intergration (95 2b part i) (1 Viewer)

denyag · May 25, 2010

(int with limits 1-0) x(1-x)^n dx

Prove this = n!/(n+2)!

thanks for looking

Tofuu · May 25, 2010

$\int_{0}^{1}x(1-x)^n dx$
by parts
$I=[\frac{x(1-x)^{n+1}}{-(n+1)}]+\int_{0}^{1}\frac{(1-x)^{n+1}}{(n+1)} dx$
$[\frac{x(1-x)^{n+1}}{-(n+1)}]=0$
$I=\int_{0}^{1}\frac{(1-x)^{n+1}}{(n+1)} dx$
$I=\frac{1}{(n+1)}\int_{0}^{1}(1-x)^{n+1}dx$
$I=\frac{1}{(n+1)}\left [ \frac{(1-x)^{n+2}}{-(n+2)} \right ]$
$I=\frac{1}{(n+1)}\left \cdot [ 0+\frac{1}{n+2}]$
$I=\frac{1}{(n+1)(n+2)}$
$but$
$\frac{n!}{(n+2)!}$
$=\frac{n!}{(n+2)(n+1)n!}=\frac{1}{(n+1)(n+2)}$

Tofuu · May 25, 2010

if its a prove question, you can kinda work backwards on it, but other wise, i wouldn't think about putting it in fatorial form

denyag · May 25, 2010

Tofuu said:
if its a prove question, you can kinda work backwards on it, but other wise, i wouldn't think about putting it in fatorial form

cheers guys

Drongoski · May 26, 2010

Another way (without ibp)

$\int \ x(1-x)^{n}\ dx\ \ =\ \ -\int (1-x-1)(1-x)^{n}\ dx\\ \\ =\ \ \int \left [ (1-x)^{n} - (1-x)^{n+1} \right ]\ dx\ \ =\ \ -\frac{(1-x)^{n+1}}{n+1}\ +\ \frac{(1-x)^{n+2}}{n+2}\ +\ C\\ \\ \therefore \ \ \int_{0}^{1}x(1-x)^{n}\ dx\ \ =\ \ \left [ -\frac{(1-x)^{n+1}}{n+1} +\frac{(1-x)^{n+2}}{n+2}\right ]^{1}_{0}\\ \\ =\ \ (\ 0\ )\ -\ ( -\frac{1}{n+1}\ +\ \frac{1}{n+2}\ )\ \ =\ \ \frac{1}{(n+1)(n+2)}\ = \ \frac{n!} {(n+2)!}$

Trebla · May 26, 2010

blackman217 · May 26, 2010

Affinity · May 27, 2010

This is a special case of the Beta function, for positive integers m,n you can prove using integration by parts that

B(m+1,n+1) = (int with limits 1-0) x^m(1-x)^n dx
= m!n!/(m+n+1)!

intergration (95 2b part i) (1 Viewer)

denyag

New Member

Tofuu

Member

Tofuu

Member

denyag

New Member

Drongoski

Well-Known Member

Trebla

Administrator

blackman217

New Member

Affinity

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)