Integral relationship

szk_kei · 2005-11-07 00:15:59

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/

MathsIsFun · 2005-11-07 08:23:47

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 ??

mathsyperson · 2005-11-07 08:57:58

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)

MathsIsFun · 2005-11-07 18:24:23

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 ...

szk_kei · 2005-11-07 22:33:14

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)]

szk_kei · 2005-11-09 00:57:06

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".

mathsyperson · 2005-11-09 06:07:18

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)

szk_kei · 2005-11-10 00:24:01

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

MathsIsFun · 2005-11-10 08:22:11

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

We could try another example: y=x²

mathsyperson · 2005-11-10 09:16:13

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)

szk_kei · 2005-11-13 23:42:26

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.

Jai Ganesh · 2005-11-14 00:06:35

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.

szk_kei · 2005-11-14 01:25:54

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

Jai Ganesh · 2005-11-14 18:20:48

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

MathsIsFun · 2005-11-14 20:07:54

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

szk_kei · 2005-11-17 02:46:11

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/

szk_kei · 2005-12-01 12:54:03

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

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

szk_kei · 2005-12-27 12:50:25

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

MathsIsFun · 2005-12-27 18:11:07

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

krassi_holmz · 2005-12-28 03:16:27

Very good site, Keiichi.

krassi_holmz · 2005-12-28 03:21:47

And we can form something like that for definite integrals:

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

krassi_holmz · 2005-12-28 03:36:18

Geometric proof

szk_kei · 2005-12-28 20:46:35

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

Thank you

szk_kei · 2006-01-06 23:20:49

>

krassi_holmz wrote:

And we can form something like that for definite integrals:
>

Please visit below
http://www.freewebs.com/keiichi_suzuki/math/sekibun0.html

krassi_holmz · 2006-01-06 23:22:43

I'll visit this... now

Math Is Fun Forum

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

Integral relationship

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

Re: Integral relationship

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

Re: Integral relationship

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

Re: Integral relationship

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

Re: Integral relationship

#6 2005-11-09 00:57:06

Re: Integral relationship

#7 2005-11-09 06:07:18

Re: Integral relationship

#8 2005-11-10 00:24:01

Re: Integral relationship

#9 2005-11-10 08:22:11

Re: Integral relationship

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

Re: Integral relationship

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

Re: Integral relationship

#12 2005-11-14 00:06:35

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#13 2005-11-14 01:25:54

Re: Integral relationship

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

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#15 2005-11-14 20:07:54

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#16 2005-11-17 02:46:11

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#17 2005-12-01 12:54:03

Re: Integral relationship

#18 2005-12-27 12:50:25

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#19 2005-12-27 18:11:07

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#20 2005-12-28 03:16:27

Re: Integral relationship

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

Re: Integral relationship

#22 2005-12-28 03:36:18

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#23 2005-12-28 20:46:35

Re: Integral relationship

#24 2006-01-06 23:20:49

Re: Integral relationship

#25 2006-01-06 23:22:43

Re: Integral relationship

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