Integration Q (1 Viewer)

[5647382910]

if (I_n) = (integral from 1 to 0){x((1 - x^3)^n)} for n>=0
show that:
(I_n) = = ((3n)/(3n +2))I_(n - 1) for n>=1

thanks in advance

 

[Drongoski]

if (I_n) = (integral from 1 to 0){x((1 - x^3)^n)} for n>=0
show that:
(I_n) = = ((3n)/(3n +2))I_(n - 1) for n>=1

thanks in advance

(I'm not good at typing out the mathematical symbols - so please excuse; to save typing I'll just say Int when Int from 0 to 1 is intended))

I-n = 1/2Int (1 - x^3)^n dx^2
= 1/2 [x^2(1-x^3)^n ] (from 0 to 1) - 1/2 Int n/2x^2 d(1 - x^3)n
= zero - n/2 Int x^2 (1 - x^3)^(n-1) (-3x^2) dx
= - 3n/2 Int (-x^4) (1 - x^3)^(n-1) dx
= - 3n/2 Int x(1 - x^3 - 1) (1 - x^3)^(n-1) dx


= -3n/2 Int {x(1 - x^3)^n - x(1 - x^3)^(n-1) } dx
= -3n/2 [ I-n - I-(n-1)]
Therefore (1 + 3n/2)I-n = 3n/2 I-(n-1)
Simplifying you get I-n = {3n/(2 + 3n)} I-(n-1)

So do you get it now ???

 

[jet]

Just so its easy to read.

 

[Drongoski]

Just so its easy to read.

Jetblack2007 - can you please explain how you specify in LaTeX the limits of integration (0 & 1 in this case) on the closing ']' Thanks

 

[jet]

just use the sub- and superscripts. For example, to type

you would type:
\left [ uv \right ] _0 ^1 [with or without the spaces] This is the same for the integral.

 

[5647382910]

Thanks heaps guys