2005 CSSA Reduction Integration Formula (1 Viewer)

[AreYouAlright?]

Hey guys, this question is probably easy but i can never figure out whether to do it by integration by parts or not, and if not i never know what to do to start it off. Lol could any of you explain to me how i work this out.

If I_n = intg(0-->1) [x(1-x)^n] dx show that I_n = n/(n+2).I_(n-1) ...? Help lease lol i got 43% in my trial ewwww....

 

[KFunk]

Are you sure there's not an x² term in there somewhere? 'Cause I keep turning up answers in terms of 'n' via straight integration. Maybe I've fallen out of practise with the recurrence stuff already but trying the substitutions (u = 1 - x) and (x = sin²θ ) I got non-recurrence answers.

Edit: Just so you know what I mean. If you let u = 1 - x you get:

∫ uⁿ - uⁿ⁺¹ dx (from 0 to 1)

= 1/(n+1) - 1/(n+2) = 1/(n² + 3n + 2)

 

[shafqat]

Substitute n - 1 for n into that integration result and you can prove the reduction formula as well.

 

[AreYouAlright?]

Are you sure there's not an x² term in there somewhere? 'Cause I keep turning up answers in terms of 'n' via straight integration. Maybe I've fallen out of practise with the recurrence stuff already but trying the substitutions (u = 1 - x) and (x = sin²θ ) I got non-recurrence answers.

Edit: Just so you know what I mean. If you let u = 1 - x you get:

∫ uⁿ - uⁿ⁺¹ dx (from 0 to 1)

= 1/(n+1) - 1/(n+2) = 1/(n² + 3n + 2)

No there are no x² terms missing.. i have no idea

It is Question 2 e) i. on CSSA 2005 word for word

 

[AreYouAlright?]

Substitute n - 1 for n into that integration result and you can prove the reduction formula as well.

How does that prove that I_n= n/(n+2).I_n-1?

 

[shafqat]

Are you sure there's not an x² term in there somewhere? 'Cause I keep turning up answers in terms of 'n' via straight integration. Maybe I've fallen out of practise with the recurrence stuff already but trying the substitutions (u = 1 - x) and (x = sin²θ ) I got non-recurrence answers.

Edit: Just so you know what I mean. If you let u = 1 - x you get:

∫ uⁿ - uⁿ⁺¹ dx (from 0 to 1)

= 1/(n+1) - 1/(n+2) = 1/(n² + 3n + 2)

So In = 1/(n+1)(n+2)

Similarly work out In-1 = 1/n(n+1)

So In = n/(n+2) In-1

 

[haboozin]

your too good shafqat

 

[who_loves_maths]

interesting question ... this is the first reduction question i've seen that actually uses an exact value of the general integral in terms of 'n' to go back and involve I(n-1) again.

albeit not difficult, maybe the question would've been slightly harder to do had it said:

Prove that I(n) = [1 - 2n*I(n-1)]/(n^2 + n - 2) for the same general integral...


this, btw, is also (another) correct reduction formula of I(n) for the same integral in question ... and 'prove' of course means to derive the formula through integration.