# fp1 IMPROPER INTEGRALS

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Thread starter 6 years ago
#1
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)

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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6 years ago
#2
Are you sure that you wrote down the question correctly? As it stands, the integral doesn't converge.
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6 years ago
#3
maybe should be x^-4 - x^5 - as atsruser sais....
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Thread starter 6 years ago
#4
(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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Thread starter 6 years ago
#5
(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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6 years ago
#6
(Original post by maggiehodgson)
Oh made a mistake with latex. The denominator should've been x^5.
So you're integrating 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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Thread starter 6 years ago
#7
(Original post by davros)
So you're integrating 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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6 years ago
#8
(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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Thread starter 6 years ago
#9
(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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