maggiehodgson
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The questions ask to decide whether the integral can be found and if so give an answer

Here is is with my workings (gave up on latex)



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The answer in the book is 3/2 but for me as P approaches 0 then integration cannot be found

It is an underlined question so I suspect there will be something dodgy that I haven't thought of so pointers/explanations would be gratefully appreciated
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atsruser
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Are you sure that you wrote down the question correctly? As it stands, the integral doesn't converge.
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Hasufel
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maybe should be x^-4 - x^5 - as atsruser sais....
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maggiehodgson
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(Original post by atsruser)
Are you sure that you wrote down the question correctly? As it stands, the integral doesn't converge.

Oh made a mistake with latex. The denominator should've been x^5.
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maggiehodgson
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(Original post by Hasufel)
maybe should be x^-4 - x^5 - as atsruser sais....

Oh made a mistake with latex. The denominator should've been x^5.
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davros
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(Original post by maggiehodgson)
Oh made a mistake with latex. The denominator should've been x^5.
So you're integrating \dfrac{x-1}{x^5} between 0 and 1? That doesn't look like it's going to converge at all!

(Is P what you're told to use as a lower limit and then let P->0 ?)
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maggiehodgson
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(Original post by davros)
So you're integrating \dfrac{x-1}{x^5} between 0 and 1? That doesn't look like it's going to converge at all!

(Is P what you're told to use as a lower limit and then let P->0 ?)

In the examples in the book it tells me that because the denominator at x=0 would give an infinite value then you replace it by P, integrate and see what happens.

For example from the book:

Integrate 1/root x between 0 and 1. This is an improper integral since the integral is undefined at x = 0. Replace the lower limit by P and integrate. to give [2x^.5] between 1 and p. This gives 2 - root p. As p approaches zero the integral approaches 2. Therefore the improper integral can be found and its value is 2.

That's what I was copying for the question posted. But because I still end up with a p denominator then as p approaches zero it's not converging to any finite number.

Have I explained myself properly?

Thanks for the time you're taking.
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davros
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(Original post by maggiehodgson)
In the examples in the book it tells me that because the denominator at x=0 would give an infinite value then you replace it by P, integrate and see what happens.

For example from the book:

Integrate 1/root x between 0 and 1. This is an improper integral since the integral is undefined at x = 0. Replace the lower limit by P and integrate. to give [2x^.5] between 1 and p. This gives 2 - root p. As p approaches zero the integral approaches 2. Therefore the improper integral can be found and its value is 2.

That's what I was copying for the question posted. But because I still end up with a p denominator then as p approaches zero it's not converging to any finite number.

Have I explained myself properly?

Thanks for the time you're taking.
Your method is absolutely fine

If the question is as you've stated it then the answer can't be 3/2 or any other finite value as you've correctly deduced.
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maggiehodgson
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(Original post by davros)
Your method is absolutely fine

If the question is as you've stated it then the answer can't be 3/2 or any other finite value as you've correctly deduced.

Well, Thank you very much. As I am only working from a text book with no teacher I often have to consult TSR for clarification. I'll carry on now.

Again, thanks.
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