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**peter010****Guest**

Hello,

Integration can be defined as:

b b

∫ f(x).dx = ∑ f(x)

a a

Now, If I want to find the integration for f(x)=10, where a=1, b=5 .. I ll got two different results!!!

(1)

5

∫ 10 = (10*5) - (10*1) = 40

1

(2)

5

∑ 10 = 10 + 10 + 10 +10 +10 = 50, or: = (10 *5) = 50

-----------

So, At first i got 40, while at second I got 50, taking i got 50.

Kindly, what do you think, where is my mistake?

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Integration can be defined as:

b b

∫ f(x).dx = ∑ f(x)

a a

Who said integration and summation were the same thing? Sometimes, we can approximate one with the other but they are not the same thing.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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**peter010****Member**- Registered: 2016-09-05
- Posts: 22

bobbym wrote:

Hi;

Integration can be defined as:

b b

∫ f(x).dx = ∑ f(x)

a aWho said integration and summation were the same thing? Sometimes, we can approximate one with the other but they are not the same thing.

http://www.mathcentre.ac.uk/resources/Engineering%20maths%20first%20aid%20kit/latexsource%20and%20diagrams/8_12.pdf

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**peter010****Member**- Registered: 2016-09-05
- Posts: 22

thickhead wrote:

Thanks

So, what if I considered it like: 0 to 1 represents: 1, 1 to 2 represents: 2, 2 to 3 represents: 3, and so on, thus I can presume the following:

5

∫10 = (10*5) - (10*0) = 50

0

What do you think ?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

As I said summation and integration can be used as approximations of each other. The discretizing trick will not always produce exact same answers.

For instance:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**peter010****Member**- Registered: 2016-09-05
- Posts: 22

bobbym wrote:

As I said summation and integration can be used as approximations of each other. The discretizing trick will not always produce exact same answers.

For instance:

Thanks

I totally agree. Its here a special matter of integrating a constant, thus I presume exact answer could be achieved.

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Indeed, the integral is sometimes compared with a series to prove that it converges (or diverges). You can read about it **here**.

peter010 wrote:

Hello,

Integration can be defined as:

b b

∫ f(x).dx = ∑ f(x)

a a

peter010 wrote:

That isn't the same as what you wrote -- you need the (representing the width of the rectangles), otherwise the equality isn't true. Their definition of the definite integral was:assuming the function is integrable on to begin with. If you leave that out, then you are essentially approximating the definite integral of the function by rectangles of width 1, so in general there will be a discrepancy between the integral and the sum itself.
http://www.mathcentre.ac.uk/resources/Engineering%20maths%20first%20aid%20kit/latexsource%20and%20diagrams/8_12.pdf

*Last edited by zetafunc (2016-09-05 22:55:21)*

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

He might do, yes, but the opening line of his post suggested that he was using that definition of an integral for a general function f.

That is correct and why I answered in the way I did.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**peter010****Member**- Registered: 2016-09-05
- Posts: 22

thickhead wrote:

I cant describe this case better than this

zetafunc wrote:

He might do, yes, but the opening line of his post suggested that he was using that definition of an integral for a general function .

Thanks

bobbym wrote:

He might do, yes, but the opening line of his post suggested that he was using that definition of an integral for a general function f.

That is correct and why I answered in the way I did.

Thanks Indeed

*Last edited by peter010 (2016-09-06 06:17:06)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Welcome to the forum.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**peter010****Member**- Registered: 2016-09-05
- Posts: 22

Here I put my code for calculating the integration by approximation method (the series),,,

//Integrating Functions

double function;

double width = .0001;

double sum = 0, j = 1;

int interval = 5;

for (double i = width; j <= (interval / width); j++, i += width)

{

function = 10;

sum += (width * function);

}

Console.WriteLine(sum);//50

Console.WriteLine("\nDone!");

///

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Did you put a lot of very thin rectangles under the curve?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**peter010****Member**- Registered: 2016-09-05
- Posts: 22

bobbym wrote:

Did you put a lot of very thin rectangles under the curve?

This code is to meet any function precisely (not specifically the function of 10 ) with a rectangular width=0.0001 (it can be definitely adjusted by the user)

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Any function precisely? No amount of rectangles can do that unless it is an infinite number of them. There are even many more precise methods of numerical integration ( exactly what you are doing ) but each except for limited cases is just an approximation.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**peter010****Member**- Registered: 2016-09-05
- Posts: 22

bobbym wrote:

Hi;

Any function precisely? No amount of rectangles can do that unless it is an infinite number of them. There are even many more precise methods of numerical integration ( exactly what you are doing ) but each except for limited cases is just an approximation.

ok

i just wanted to share "my code" of series "approximation" method with ppl.

noth. more,,

Regards.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

That is fine! You are making discoveries for yourself and thinking about how to get your computer to do them. The epitome of EM. I was just giving you some of what I do in that field. My favorites are numerical integration and curve fitting.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**peter010****Member**- Registered: 2016-09-05
- Posts: 22

bobbym wrote:

That is fine! You are making discoveries for yourself and thinking about how to get your computer to do them. The epitome of EM. I was just giving you some of what I do in that field. My favorites are numerical integration and curve fitting.

That is a lovely field.

And I appreciate and welcoming your advices, which is the reason why I come to this forum

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Many integrals are what we call pathogenic ( diseased or sick ) in numerical work it means they do not respond well to numerical integration. One of these is

It is easy to do using calculus but not so easy using numerical methods. You might like to try it when you have some spare time.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**peter010****Member**- Registered: 2016-09-05
- Posts: 22

bobbym wrote:

Many integrals are what we call pathogenic ( diseased or sick ) in numerical work it means they do not respond well to numerical integration. One of these is

It is easy to do using calculus but not so easy using numerical methods. You might like to try it when you have some spare time.

hahaha I just did..

And I found yes it is diseased

*Last edited by peter010 (2016-09-06 10:37:12)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Did you get anything close to 2 / 3 = .66666666...?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**peter010****Member**- Registered: 2016-09-05
- Posts: 22

bobbym wrote:

Did you get anything close to 2 / 3 = .66666666...?

= 0.266

but this still does not mean that i cannot customizing the solution/code to fit this kind of functions/

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

You can make particular formulas for any function but it appears there will always be some that will not like your formula.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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