Question

Integration question: Show that

Answer 1

\[\int\limits_{0}^{1}x ^{m}(1-x)^{n}dx=\frac{ n }{ m+1 }\int\limits_{0}^{1}x ^{m+1}(1-x)^{n-1}dx\]

Answer 2

for m>0 and n>0. Hence or otherwise, show that\[\int\limits_{0}^{1}x ^{m}(1-x)^{n}dx=\frac{ m!n! }{ (m+n+1)! }\]

Answer 3

\[
I=\int_0^1x^m(1-x)^ndx\\
\text{integrating by parts:}\\
part1=(1-x)^n\qquad part2=x^m\\
I=\left[(1-x)^n\int x^mdx\right]_0^1-\int_0^1\left[{d\over dx}(1-x)^n\int x^m dx\right]dx\\
I=\left[(1-x)^nx^{m+1}\over m+1\right]_0^1-\int_0^1(-n)(1-x)^{n-1}{x^{m+1}\over m+1}dx\\
I=0+{n\over m+1}\int_0^1x^{m+1}(1-x)^{n-1}dx\\
\]QED

Answer 4

also, this is the definition of Beta Function

Answer 5

for the second part, you keep reducing the integrand this way and you'd see this is what it'd beome:
\[
I={n\over m+1}\int_0^1x^{m+1}(1-x)^{n-1}dx\\
I={n\over m+1}{n-1\over m+2}\int_0^1x^{m+2}(1-x)^{n-2}dx\\
\vdots
\]

Answer 6

this would reduce "n" times

Answer 7

you'd end up with
\[
I={n!\over (m+1)(m+2)\dots (m+n)}\int _0^1x^{m+n}dx
\]

Answer 8

simplify a bit and you get the answer

Answer 9

@electrokid thanks for you explanations! It's very understandable.

Answer 10

you are welcome.

Answer 11

is there a specific question here ?

Answer 12

\[\frac{ n! }{ (m+1)(m+2)........(m+n) }=\frac{ n! }{ m ^{n}+n!+? }\]

Answer 13

I can't convert the integral into m!n!/(m+n+1)! form

Answer 14

Understood the integral process but could not simplify

Answer 15

if we start with
\[ I={n!\over (m+1)(m+2)\dots (m+n)}\int _0^1x^{m+n}dx \]
the integral gives
\[ \frac{1}{m+n+1} \]
so you get
\[ I={n!\over (m+1)(m+2)\dots (m+n)(m+n+1)} \]

we can replace the denominator with
\[ \frac{ 1\cdot 2 \cdot 3 ... (m-1)m}{ 1\cdot 2 \cdot 3 ... (m-1)m(m+1)...(m+n)(m+n+1)}\]

in other words (m+n+1)! / m! gives (m+1)(m+2)....(m+n+1)

Finally, we have
\[ I = \frac{m! \ n!}{(m+n+1)!} \]

Answer 16

@phi Oh! So I have to times both nominator and denominator with m! thanks!