1. ## Integral problem

Need help with some aspects of this problem (particularly the last part). Would greatly appreciate any help offered.

'For $n \geq 0$, let

$I_n=\int\limits_0^1{x}^{n}(1-x)^n\, \mathrm{d}x$

For $n \geq 1$, show, by means of a substitution, that

$\int\limits_0^1{x}^{n-1}(1-x)^n\, \mathrm{d}x=\int\limits_0^1{x}^n(1-x)^{n-1}\, \mathrm{d}x$

and deduce that

$2\int\limits_0^1{x}^{n-1}(1-x)^n\, \mathrm{d}x=I_{n-1}$

Show also, for $n \geq 1$, that

$I_n=\frac{n}{n+1}\int\limits_0^1{x}^{n-1}(1-x)^{n+1}\, \mathrm{d}x$

and hence that

$I_n=\frac{n}{2(2n+1)}I_{n-1}$.'

2. ## Re: Integral problem

A fairy obvious substitution is u= 1- x. Of course the dummy variable of integration on the right would then be "u" which you could change to "x".

3. ## Re: Integral problem

Yes, thanks Halls, no problem with the first.
Would this be correct for the next?:

$2\int\limits_0^1{x}^{n-1}(1-x)^{n}\, \mathrm{d}x = \int\limits_0^1{x}^{n-1}(1-x)^n\, \mathrm{d}x + \int\limits_0^1{x}^n(1-x)^{n-1}\, \mathrm{d}x = \int\limits_0^1{x}^{n-1}(1-x)^{n} + {x}^{n}(1-x)^{n-1}\, \mathrm{d}x = \int\limits_0^1{x}^{n-1}(1-x)^{n-1}(1-x+x) = I_{n-1}$

4. ## Re: Integral problem

Ok, I'm alright with this problem apart from the last part. Can anybody help?

5. ## Re: Integral problem

Originally Posted by jimi
Ok, I'm alright with this problem apart from the last part. Can anybody help?
Integration by parts

$\displaystyle u=x^n$

$\displaystyle dv=(1-x)^ndx$