1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: More reduction formulae troubles
  1. Reduction Formula (Replies: 9)

  2. Reduction formula (Replies: 3)

  3. Reduction Formulae (Replies: 1)

  4. Reduction formula (Replies: 5)

Loading...