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Math Help - Improper integral question

  1. #1
    Junior Member
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    Feb 2006
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    Victoria, Australia
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    36

    Improper integral question

    I got this question and im having trouble with, how do i go about doing it.\

    question 7



    and also

    question 10




    thanks
    Last edited by sterps; March 24th 2006 at 05:12 PM.
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  2. #2
    Member
    Joined
    Nov 2005
    From
    Wethersfield, CT
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    92
    Hi:

    As the present time is 2:14 AM, I will get you started with the first problem, and return in a few hours to finish up (hopefully).

    First, we interpret the improper integral, i.e., int[f(x) dx] ; [a,inf], as the limit of the definite integral on interval [a,n] as n --> inf. In the present case the lower limit of integration is 3, making the interval [3,n].

    Now, if we complete the square on denominator x^2+2x+10, the result is (x^2+2x+1) + 9, or (x+1)^2 + 9. So we have:

    Lim_n->inf [12 int (dx / ((x+1)^2 + 9))]. Yes, I know it's time to learn latex.
    Letting x+1 = 3tan() gives dx = 3sec^2() * d. Hence the integral becomes:

    12*3 int [sec^2() * d / (9tan^2() + 9)] = 4 int [sec^2() * d / (tan^2() + 1)].

    Applying the identity tan^2() + 1 = sec^2() leaves 4 int [sec^2() * d / sec^2() ] = 4 int (d) = 4.

    Now, recall the substitution x+1 = 3tan(). Because the interval on x is [3,n], we can derive an equivalent interval on by solving for in,

    1) 3+1 = 3tan() and 2) n+1 = 3tan(). Thus the new interval is [arctan(4/3), arctan((n+1)/3)]. And the integral, i.e., 4 ; [x=(3,n)] is evaluated as:

    4 ; [arctan(4/3), arctan((n+1)/3)] = 4 * [arctan((n+1)/3) - arctan(4/3)].

    I leave the finishing details to you. I'm off to catch up on some sleep.

    Regards,

    Rich B.
    Last edited by Rich B.; March 25th 2006 at 03:30 AM.
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