ahh, sorry my working out must be really bad
I'll try to explain it better then..
I_n=int. x(1-x)^n dx {from x=0 to x=1}
Using the technique of integration by parts...
let u=(1-x)^n and dv\dx=x
du\dx=-n(1-x)^(n-1) and v=(1\2)x²
I_n=uv-int. v*(du\dx)*dx
=[(1-x)^n * (1\2)x²]-->(x=0 to x=1) - int. (1\2)x²*(-n(1-x)^(n-1) ) dx (x=0 to x=1)
I_n=[(1\2)x² (1-x)^n]-->(x=0 to x=1) +n\2 int.x²(1-x)^(n-1) dx (x=0 to x=1)
So the first bracket cancels if you sub in the terminals, this leaves you with:
I_n=n\2 int.x²(1-x)^(n-1) dx {x=0 to x=1}
Now I decided to put back the negative in the integral,
I also split up the x² into x.x
I_n=-(n\2) int. -x. x(1-x)^(n-1) dx {x=0 to x=1}
Now I made it (1-x) there... but you gota take away any extra things in the process of doing that... that's why I take away the x(1-x)^(n-1) at the end:
=(-n\2)int. [ (1-x)x(1-x)^(n-1)-x(1-x)^(n-1) ] dx {x=0 to x=1}
=(-n\2) int. x(1-x)^n dx +(n\2)int. x(1-x)^(n-1) dx {x=0 to x=1}
=(-n\2)I_n+(n\2) I_(n-1)
.: I_n(1+n\2)=(n\2) I_(n-1)
I_n=(n\2)\(1+n\2) * I_(n-1)
=n\(2+n) I_(n-1)
Lets c if this will explain it...
I'm not very good at explaining lol...
