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Definite integral approaches infinity

  1. Feb 17, 2014 #1
    1. The problem statement, all variables and given/known data

    [itex]180\int_5^\propto \frac{2}{(4+x^{2})^{3/2}} dx[/itex]


    2. Relevant equations

    Trigonometric Substitutions: (x=2 tan z).

    3. The attempt at a solution

    I've computed the integral and ended up with [itex]180 [\frac{x}{2(4+x^2)^{1/2}}][/itex] from 5 to infinity.

    I could've easily computed the limit of 5, but couldn't have found the limit approaches infinity, as I'm always ending up with 0, although my textbook and symolab calculator says that it's 1/2.
     
  2. jcsd
  3. Feb 17, 2014 #2

    Ray Vickson

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    Either use l'Hospital's rule or else note that for x > 0 we have
    [tex] \frac{x}{\sqrt{x^2+4}} = \frac{x}{x \sqrt{1 + 4x^{-2}}} = \frac{1}{\sqrt{1+4 x^{-2}}} \to 1 [/tex]
    as ##x \to \infty##.
     
  4. Feb 17, 2014 #3

    SteamKing

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    [itex]\propto[/itex] means 'is proportional to' and should not be used to indicate infinity [itex]\infty[/itex]
     
  5. Feb 18, 2014 #4
    If it is infinity and not saying proportional to. Then you can calculate the limit to infinity pretty easily.

    Take [itex]\frac{x}{2\sqrt{4+x^{2}}}[/itex] and then make the numerator [itex]\sqrt{x^{2}}[/itex] and pull out the [itex]\frac{1}{2}[/itex]

    So now you can call the whole thing [itex]\frac{1}{2}[/itex][itex]\sqrt{\frac{x^{2}}{4+x^{2}}}[/itex]

    Now apply L'Hopital's rule.

    [itex]\frac{1}{2}[/itex][itex]\sqrt{\frac{2x}{2x}}[/itex]

    this way, the 2x and 2x cancel to give you [itex]\sqrt{1}[/itex] multiplied by [itex]\frac{1}{2}[/itex]. Which was your answer.
     
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