Show that ∫(x-x^2)^k {nx}dx = (k!)^2/[2(2k+1)]?

0 ⤊

Show that ∫(x-x^2)^k {nx}dx = (k!)^2/[2(2k+1)]?

where n is a natural number .. k is a non-negative integral .. and {nx}=nx-[nx]

2007-07-14 05:52:53 · 2 answers · asked by Hadeel 2 in Science & Mathematics ➔ Mathematics

oh sorry .. yeas... non -negative integer

2007-07-14 06:48:43 · update #1

2 answers

Do you mean k is a non-negative integer?

I'm not sure I understand your notation concerning n. Could you clarify it? It seems that n ought to appear in the answer.

What are the limits of integration? Is this somehow provided by {nx} = nx - [nx]?

2007-07-14 06:41:05 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋

im good in math
u + i = 69

2007-07-14 13:28:25 · answer #2 · answered by beentherestillthere 2 · 0⤊ 4⤋