Induction to prove an expanded integral

[skoomafiend]

Homework Statement

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

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, I am 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!

 

[SammyS]

Homework Statement

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

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, I am 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 = x^k+1 , dv = (1-x)^r dx .

Then do some algebra.

 

[skoomafiend]

i think this is what I'm having trouble with. how do i deal with expanded integral?

 

[SammyS]

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?

 

[skoomafiend]

$\frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\int(1-x)^{(r+1)} x^k$


CORRECTION:
restriction on r as stated in problem is,
$r>-1$

 

[SammyS]

$\frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\int(1-x)^{(r+1)} x^k$


CORRECTION:
restriction on r as stated in problem is,
$r>-1$

Use " \displaystyle " in LaTeX to make the fractions & ∫ sign larger.

$\displaystyle\frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\int(1-x)^{(r+1)} x^kdx$​

I also inserted the dx .

The following equation has been edited.
That integral can be broken up as follows:

$\displaystyle\int(1-x)(1-x)^{r} x^k dx=\int(1-x)^{r} x^kdx-\int\,x(1-x)^{r} x^{k}dx$​

And what you're trying to evaluate is the integral $\displaystyle\int(1-x)^{r} x^{k+1}dx\,,$ 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 I_k as $\displaystyle I_k=\int(1-x)^{r} x^{k}dx\,.$

Solve for I_k+1 .

Added in Edit:
Fixed typo above. I messed that all up with cutting & pasting !

 

[skoomafiend]

I've applied integration by parts to I_k+1

$\int(1-x)^{r} x^{(k+1)}dx$

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

$\displaystyle\frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\int(1-x)^{(r+1)} x^kdx$​

im confused about where to proceed from here.

 

[SammyS]

I've applied integration by parts to I_k+1

$\int(1-x)^{r} x^{(k+1)}dx$

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

$\displaystyle\frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\int(1-x)^{(r+1)} x^kdx$​

I'm confused about where to proceed from here.

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

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

Rewrite your integration by parts result as:

$\displaystyle I_{k+1}= \frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\left(I_k - I_{k+1}\right)$​

Solve that for I_k+1. (The first term is zero when evaluated at x=0 and x=1.)

 

[skoomafiend]

i still feel as if I am integrating in circles and i don't see the bigger picture.

EDIT:
just saw your most recent post. i'll try your suggestion.
thanks!

 

[skoomafiend]

Rewrite your integration by parts result as:

$\displaystyle I_{k+1}= \frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\left(I_k - I_{k+1}\right)$​

Solve that for I_k+1. (The first term is zero when evaluated at x=0 and x=1.)

Do you mean solving for I_k+1 and then evaluating the integral between x=0 and x=1?

i'm ending up with I_k+1 = 0

 

[SammyS]

Do you mean solving for I_k+1 and then evaluating the integral between x=0 and x=1?

I'm ending up with I_k+1 = 0

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

Remember that $\displaystyle I_k=\frac{k\,!}{(r+1)(r+2)...(r+k+1)}$ when evaluated from x=0 to x=1.

 

[skoomafiend]

EDIT:
this post was confusing. sorry!
got it now though.

 

[skoomafiend]

Hey!
I actually just got it.

Thank you so much!