integration question (1 Viewer)

[Noodles]

could someone point me in the right direction for this q, i've had a few attempts

boundries = a & 0

I n = (a^2 - x^2)^n .dx for n>=0

show I n = ((2a^2n)/(2n+1)) *I n-1 for n>=1

thanks

 

[ND]

Hmmm i managed to prove that I_n = (a^2n)/(1+n)*I_(n-1), guess i must've made a mistake somewhere...

 

[Noodles]

how did u go about it? I've tried spliting it into (a+x)^n * (a-x)^n then integration by parts, but i can't get it out. Should i use that special property: integral f(x).dx = integral f(a-x).dx (both with boundries a & 0)???

 

[spice girl]

Took a long time, but here it is:

I_n = {I: 0->a} (a^2 - x^2)^n dx
= a^2 * {I: 0->a}(a^2 - x^2)^(n-1) dx - {I: 0->a}x^2 * (a^2 - x^2)^(n-1) dx (*)

Now, integrating by parts: {I: 0->a}x^2 * (a^2 - x^2)^(n-1) dx
= (-x/2n)(a^2 - x^2)^n |(0->a) + {I: 0->a}(1/2n)(a^2 - x^2)^n
= 0 + (1/2n)I_n

So, back to (*)
I_n = (a^2)I_(n-1) - (1/2n)I_n
(2n+1)/(2n)I_n = (a^2)I_(n-1)
I_n = 2na^2/(2n+1) I_(n-1)