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Question Number 127090 by physicstutes last updated on 26/Dec/20

Given that

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

show that (n + 2) I_{n} = {n I}_{n - 1}, n ⩾ 1 .

Answered by Dwaipayan Shikari last updated on 26/Dec/20

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

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

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

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

Answered by Olaf last updated on 26/Dec/20

I_{n} = \int_{0}^{1} x {(1 - x)}^{n} d x, n \in Z^{+}

Let u = 1 - x

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

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

and \frac{I_{n}}{I_{n - 1}} = \frac{n}{n + 2}

Commented by physicstutes last updated on 26/Dec/20

thanks sirs, but how about a method using

integration by parts

by letting u = x and d v = {(1 - x)}^{n} d x

Answered by Ar Brandon last updated on 26/Dec/20

I_{n} = \int_{0}^{1} x {(1 - x)}^{n} dx, n \in Z^{+}

{\begin{cases} u (x) = x \\ v^{'} (x) = {(1 - x)}^{n} \end{cases} \Rightarrow {\begin{cases} u^{'} (x) = 1 \\ v (x) = - \frac{{(1 - x)}^{n + 1}}{n + 1} \end{cases}

I_{n} = u (x) \cdot v (x) - \int v (x) \cdot u^{'} (x) dx

= {[- \frac{x {(1 - x)}^{n + 1}}{n + 1}]}_{0}^{1} + \int_{0}^{1} \frac{{(1 - x)}^{n + 1}}{n + 1} dx

= - {[\frac{{(1 - x)}^{n + 2}}{(n + 1) (n + 2)}]}_{0}^{1} = \frac{1}{(n + 1) (n + 2)}

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

(n + 2) I_{n} = n I_{n - 1}

Commented by Ar Brandon last updated on 27/Dec/20

D e n a d a!

Commented by physicstutes last updated on 27/Dec/20

excellento! gracias

Answered by Ar Brandon last updated on 26/Dec/20

I_{n} = \int_{0}^{1} x {(1 - x)}^{n} dx, n \in Z^{+}

{\begin{cases} u (x) = {(1 - x)}^{n} \\ v^{'} (x) = x \end{cases} \Rightarrow {\begin{cases} u^{'} (x) = - n {(1 - x)}^{n - 1} \\ v (x) = \frac{x^{2}}{2} \end{cases}

I_{n} = {[\frac{x^{2}}{2} \cdot {(1 - x)}^{n}]}_{0}^{1} + \frac{n}{2} \int_{0}^{1} x^{2} {(1 - x)}^{n - 1} dx

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

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

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

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

(n + 2) I_{n} = n I_{n - 1}