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

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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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

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

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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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

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

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

You can find those relations in all calculus books.

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

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

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

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

My definition is only differentiable.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

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

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

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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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,394

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.

Character is who you are when no one is looking.

**Online**

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

I'm not mathematician.

I'm working for a manufacturer.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,394

I'm not a mathematician either.

I work taxing manufacturers (collecting taxes).

Character is who you are when no one is looking.

**Online**

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

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

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

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

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

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

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

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

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

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

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

Very good site, Keiichi.

IPBLE: Increasing Performance By Lowering Expectations.

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

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

Geometric proof

IPBLE: Increasing Performance By Lowering Expectations.

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

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

Thank you

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