clalculate-A-n-0-1-t-2n-1-t-n-dt-with-n-integr-natural-

Question Number 57236 by maxmathsup by imad last updated on 31/Mar/19

c l a l c u l a t e A_{n} = \int_{0}^{1} t^{2 n} {(1 - t)}^{n} d t w i t h n i n t e g r n a t u r a l .

Commented by maxmathsup by imad last updated on 04/Apr/19

l e t f i n d A_{n, p} = \int_{0}^{1} t^{n} {(1 - t)}^{p} d t b y p a r t s u^{^{'}} = t^{n} a n d v = {(1 - t)}^{p}

A_{n, p} = {[\frac{1}{n + 1} t^{n + 1} {(1 - t)}^{p}]}_{0}^{1} - \int_{0}^{1} \frac{1}{n + 1} t^{n + 1} (- p) {(1 - t)}^{p - 1} d t

= \frac{p}{n + 1} \int_{0}^{1} t^{n + 1} {(1 - t)}^{p - 1} = \frac{p}{n + 1} A_{n + 1, p - 1} \Rightarrow

A_{n, p} = \frac{p (p - 1)}{(n + 1) (n + 2)} A_{n + 2, p - 2} = \frac{p (p - 1) \dots (p - k + 1)}{(n + 1) (n + 2) \dots . (n + k)} A_{n + k, p - k} . =_{k = p} \frac{p!}{(n + 1) (n + 2) \dots . (n + p)} A_{n + p, o}

A_{n + p, 0} = \int_{0}^{1} t^{n + p} d t = \frac{1}{n + p + 1} \Rightarrow A_{n, p} = \frac{p!}{(n + 1) (n + 2) \dots (n + p + 1)} \Rightarrow

A_{n, p} = \frac{p!}{(n + p + 1)!} \times n! = \frac{n! . p!}{(n + p + 1)!} \Rightarrow A_{n} = A_{2 n, n} = \frac{(2 n)! (n!)}{(3 n + 1)!}

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Apr/19

\int_{0}^{1} t^{2 n + 1 - 1} {(1 - t)}^{n + 1 - 1} = \frac{⌈ (2 n + 1) ⌈ (n + 1)}{⌈ (2 n + 1 + n + 1)}

b e t a f u n c t i o n \int_{0}^{1} x^{m - 1} {(1 - x)}^{n - 1} d x = \frac{⌈ (m) ⌈ n)}{⌈ (m + n)}

Commented by maxmathsup by imad last updated on 04/Apr/19

s i r T a n m a y l o o k t h a t Γ (2 n + 1) = (2 n)!, Γ (n + 1) = n!, Γ (3 n + 2) = (3 n + 1)!

\Rightarrow A_{n} = \frac{(2 n)! (n!)}{(3 n + 1)!} .