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More reduction formulae troubles

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\int[/itex]x[itex]^{n}[/itex]2[itex]^{x}[/itex]dx = [itex]\frac{2}{ln 2}[/itex] - [itex]\frac{n}{ln 2}[/itex] [itex]\int[/itex]x[itex]^{n-1}[/itex]2[itex]^{x}[/itex]dx

    For n is greater or equal to 1, find [itex]\int[/itex]x[itex]^{3}[/itex]2[itex]^{x}[/itex]dx

    This is a definite integral from 0 to 1

    3. The attempt at a solution

    My first question is, after reducing this once, you are left with x[itex]^{2}[/itex], and attempting to reduce that again means the constant outside the integral will effectively become 0, therefore the constant term outside the integral will be 0 no matter how many more times you reduce it, and after you do the integral, 0 times the integrated result will be zero. What am I not seeing properly here?

    Ignoring that and continuing to do the reduction, I get [itex]\frac{1}{ln 2}[/itex]*[itex]\frac{1}{ln 2}[/itex] which doesn't seem right to ignore it anyways.

    That is to say, when it becomes x[itex]^{1}[/itex], and you apply the formula again, you'll get [itex]\frac{1}{ln 2}[/itex] outside the integral, then you're left with x[itex]^{0}[/itex] which is 1, and effectively you just integrate 2[itex]^{x}[/itex] which becomes [itex]\frac{1}{ln 2}[/itex]2[itex]^{x}[/itex], substituting 0 and 1, and doing the maths, you're left with [itex]\frac{1}{ln 2}[/itex]*[itex]\frac{1}{ln 2}[/itex] as the answer.
  2. jcsd
  3. Aug 15, 2011 #2
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