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!

Induction to prove an expanded integral

  1. Jan 20, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\int_0^1{x^n (1-x)^r}\,dx = \frac{n!}{(r+1)(r+2)...(r+n+1)}[/itex]

    2. Relevant equations
    n≥0, r≥0

    3. The attempt at a solution

    i solved it for the case n=0, where RHS and LHS are the same. however, im having trouble integrating the LHS for the induction step where n=k+1.

    i'm trying to do it by parts, but since the integral keeps expanding i am having trouble with it. any help would be greatly appreciated. thanks!
     
  2. jcsd
  3. Jan 20, 2012 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I suspect you will have to use integration by parts.

    Let u = xk+1 , dv = (1-x)r dx .

    Then do some algebra.
     
  4. Jan 20, 2012 #3
    i think this is what i'm having trouble with. how do i deal with expanded integral?
     
  5. Jan 20, 2012 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    What do you get after applying integration by parts?
     
  6. Jan 20, 2012 #5
    [itex] \frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\int(1-x)^{(r+1)} x^k[/itex]


    CORRECTION:
    restriction on r as stated in problem is,
    [itex]r>-1[/itex]
     
  7. Jan 20, 2012 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Use " \displaystyle " in LaTeX to make the fractions & ∫ sign larger.
    [itex] \displaystyle\frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\int(1-x)^{(r+1)} x^kdx[/itex]​
    I also inserted the dx .

    The following equation has been edited.
    That integral can be broken up as follows:
    [itex]\displaystyle\int(1-x)(1-x)^{r} x^k dx=\int(1-x)^{r} x^kdx-\int\,x(1-x)^{r} x^{k}dx[/itex]​

    And what you're trying to evaluate is the integral [itex]\displaystyle\int(1-x)^{r} x^{k+1}dx\,,[/itex] which is the same as the second integral on the right hand side of the equation immediately above..

    To make the algebra a bit easier, define Ik as [itex]\displaystyle I_k=\int(1-x)^{r} x^{k}dx\,.[/itex]

    Solve for Ik+1 .

    Added in Edit:
    Fixed typo above. I messed that all up with cutting & pasting !
     
    Last edited: Jan 20, 2012
  8. Jan 20, 2012 #7
    I've applied integration by parts to Ik+1

    [itex]\int(1-x)^{r} x^{(k+1)}dx[/itex]

    by setting u=xk+1 and dv=(1-x)r and the result was,

    [itex] \displaystyle\frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\int(1-x)^{(r+1)} x^kdx[/itex]​


    im confused about where to proceed from here.
     
  9. Jan 20, 2012 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I had some cut/paste errors in my previous post. :redface: Sorry about that !!! :redface:

    So, my previous post basically says ... The integral you get in the integration by parts is [itex]\displaystyle \int(1-x)^{(r+1)} x^kdx = I_k - I_{k+1}.[/itex]

    Rewrite your integration by parts result as:
    [itex] \displaystyle I_{k+1}= \frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\left(I_k - I_{k+1}\right)[/itex]​
    Solve that for Ik+1. (The first term is zero when evaluated at x=0 and x=1.)
     
  10. Jan 20, 2012 #9
    i still feel as if im integrating in circles and i don't see the bigger picture.

    EDIT:
    just saw your most recent post. i'll try your suggestion.
    thanks!
     
  11. Jan 20, 2012 #10
    Do you mean solving for Ik+1 and then evaluating the integral between x=0 and x=1?

    i'm ending up with Ik+1 = 0
     
  12. Jan 20, 2012 #11

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes, solve for Ik+1. It doesn't give Ik+1 = 0 .

    Remember that [itex]\displaystyle I_k=\frac{k\,!}{(r+1)(r+2)...(r+k+1)}[/itex] when evaluated from x=0 to x=1.
     
  13. Jan 20, 2012 #12
    EDIT:
    this post was confusing. sorry!
    got it now though.
     
    Last edited: Jan 20, 2012
  14. Jan 20, 2012 #13
    Hey!
    I actually just got it.

    Thank you so much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Induction to prove an expanded integral
  1. Prove by induction (Replies: 3)

  2. Prove by induction (Replies: 5)

  3. Prove by induction (Replies: 5)

  4. Proving by induction (Replies: 5)

Loading...