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How do I evaluate this definite integral in Calc 1?

  1. Nov 25, 2005 #1
    I am trying to evaluate this integral,

    [tex]\[
    \int_{ - r}^r {\left( {s\sqrt {1 + \frac{{x^2 }}{{r^2 - x^2 }}} } \right)} {\rm }dx\] [/tex]

    Is it a valid integral?

    If I evaluate it by plugging in r and -r, it becomes undefined.
    How else can I evaluate this integral? I'm in calculus 1 and I am guessing
    this may be a topic covered in later calculus classes?


    EDIT: I had the integral wrong. It should be,

    [tex]\[
    \int_{ - r}^r {\left( {s\sqrt {1 + \frac{{r^2 }}{{r^2 - x^2 }}} } \right)} {\rm }dx\] [/tex]

    Thanks
     
    Last edited: Nov 25, 2005
  2. jcsd
  3. Nov 25, 2005 #2

    Physics Monkey

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    The integral is well defined (the singularity you observed is called integrable because it doesn't bother the integral), and you can calculate it really easily. Hint: think about symmetry.

    Edit: Haha. Ok Halls, fairly easily. :smile:
     
    Last edited: Nov 25, 2005
  4. Nov 25, 2005 #3

    HallsofIvy

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    Well, not all that "really easily". Opticaltempest, have you covered "trigonometric substitutions" yet?
     
  5. Nov 25, 2005 #4
    I got the original integral wrong. I updated it in an edit. Sorry

    I am allowed to use technology right now to evaluate the above integral.

    Using Maple 10 I find the above integral evaluates to

    [tex] \[
    - \sqrt r \cdot \sqrt { - r} \cdot s\left( {\ln ( - r) - \ln (r)} \right)
    \] [/tex]

    Which simplifies to,

    [tex] \[
    - r \cdot s \cdot i(\pi \cdot i)
    \]
    [/tex]

    Which simplified to,

    [tex] \[
    r \cdot s \cdot \pi
    \]
    [/tex]


    This was what I was looking for :) I haven't covered using trigonometric substitution yet.
    I was just curious how I would integrate that without using a CAS.
     
  6. Nov 26, 2005 #5

    GCT

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    The form seems familiar, from what I remember it may have something to do with the arc length derivation.
     
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