find-the-value-of-I-n-0-1-x-n-1-x-dx-

Question Number 36410 by abdo.msup.com last updated on 01/Jun/18

f i n d t h e v a l u e o f I_{n} = \int_{0}^{1} x^{n} \sqrt{1 - x} d x

Commented by abdo.msup.com last updated on 04/Jun/18

c h s n g e m e n t \sqrt{1 - x} = t g i v e

1 - x = t^{2} \Rightarrow x = 1 - t^{2} a n d

I_{n} = \int_{0}^{1} {(1 - t^{2})}^{n} t (2 t) d t

= 2 \int_{0}^{1} t^{2} (\sum_{k = 0}^{n} {(- t^{2})}^{k}) d t

= 2 \sum_{k = 0}^{n} {(- 1)}^{k} \int_{0}^{1} t^{2 k + 2} d t

= 2 \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{2 k + 3} c h a n g e m e n t o f i n d i c e

k + 1 = p g i v e

I_{n} = 2 \sum_{p = 1}^{n + 1} \frac{{(- 1)}^{p - 1}}{2 p + 1} \Rightarrow

I_{n} = 2 \sum_{k = 1}^{n + 1} \frac{{(- 1)}^{k - 1}}{2 k + 1} .