MHB Calculate Integral $\int_{0}^{1}x^{2014}\cdot \left(1-x\right)^{2014}dx$

AI Thread Summary
The integral $\int_{0}^{1} x^{2014} (1-x)^{2014} dx$ can be calculated using integration by parts, leading to the result $\frac{(2014!)^2}{4029!}$. The process involves recursive integration, simplifying the integral step by step. The final expression is derived from the factorial representation, confirming the relationship between the integral and the factorials. This method highlights the efficiency of integration techniques in solving complex integrals. The discussion emphasizes the correctness of the solution provided by multiple contributors.
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Calculation of Integral $\displaystyle \int_{0}^{1}x^{2014}\cdot \left(1-x\right)^{2014}dx$
 
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If we use the beta function:
\[
B(a+1,b+1) = \int_0^1 x^a (1-x)^b dx,
\]
and the property of
\[
B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}
\]
and the fact that for integer $n$, $\Gamma(n) = (n-1)!$, we have:
\[
B(a,b) = \frac{(a-1)!(b-1)!}{(a+b-1)!}.
\]

For this problem $a = b = 2013$, so $B(2013,2013) = \frac{(2014!)^2}{4029!}$.
 
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jacks said:
Calculation of Integral $\displaystyle \int_{0}^{1}x^{2014}\cdot \left(1-x\right)^{2014}dx$

[sp]Is...

$\displaystyle \int_{0}^{1} x^{2014}\ (1-x)^{2014}\ dx = \frac{1}{2015}\ (|x^{2015}\ (1-x)^{2014}|_{0}^{1} + 2014\ \int_{0}^{1} x^{2015}\ (1-x)^{2013}\ d x ) = \frac{2014}{2015} \int_{0}^{1} x^{2015}\ (1-x)^{2013}\ dx\ (1) $

Proceeding in this way we obtain...

$\displaystyle \int_{0}^{1} x^{2014}\ (1-x)^{2014}\ dx = \frac{2014 \cdot 2013 ... 2 \cdot 1}{2015 \cdot 2016 ... 4028 \cdot 4029} = \frac{(2014!)^{2}}{4029!}\ (2)$

[/sp]

Kind regards

$\chi$ $\sigma$
 
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Thanks http://mathhelpboards.com/members/magneto/ and chisigma Nice solution.

My Solution is same as Chisigma(Integration by parts):

Let $\displaystyle I(m,n) = \int_0^1 \! x^m(1-x)^n \, dx$.

We have, for $n \ge 1$,

\[ \begin{eqnarray}
I(m,n) &=& \left. -\frac{x^{m+1}(1-x)^n}{m+1}\right]_0^1 + \int_0^1 \frac{n}{m+1}x^{m+1}(1-x)^{n-1} \, dx \nonumber \\
&=& \frac{n}{m+1}I(m+1,n-1) \nonumber
\end{eqnarray}\]
Repeating this gives us
\[ \begin{eqnarray}
I(m,n) &=& \frac{n}{m+1}\cdot\frac{n-1}{m+2}\cdot\cdots\cdot\frac{1}{m+n}I(m+n,0) \nonumber \\
&=& \frac{n!}{\frac{(m+n)!}{m!}}\cdot\frac{1}{m+n+1} \nonumber \\
&=& \frac{1}{m+n+1}\cdot\frac{1}{\binom{m+n}{m}} \nonumber
\end{eqnarray} \]

Now Put $m=2014$ and $n=2014$

So $\displaystyle I(2014,2014) = \int_{0}^{1}x^{2014}\cdot (1-x)^{2014}dx = \frac{1}{2014+2014+1}\cdot \frac{1}{\binom{2014+2014}{2014}} = \frac{1}{4029}\cdot \frac{1}{\binom{4028}{2014}}$
 
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