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Find a and b

  1. Jun 25, 2013 #1

    utkarshakash

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    Gold Member

    1. The problem statement, all variables and given/known data
    If [itex] \displaystyle \int^b_a \dfrac{x^n dx}{x^n + (16-x)^n} = 6 [/itex] and a+b=16, then find a and b.

    3. The attempt at a solution

    [itex] \displaystyle \int^b_a \dfrac{dx}{1 + (16/x - 1)^n} = 6 [/itex]

    I tried substitution but it did not work.
     
  2. jcsd
  3. Jun 25, 2013 #2

    Curious3141

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    Homework Helper

    What sub did you try? Hint: try ##x = 16-y## on the original integral.

    This is a simple algebra problem, more or less. Very little actual integration need be done.
     
  4. Jun 25, 2013 #3

    pasmith

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    Hint:
    [tex]
    \int_a^b\frac{x^n}{x^n + (16-x)^n}\,\mathrm{d}x = \int_a^b 1\,\mathrm{d}x - \int_a^b\frac{(16-x)^n}{x^n + (16-x)^n}\,\mathrm{d}x
    [/tex]
    Can you find a substitution which turns the integrand of the second integral on the right into the integrand of the integral on the left?
     
  5. Jun 26, 2013 #4
    Why do you need substitution right now ?

    Apply the property:

    I= [itex] \displaystyle \int^b_a \dfrac{x^n dx}{x^n + (16-x)^n} [/itex]

    And I= [itex] \displaystyle \int^b_a \dfrac{(16-x)^n dx}{(16-x)^n + (16-(16-x))^n}[/itex]

    Now add the two to get,

    2I= ab∫dx

    Can you proceed? You already know that I=6...
     
  6. Jun 26, 2013 #5

    utkarshakash

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    Oh that was so easy. I first thought of applying property but somehow couldn't notice that a+b=16 was already given.
     
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