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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

integral of 1/(x^4+x^2+1):D:D

*Last edited by Yusuke00 (2014-02-05 23:00:34)*

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

ok i understand....thank you!

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

integral of sqrt(x-x^2)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,415

Hi;

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,316

Hi Yusuke00

The first can also be solved using x^4+x^2+1=(x^2-x+1)(x^2+x+1) without complex numbers.

For the second one, I am getting

*Last edited by anonimnystefy (2014-02-06 03:00:07)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,415

Hi;

Did you test that second answer?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

f:[0,3]->R f(x)=max{3-x,2x+[x]} . Show that f is integrable on[0,3] and calculated integral of f(x)

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

integral of 1/(sin^4x+cos^4x)dx

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Yusuke00 wrote:

f:[0,3]->R f(x)=max{3-x,2x+[x]} . Show that f is integrable on[0,3] and calculated integral of f(x)

f is integrable because it is piecewise continuous (except at the right-hand end point), i.e. it is continuous over the subintervals [0, 1), [1, 2), and [2, 3). (It doesnt matter if its not continuous at an end point.)

*Last edited by Nehushtan (2014-02-08 03:48:30)*

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**gAr****Member**- Registered: 2011-01-09
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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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bobbym wrote:

Hi;

Did you test that second answer?

Yeah.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,415

Hi;

When you differentiate back, do you get the same function? I do not.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,316

I do. What do you get?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,415

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,316

Oh, sorry, I read it as 1/sqrt(...). That makes it different, sorry.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,415

That is okay, happens to me too.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

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