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    dks2114's Avatar
    dks2114 Posts: 32, Reputation: 2
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    #1

    Aug 14, 2007, 07:49 PM
    Improper integrals
    Can anyone tell me if the answer to this is: converges to -1/2
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    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #2

    Aug 14, 2007, 09:22 PM
    That depends. What are the limits? -infinity to 0? If so, then yes, it's -1/2.
    dks2114's Avatar
    dks2114 Posts: 32, Reputation: 2
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    #3

    Aug 14, 2007, 09:53 PM
    What if it is from 2 to + inphinity??
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #4

    Aug 15, 2007, 03:47 AM
    In that event, no, it's not -1/2. Show me some of your workings and I'll be glad to point you in the right direction. BTW, I like your spelling of 'infinity'. That's the way it should be spelled.:) :)

    Is this it?:

    dks2114's Avatar
    dks2114 Posts: 32, Reputation: 2
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    #5

    Aug 16, 2007, 09:46 AM
    Haha wow I didn't even notice I wrote "infinity" that way. I'm so use to writing "inphinity" because "Inphinity" he is a great Euro Dj.

    yes the problem does look like that. I think I realized where I went wrong. I someplace lost the squared (1+ x^2) ^2 . Therefore this time I get the answer to be - 1/10

    Here is the new work:


    Name:  problem 5.bmp
Views: 85
Size:  220.8 KB

    I used u-sub to get the general integral of the problem and that I got -1/ 2(1+ x^2) dx that is where I got the third line numbers from. U = 1+ x^2 du= 2x dx 1/2 du = x dx

    If I'm wrong anywhere where did I go wrong and why?
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #6

    Aug 16, 2007, 11:02 AM
    You did good. That's correct.

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