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Math Help - Improper Integrals

  1. #1
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    Improper Integrals

    Can someone please help me with these? I need some guidance. I would really appreciate it.

    1. int {1/[radical(x-1)]} from 2 to infinity. 2 is the lower limit, infinity is the upper limit. i put it converges because i looked at the graph and i think it has a limit?



    2. int {e^2x} dx from -infinity to 0

    i put it diverges because the limit does not exist. i got [(e^2x)/2] 0 to N. then i put (e^0 / 2) - (e^2N / 2)
    and then i looked at the graph of the integral.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by darkblue View Post
    Can someone please help me with these? I need some guidance. I would really appreciate it.

    1. int {1/[radical(x-1)]} from 2 to infinity. 2 is the lower limit, infinity is the upper limit. i put it converges because i looked at the graph and i think it has a limit?
    this diverges. we only have a problem at x = 1, so since that is not between our limits of integration we dont have to split the integral up.

    \int_{2}^{ \infty} \frac {1}{ \sqrt {x - 1}} dx = \lim_{N \to \infty} \int_{2}^{N} (x - 1)^{- \frac {1}{2}}

    = \lim_{N \to \infty} \left[ 2(x - 1)^{ \frac {1}{2}} \right]_{2}^{N}

    = \lim_{N \to \infty} \left[ 2(N - 1)^{ \frac {1}{2}} - 2(2 - 1)^{ \frac {1}{2}} \right]

    = \infty

    So the integral diverges


    looking at the graph and seeing that is has a limit means nothing. look at the graph of \frac {1}{x}, it's limit is 0 as x goes to \infty, but its improper integral diverges

    2. int {e^2x} dx from -infinity to 0

    i put it diverges because the limit does not exist. i got [(e^2x)/2] 0 to N. then i put (e^0 / 2) - (e^2N / 2)
    and then i looked at the graph of the integral.
    this converges.


    \int_{- \infty}^{0} e^{2x}dx = \lim_{N \to - \infty} \int_{N}^{0} e^{2x} dx

    = \lim_{N \to - \infty} \left[ \frac {1}{2} e^{2x} \right]_{N}^{0}

    = \lim_{N \to - \infty} \left[ \frac {1}{2} e^0 - \frac {1}{2} e^{N} \right]

    = \frac {1}{2} - 0 < \infty

    Thus the integral converges


    remember, when we raise something to a negative exponent, it's the same as putting 1 over the something to the positive exponent. So e^{- \infty} = \frac {1}{e^{ \infty}} = \frac {1}{ \infty} = 0
    Last edited by Jhevon; May 28th 2007 at 10:11 PM.
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