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#1 2008-02-15 15:22:31
- mathminor88
- Member
- Registered: 2007-09-16
- Posts: 12
improper integral type 2
Can something please explain to me how I can finish this. I thought that for t being 6, I would get 1/0 = infinity. Then it would be positive infinity * a negative number and I would get zero, yet somehow the answer is suppose to be infinity.
Last edited by mathminor88 (2008-02-15 15:25:00)
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#2 2008-02-15 15:33:10
- mathminor88
- Member
- Registered: 2007-09-16
- Posts: 12
Re: improper integral type 2
Alright, I don't know what I was doing here...
Last edited by mathminor88 (2008-02-15 15:33:59)
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#3 2008-02-15 15:52:08
- JaneFairfax
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- Registered: 2007-02-23
- Posts: 6,868
Re: improper integral type 2
Given ε > 0, let
Then for all t such that 6 < t < 6+δ, we have
This proves that
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#4 2008-02-15 16:21:34
- John E. Franklin
- Member
- Registered: 2005-08-29
- Posts: 3,588
Re: improper integral type 2
What is this technique called?
I'd love to read about it!
igloo myrtilles fourmis
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