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Homework Help: Simple maths stuff

  1. Sep 23, 2004 #1
    This seems so simple, but I can't for the life of me work it out. I've forgotten a lot of maths over the years, so I need a little help.

    The question goes:

    Using the substitution


    Show that :

    [tex]\int_R^{R_1} \sqrt{\frac{R_1}{r}-1} dr = \int_\frac{R}{R_1}^1 \sqrt{\frac{1}{u}-1} du[/tex]

    So, the substitution under the root is easy enough, it's just the changing of the limits that I can't seem to figure out (or perhaps remember).

    I got



    So substituting

    [tex]\int_R^{R_1} \sqrt{\frac{R_1}{r}-1} dr = R_1\int_R^{R_1}\sqrt{\frac{1}{u}-1} du[/tex]

    I can see that if I divide both limits by the factor outside the integral, [itex] R_1[/itex], I'll get the right answer, but surely I can't just do that.
    Any hints?

  2. jcsd
  3. Sep 23, 2004 #2
    The limits are r = R and r = R1. The corresponding limits after the substitution are u = R/R1 and u = 1 (since uR1 = r). The factor of R1 remains though (in your last integral). I don't see how it could disappear, but who knows...
  4. Sep 23, 2004 #3
    thanks, I knew it was reasonably trivial, I just couldn't see it.

    That factor of R1 is still supposed to be there though right?
    Maybe another mistake on this assignment....

    I also need help with the next part, which I don't even know where to start on.

    Show that in the limit [itex]\frac{R}{R_1}\rightarrow0[/itex]

    [tex]\int_{\frac{R}{R_1}}^1\sqrt{\frac{1}{u}-1}du \approx \frac{\pi}{2}-\int_0^{\frac{R}{R_1}}\sqrt{\frac{1}{u}}du[/tex]

    I have no bloody idea where to start, it's doing my head in! It can't be that difficult!!
    Any hints?
  5. Sep 23, 2004 #4


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    [tex]\int _x^1 \sqrt {\frac {1}{u} -1} du = \int _0^1\sqrt {\frac {1}{u} -1} du - \int_0^x \sqrt {\frac {1}{u} -1} du[/tex]

    The first integral is [itex]\frac {\pi}{2}[/itex] and for the second observe that
    [tex]\sqrt {\frac {1}{u} - 1} = \sqrt {\frac { 1-u}{u}}[/tex]
    and use the binomial expansion on the numerator. I think there's a factor of [itex]\frac {1}{2}[/itex] missing in the last term of your equation.
  6. Sep 23, 2004 #5
    How do you just know that the first integral is [itex]\frac{\pi}{2}[/itex]?
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