Given-that-I-n-0-1-x-1-x-n-dx-obtain-a-reduction-formulae-for-I-n-in-terms-of-I-n-2-Hence-evaluate-0-1-x-1-x-5-dx-

Question Number 122166 by physicstutes last updated on 14/Nov/20

Given that I_{n} = \underset{0}{\int^{1}} x {(1 - x)}^{n} d x

obtain a reduction formulae for I_{n} in terms

of I_{n - 2} Hence evaluate \underset{0}{\int^{1}} x {(1 - x)}^{5} d x .

Answered by liberty last updated on 14/Nov/20

{\underset{0}{\int^{1}} x {(1 - x)}^{5} dx = - \frac{x {(1 - x)}^{6}}{6}]}_{0}^{1} + \underset{0}{\int^{1}} \frac{{(1 - x)}^{6}}{6} dx

= - \frac{1}{42} {[(1 - x)]}_{0}^{1} = \frac{1}{42}

Commented by physicstutes last updated on 14/Nov/20

from your method i deduce that :

I_{n} = - {[\frac{x {(1 - x)}^{n + 1}}{n + 1}]}_{0}^{1} + \underset{0}{\int^{1}} \frac{{(1 - x)}^{n + 1}}{n + 1} d x

I_{n} = \frac{1}{n + 1} \underset{0}{\int^{1}} {(1 - x)}^{n + 1} d x

Thanks

Commented by Dwaipayan Shikari last updated on 14/Nov/20

\int_{0}^{1} x {(1 - x)}^{n} d x = \int_{0}^{1} (1 - x) x^{n} d x = \frac{1}{n + 1} - \frac{1}{n + 2}

h e r e n = 5 s o, \frac{1}{42}

Answered by Dwaipayan Shikari last updated on 14/Nov/20

\int_{0}^{1} x {(1 - x)}^{n} d x

= β (2, n + 1) = \frac{Γ (2) Γ (n + 1)}{Γ (n + 3)} = \frac{Γ (n + 1)}{Γ (n + 3)}

S o

\int_{0}^{1} x {(1 - x)}^{5} d t = \frac{Γ (5 + 1)}{Γ (5 + 3)} = \frac{5!}{7!} = \frac{1}{42}

Commented by physicstutes last updated on 14/Nov/20

Magnifcent!

Answered by mathmax by abdo last updated on 14/Nov/20

I_{n} = \int_{0}^{1} x {(1 - x)}^{n} dx we know B (p, q) = \int_{0}^{1} x^{p - 1} {(1 - x)}^{q - 1} dx \Rightarrow

B (p, q) = \frac{Γ (p) . Γ (q)}{Γ (p + q)} \Rightarrow I_{n} = \int_{0}^{1} x^{2 - 1} {(1 - x)}^{n + 1 - 1} dx = B (2, n + 1)

= \frac{Γ (2) . Γ (n + 1)}{Γ (2 + n + 1)} = \frac{1! n!}{(n + 2)!} = \frac{n!}{(n + 2) (n + 1) n!} = \frac{1}{(n + 1) (n + 2)}

\frac{I_{n}}{I_{n - 2}} = \frac{1}{(n + 1) (n + 2)} . (n - 1) n = \frac{n (n - 1)}{(n + 1) (n + 2)} \Rightarrow

I_{n} = \frac{n (n - 1)}{(n + 1) (n + 2)} I_{n - 2} (n ⩾ 2)

\int_{0}^{1} x {(1 - x)}^{5} dx = I_{5} = \frac{1}{6.7} = \frac{1}{42}