Exercise 5.5 Cambridge Question 20 (1 Viewer)

[.ben]

hi can someone do this question for me. i know how to do the bit by parts but then manipulating it to get the I_n expression is weird. Does it have something to do with expalnding it?

 

[Rax]

I got a different Cambridge....so I can't help ya.

Looking through it but cant find the actual question you are talking about

 

[.ben]

yes, sorry i forgot to post the question

 

[Trev]

It's to do with expanding it and finding a pattern.
I₁=3/5
I₂=[(3.2)/(3.2+2)].I₁=(6/8).(3/5)
I₃=[(3.3)/(3.3+2)].I₂=(9/11).(6/8).(3/5)
Etc.
You can see that the numerator will be 3.6.9.12...etc, which is 3(1.2.3.4...); 3(n!).
You then need to get the denominator in terms of n.

 

[shimmerz_777]

doing it by parts is tricky, i got stuck but eventually got it, dont ask how, heres the working, note that l represents the intergral, and it will be between 0 and 1 for all uses, as i dont know how to type it up neatly on the cpu


l x(1-x^3)^n dx dv=x v=1/2 x^2 u= (1-x^3)^n du= (-3nx^2)(1-x^3)^n-1

=[ v*u] (which equals 0 between 1 and 0) + 3n/2 *l (x^4)(1-x^3)^n-1 dx

this bit gets really tricky, i knew i was on the right track because i had a 3n/2, which was kinda like the answer

x^4 becomes rearranged to x -x(1-x^3), then sub it back in and u get:

(3n/2)* l [x -x(1-x^3)](1-x^3)^n-1 dx =
(3n/2)* l x(1-x^3)^(n-1) - x(1-x^3)^n dx
so you get

In= 3n/2* [I(n-1) - In]
In*(1 + 3n/2)= 3n/2* I(n-1)
which simplifies to In= (3n/(2+3n))* I(n-1)


i had help from my tutor at the x^4 bit, but its all there, may be a bit hard to follow though due to the typing