needs to. I tried it out though, and it seems to work.

y = x

∫ydx = 0.5x² + c
∫xdy = 0.5y² + c

∫ydx + ∫xdy = 0.5x² + 0.5y² + c

x = y ∴ x² = y² ∴ 0.5x² + 0.5y² + c = x² + c
x = y ∴ x² + c = xy + c

Last edited by mathsyperson (2005-11-07 08:58:09)


Why did the vector cross the road?
It wanted to be normal.

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#4 2005-11-07 18:24:23

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,535

Re: Integral relationship

Graphically ∫ydx is the area "below" the curve to the x axis, while ∫xdy is the area to the "left" of the curve to the y-axis, so together they form a square (well they do if the limits are 0 to some value).

So this probably applies to y = x², too, and many more functions ...


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#5 2005-11-07 22:33:14

szk_kei
Member
Registered: 2005-11-04
Posts: 21

Re: Integral relationship

y=f(x):differentiable

(xy)'=y+x(dy/dx)
xy+C=∫ydx + ∫x(dy/dx)dx
        =∫y dx + ∫x dy

If you want to sea its detail, please visit my site http://www.freewebs.com/keiichi_suzuki/
->[The Study (Mathematics created by myself)]

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#6 2005-11-08 06:22:48

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: Integral relationship

You can find those relations in all calculus books.

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#7 2005-11-09 00:57:06

szk_kei
Member
Registered: 2005-11-04
Posts: 21

Re: Integral relationship

When I show an expression without proof, they say always "It can never hold".
When I show an expression with proof, someone says "It is only an application of integration by parts",
someone says "It is simple and beautiful".

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#8 2005-11-09 06:07:18

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Integral relationship

MathsIsFun wrote:

Graphically ∫ydx is the area "below" the curve to the x axis, while ∫xdy is the area to the "left" of the curve to the y-axis, so together they form a square (well they do if the limits are 0 to some value).

That's a very nice way of putting it without getting involved in lots of heavy maths. It actually forms a rectangle though.

Last edited by mathsyperson (2005-11-09 06:07:26)


Why did the vector cross the road?
It wanted to be normal.

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#9 2005-11-10 00:24:01

szk_kei
Member
Registered: 2005-11-04
Posts: 21

Re: Integral relationship

Your definition is monotone increasing function passing through the origin and  differentiable.
My definition is only differentiable.

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#10 2005-11-10 08:22:11

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,535

Re: Integral relationship

The y=x function was only an example, I think

We could try another example: y=x²


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#11 2005-11-10 09:16:13

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Integral relationship

Yes, it was. I'll do the y = x², because it gives me a chance to practice the new code things.

y = x²
∫ydx = x³/3 + c

x=√y
∫xdy = 2/3√(y³) + d

x³/3 + c + 2/3√(y³) + d = x³/3 + 2/3x³ + c + d = x³ + c + d

x³ = x*x² = xy + c + d

c and d are both arbitrary constants, so they can be combined into an arbitrary constant C.

∫xdy + ∫ydx = xy + C. It still works!

Last edited by mathsyperson (2005-12-02 04:49:16)


Why did the vector cross the road?
It wanted to be normal.

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#12 2005-11-13 23:42:26

szk_kei
Member
Registered: 2005-11-04
Posts: 21

Re: Integral relationship

My name is Keiichi Suzuki.
I'm a 48-year-old male.
I'm a Japanese.

I publish online mathematics created by myself,
including arithmetic (mathematics) for children.
http://www.freewebs.com/keiichi_suzuki/

My site is listed in
http://dir.yahoo.com/Science/Mathematics/Calculus/

Please visit my site.

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#13 2005-11-14 00:06:35

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,428

Re: Integral relationship

Hi Mr.Keiichi Suzuki,
Welcome to the forum.
Your website is cool, your proof is elgeant and precise.
I do have  great respect for Japanese Mathematicians, I had read about Shimura and Taniyama before. smile


Character is who you are when no one is looking.

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#14 2005-11-14 01:25:54

szk_kei
Member
Registered: 2005-11-04
Posts: 21

Re: Integral relationship

I'm not mathematician.
I'm working for a manufacturer.

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#15 2005-11-14 18:20:48

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,428

Re: Integral relationship

I'm not a mathematician either.
I work taxing manufacturers (collecting taxes).


Character is who you are when no one is looking.

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#16 2005-11-14 20:07:54

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,535

Re: Integral relationship

LOL! Thank goodness you guys live in different countries!!


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#17 2005-11-15 05:07:17

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: Integral relationship

fortunately for me, I'm still a student. : )

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#18 2005-11-17 02:46:11

szk_kei
Member
Registered: 2005-11-04
Posts: 21

Re: Integral relationship

There are some topics in my site.

Taylor Series.
What is the necessary and sufficient condition for the intervals of n of  the degree as expressed below ?
f(x)=A0 + An(x-a)^n + A2n(x-a)^2n + A3n(x-a)^3n + A4n(x-a)^4n + +

Please visit my site
http://www.freewebs.com/keiichi_suzuki/

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#19 2005-12-01 12:54:03

szk_kei
Member
Registered: 2005-11-04
Posts: 21

Re: Integral relationship

Can you see the full explanation of integral relation and property of laurent expansion with an acrobat reader.

http://www.freewebs.com/keiichi_suzuki/

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#20 2005-12-27 12:50:25

szk_kei
Member
Registered: 2005-11-04
Posts: 21

Re: Integral relationship

Can you see the full explanation of integral relation and property of laurent expansion with an acrobat reader.

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#21 2005-12-27 18:11:07

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,535

Re: Integral relationship

Looks good, szk_kei, though I haven't checked if it is right or not.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#22 2005-12-28 03:16:27

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Integral relationship

Very good site, Keiichi.


IPBLE:  Increasing Performance By Lowering Expectations.

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#23 2005-12-28 03:21:47

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Integral relationship

And we can form something like that for definite integrals:

Last edited by krassi_holmz (2005-12-28 03:25:27)


IPBLE:  Increasing Performance By Lowering Expectations.

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#24 2005-12-28 03:36:18

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Integral relationship

Geometric proof

View Image: rule.GIF

IPBLE:  Increasing Performance By Lowering Expectations.

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#25 2005-12-28 20:46:35

szk_kei
Member
Registered: 2005-11-04
Posts: 21

Re: Integral relationship

>Looks good, szk_kei, though I haven't checked if it is right or not.

Thank you

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