Induction to prove an expanded integral

  • #1

Homework Statement



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

Homework Equations


n≥0, r≥0

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!
 

Answers and Replies

  • #2
SammyS
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Homework Statement



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

Homework Equations


n≥0, r≥0

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!
I suspect you will have to use integration by parts.

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

Then do some algebra.
 
  • #3
i think this is what i'm having trouble with. how do i deal with expanded integral?
 
  • #4
SammyS
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i think this is what I'm having trouble with. how do i deal with expanded integral?
What do you get after applying integration by parts?
 
  • #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]
 
  • #6
SammyS
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[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]
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:
  • #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.
 
  • #8
SammyS
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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]​

I'm confused about where to proceed from here.
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.)
 
  • #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!
 
  • #10
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.)
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
 
  • #11
SammyS
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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
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.
 
  • #12
EDIT:
this post was confusing. sorry!
got it now though.
 
Last edited:
  • #13
Hey!
I actually just got it.

Thank you so much!
 

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