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Math Help - Integral from Cambridge STEP Paper.

  1. #1
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    Integral from Cambridge STEP Paper.

    Show, by means of a suitable change of variable, or otherwise, that

    <br />
\int_0^\infty f((x^2+1)^{1/2} +x)  dx = \frac{1}{2}\int_1^\infty (1+t^{-2})f(t) dt<br />

    Hence, or otherwise, show that

    <br />
\int_0^\infty ((x^2+1)^{1/2} +x)^{-3}  dx = \frac{3}{8}<br />

    This is taken directly from STEP 1, 1998. Now the challenge isn't finding a solution, that's cake for many of you, the challenge is explaining it to me. A maths student in the first year of sixth-form (I think you call it high-school in America.) I've never seen integrals like these before, the most advanced form of integration I've used is integrations by parts.

    Please explain EVERYTHING!
    Last edited by AAKhan07; May 27th 2008 at 06:30 AM.
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  2. #2
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    Quote Originally Posted by AAKhan07 View Post
    Show, by means of a suitable change of variable, or otherwise, that

    <br />
\int_0^\infty f((x^2+1)^{1/2} +x)  dx = \frac{1}{2}\int_1^\infty (1+t^{-2})f(t) dt<br />

    Hence, or otherwise, show that

    <br />
\int_0^\infty ((x^2+1)^{1/2} +x)^{-1/3}  dx = \frac{3}{8}<br />

    This is taken directly from STEP 1, 1998. Now the challenge isn't finding a solution, that's cake for many of you, the challenge is explaining it to me. A maths student in the first year of sixth-form (I think you call it high-school in America.) I've never seen integrals like these before, the most advanced form of integration I've used is integrations by parts.

    Pleas explain EVERYTHING!
    The questions requires a simple change of variable (which you do learn in high school). you start by letting t = (x^2+1)^{1/2} +x as x becomes large so does t. but for x = 0 t = 1 hence the limits change.

    dt/dx = x(x^2+1)^{-1/2} + 1 do dx = \frac{dt}{x(x^2+1)^{-1/2} + 1} now you need to show that \frac{dt}{x(x^2+1)^{-1/2} + 1} is the same as 1/2(1+t^{-2}) and the algebra involves in this is quiet fiddly.

    firstly  \frac{1}{x(x^2+1)^{-1/2} + 1} = \frac{(x^2+1)^{1/2}}{x + (x^2+1)^{1/2}} = \frac{(x^2+1)^{1/2}}{t} now multiply top and bottom by t \frac{(x^2+1)^{1/2}( (x^2+1)^{1/2} +x)}{t^2} I'm going to ommit a bit of algebra here but you should be able to work it out as 1/2 (\frac{1 + t^2}{t^2}) = 1/2 (1+t^{-2}).


    For the next part you use the pervious result.

    let f = x^{-3}
    you integral is transformed to 1/2 \int_1^\infty (1+t^{-2})(t)^{-3}  dt \ \ \Rightarrow 1/2 \int_1^\infty (t^{-3}+t^{-5})  dt

    = 1/2 \left| -(1/2) t^{-2} - (1/4) t^{-4} \right| _1^\infty  = (1/2)( 0 + 1/2 + 1/4 ) = 3/8

    note you don't need any advance knowledge of convergence / divergence to state that  \frac{1}{ \infty} = 0

    Bobak
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  3. #3
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    OK, understood.

    It all looks pretty clear, it's going to take a bit of brain training before I can do that kind of work independently, but it's nothing too hard. Thanks Bobak.
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