2005 CSSA - Q2(e) (1 Viewer)

[Stefano]

I know I need to use Integration by Parts, but I can't seem to get this out. I feel bad about this because it's only Q2 :S

I've let u'=x & v=(1-x)^n - correct so far ?

I end up with this:

I_n = n/2 Int[x^2(1-x)^n-1] .dx -yes? no ?

I don't know where to take it from there. Is that integral I_n-1 ?


Help !

 

[KFunk]

Both add and subtract n/2∫x(1-x)^n-1 dx. Take out a common factor of (1-x) after combining the negative one with what you already had and you will get I_n-1 and I_n.

 

[Stefano]

Both add and subtract n/2∫x(1-x)^n-1 dx. Take out a common factor of (1-x) after combining the negative one with what you already had and you will get I_n-1 and I_n.

I don't understand any of that. You refer to 'both', for example; thats I_n and ?

 

[香港!]

I know I need to use Integration by Parts, but I can't seem to get this out. I feel bad about this because it's only Q2 :S

I've let u'=x & v=(1-x)^n - correct so far ?

I end up with this:

I_n = n/2 Int[x^2(1-x)^n-1] .dx -yes? no ?

I don't know where to take it from there. Is that integral I_n-1 ?


Help !

I_n=int. x(1-x)^n dx {from x=0 to x=1}
let u=(1-x)^n and dv\dx=x
du\dx=-n(1-x)^(n-1) and v=(1\2)x²
I_n=[(1\2)x² (1-x)^n]+n\2 int.x²(1-x)^(n-1) dx both {x=0 to x=1}
=n\2 int.x²(1-x)^(n-1) dx {x=0 to x=1}
=-(n\2) int. -x. x(1-x)^(n-1) dx
=(-n\2)int. [ (1-x)x(1-x)^(n-1)-x(1-x)^(n-1) ] dx
=(-n\2) int. x(1-x)^n dx +(n\2)int. x(1-x)^(n-1) dx
=(-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)

 

[KFunk]

I don't understand any of that. You refer to 'both', for example; thats I_n and ?

Sorry, I should have put a bit more. It's a pretty similar thing to what 香港! did. Going from where you got up to:

I_n = n/2∫x²(1-x)^n-1dx

I_n = n/2∫x²(1-x)^n-1dx + n/2∫x(1-x)^n-1dx - n/2∫x(1-x)^n-1dx

I_n = n/2∫x(1-x)^n-1dx - (n/2)∫x(1-x)(1-x)^n-1dx

I_n = (n/2)I_n-1 - (n/2)I_n etc...

What I meant was that from where you got up to you could add and subtract n/2∫x(1-x)^n-1 dx and take out a common factor of (1-x) to get your I terms. As above.

 

[Stefano]

I_n=int. x(1-x)^n dx {from x=0 to x=1}
let u=(1-x)^n and dv\dx=x
du\dx=-n(1-x)^(n-1) and v=(1\2)x²
I_n=[(1\2)x² (1-x)^n]+n\2 int.x²(1-x)^(n-1) dx both {x=0 to x=1}
=n\2 int.x²(1-x)^(n-1) dx {x=0 to x=1}
=-(n\2) int. -x. x(1-x)^(n-1) dx ---- line6
=(-n\2)int. [ (1-x)x(1-x)^(n-1)-x(1-x)^(n-1) ] dx ---line7
=(-n\2) int. x(1-x)^n dx +(n\2)int. x(1-x)^(n-1) dx
=(-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)

HOLY SHIT!

To go from line6 to line7 is tricky shit man. How the hell did you think of that?? You must be thinking like 4 steps ahead at all times...


But thanks for the solution.

 

[香港!]

it's a "technique" that's usually taught at school or maybe even txt books @_@

 

[Stefano]

I've never EVER seen or heard it before.

Why am I saying that with 19 days till the HSC ffs!? Waste of private school education.

 

[香港!]

I've never EVER seen or heard it before.

Why am I saying that with 19 days till the HSC ffs!? Waste of private school education.

hahaha
don't worry there's not too many of these 'techniques'... so ur not missing a lot=P

 

[Stefano]

hahaha
don't worry there's not too many of these 'techniques'... so ur not missing a lot=P

I certainly hope not

So how do you know when this technique is needed? Does it crop up often in HSC exams or just CSSA trials ?

 

[香港!]

I certainly hope not

So how do you know when this technique is needed? Does it crop up often in HSC exams or just CSSA trials ?

u use it when you see u need to.......
dunno which exams they come up in...

 

[Vampire]

I wouldn't compare integration skillz with 香港! lol...he's awesome at that stuff.

 

[haboozin]

I wouldn't compare integration skillz with 香港! lol...he's awesome at that stuff.

hes got a challenge with final fantacy.

 

[Vampire]

ahahahaha...he is final fantasy

 

[香港!]

I'm not Final Fantasy!!

 

[haboozin]

I'm not Final Fantasy!!

Well he stopped talking 3 days after you joined and he is from hong kong

 

[香港!]

hahaha, but there's plenty of Hong Kong people here

 

[Stefano]

I still can't do this damn question.
This thread is now off topic.

 

[香港!]

I still can't do this damn question.
This thread is now off topic.

what part of my solution u didn't get??

 

[~ ReNcH ~]

Yea. I wonder where FinalFantasy is... This seems like the type of question he would answer. He must be dedicating his time to studying - after all, from what I can see on BoS he's definitely heading for a 95+ in MX2.