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Math Help - 'Reduction Formula' problem

  1. #1
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    'Reduction Formula' problem

    I'm just not sure how to go about solving this:

    In= ∫(x^n).(1-x)^(3/2)dx
    The integral has limits from 1 -> 0

    And I need to show :

    In = (2n/2n+5) I(n-1)


    Thanks in advance for any help
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  2. #2
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    Re: 'Reduction Formula' problem

    Quote Originally Posted by Mrhappysmile View Post
    I'm just not sure how to go about solving this:

    In= ∫(x^n).(1-x)^(3/2)dx
    The integral has limits from 1 -> 0

    And I need to show :

    In = (2n/2n+5) I(n-1)
    Integrate by parts, using the fact that \int(1-x)^{3/2} = -\tfrac25(1-x)^{5/2}. Then

    \begin{aligned}I_n &= \Bigl[-\tfrac25(1-x)^{5/2}x^n\Bigr]_0^1 + \int_0^1nx^{n-1}\tfrac25(1-x)^{5/2}dx \\ &= \tfrac{2n}5\int_0^1 x^{n-1}(1-x)(1-x)^{3/2}dx = \tfrac{2n}5I_{n-1} - \tfrac{2n}5I_n. \end{aligned}
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  3. #3
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    Re: 'Reduction Formula' problem

    Quote Originally Posted by Opalg View Post
    Integrate by parts, using the fact that \int(1-x)^{3/2} = -\tfrac25(1-x)^{5/2}. Then

    \begin{aligned}I_n &= \Bigl[-\tfrac25(1-x)^{5/2}x^n\Bigr]_0^1 + \int_0^1nx^{n-1}\tfrac25(1-x)^{5/2}dx \\ &= \tfrac{2n}5\int_0^1 x^{n-1}(1-x)(1-x)^{3/2}dx = \tfrac{2n}5I_{n-1} - \tfrac{2n}5I_n. \end{aligned}
    Of course! I'd done exactly what you'd done but I didn't seen the simple bit i.e. dividing through by 5+2n to give the answer.
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  4. #4
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    Re: 'Reduction Formula' problem

    Thanks
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