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Evaluation of a reduction formula

  1. Aug 14, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\int[/itex] (1 - x^2)^n dx = [itex]\frac{2n}{2n+1}[/itex] [itex]\int[/itex] (1 - x^2)^n-1 dx

    for n greater or equal to 1, find [itex]\int[/itex] (1 - x^2)^4 dx

    The integrals go from 0 to 1

    2. Relevant equations



    3. The attempt at a solution

    Well what I did was to keep doing n - 1 whilst pulling a fraction outside the integral and multiplying each subsequent fraction.

    So I ended up with 8/9 * 6/7 * 4/5 and the integral of (1 - x) which became x - x^2/2

    Substituting in 1 and 0, subtracting and the multiplying by the fractions I got 192/305*1/2 which simplifies to 96/305.

    Is this the correct procedure for evaluating reduction formulae, and was my answer correct?
     
  2. jcsd
  3. Aug 14, 2011 #2
    I get:

    [itex]\int_0^1 (1 - x^2)^4 dx = \frac{8}{9}\frac{6}{7}\frac{4}{5}\frac{2}{3}\int_0^1 1 dx[/itex]

    So the answer is: [itex] \frac{(8)(6)(4)(2)}{(9)(7)(5)(3)} = \frac{2^7}{(3^2)(5)(7)} = \frac{128}{315} [/itex]

    Wolframe Alpha agrees with the above answer.
     
  4. Aug 15, 2011 #3
    Ahh, I see my mistake now, for n greater or equal to 1, I needed to have reduced it one more time to get the 2/3.

    Thanks
     
  5. Aug 15, 2011 #4
    I'd like to know if possible what page you used to cross check that, I'm looking at Wolfram Alpha and can't really find anything to evaluate reduction formulae, though I can see the step-by-step integrator.
     
  6. Aug 15, 2011 #5
    You don't need to give WA the correct formulas with which to evaluate an integral...
    "int from 0 to 1 (1-x^2)^4"
    returns 128/315
    See: http://www.wolframalpha.com/input/?i=int+from+0+to+1+(1-x^2)^4
    It will do infinite sums, limits and derivatives too. I find it really useful for checking work.
     
  7. Aug 15, 2011 #6
    Awesome, thanks for that.
     
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