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How to integrate the following? $$\int_0^1 \binom{n}{y} x^y(1-x)^{n-y}dx$$ Can you show me some detailed steps? Thank you in advance.

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  • $\begingroup$ Is this the full question that you were given? $\endgroup$ – Kenny Lau Sep 9 '17 at 10:15
  • $\begingroup$ @KennyLau yes. what's wrong? $\endgroup$ – Roger Sep 9 '17 at 10:18
  • $\begingroup$ Because it would be easier if it was a summation where $y$ goes from $0$ to $n$ $\endgroup$ – Kenny Lau Sep 9 '17 at 10:18
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    $\begingroup$ Since $n$ and $y$ are constants during the integration, the integrand is just a polynomial in $x$. What is your difficulty, where are you stuck, and what have you tried so far? $\endgroup$ – Rory Daulton Sep 9 '17 at 10:19
  • $\begingroup$ @RoryDaulton I tried to integrate by part, but I did not succeed. $\endgroup$ – Roger Sep 9 '17 at 10:22
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I am assuming $y$ to be positive integer. For $s>0$ \begin{align*} \int_0^1(sx+1-x)^ndx=\int_0^1\sum_{y=0}^n\binom{n}{y}s^yx^y(1-x)^{n-y}dx=\sum_{y=0}^n\binom{n}{y}\left(\int_0^1x^y(1-x)^{n-y}dx\right) s^k, \end{align*} Substitute $u=sx+1-x$ $$\int_1^s\frac{u^n\ du}{s-1}=\frac1{n+1}\left(\frac{s^{n+1}-1}{s-1}\right)=\frac1{n+1}\sum_{k=0}^ns^k. $$ $$ \implies \frac1{n+1}\sum_{k=0}^ns^k=\sum_{y=0}^n\binom{n}{y}\left(\int_0^1x^y(1-x)^{n-y}dx\right) s^k $$ Equation the coefficients, we get $$ \frac{1}{n+1}=\binom{n}{y}\left(\int_0^1x^y(1-x)^{n-y}dx\right) $$ Therefore, $$\int_0^1 \binom{n}{y} x^y(1-x)^{n-y}dx=\frac{1}{n+1}$$

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  • $\begingroup$ please change $t$ to $s$ $\endgroup$ – Roger Sep 9 '17 at 11:25

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