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I have not frequented the board lately due to being very busy. But, I ran across what I think is an interesting problem.

$\displaystyle \int_{0}^{k}(tx+1-x)^{n}dx$, where $n\in \mathbb{Z}^{+}$ and $t$ is a parameter independent of $x$. $k=0,1,2,3,\ldots, n$

Then, use this to show that $\displaystyle \int_0^k x^{k}(1-x)^{n-k} \; dx=\frac{1}{\binom{n}{k}(n+1)}$

I managed to integrate the former by making the sub $u=x(t-1), \;\ dx=\frac{du}{t-1}$

This gives $\displaystyle \frac{1}{t-1}\int_0^{t-1}(u+1)^n \; du=\frac{t^{n+1}-1}{t-1}\cdot \frac{1}{n+1}$

This appears to be the partial sum of the geometric series $\displaystyle \sum_{k=0}^n t^k=\frac{t^{n+1}-1}{t-1}$

Now, how can we relate this to $\frac{1}{\binom{n}{k}}$?.

Since $\displaystyle \frac{1}{(n+1)\binom{n}{k}}=B(n-k+1,k+1)$, the latter integral looks like some sort of incomplete beta function. It's that upper k limit that I am kind of wondering about.

After all, $\displaystyle \frac{1}{(n+1)\binom{n}{k}}=\frac{k!(n-k)!}{(n+1)!}=\frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+2)}=B(n-k+1,k+1)$.

I assume it is right in front of me, but how does solving the first integral, which evaluates to $\displaystyle \frac{t^{n+1}-1}{t-1}\cdot \frac{1}{n+1}=\frac{1}{n+1}\sum_{k=0}^n t^k$ relate to the latter Beta-type integral which apparently evaluates to $\displaystyle \frac{1}{\binom{n}{k}(n+1)}$?.

Thank you all very much for any responses. I will be sure to leave credit and thanks to everyone. If I have not, it is simply because it was overlooked. Not out of disregard.

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Nevermind. I think I have it. It would appear there is a typo on that upper limit. I think It should be a 1 instead of a k.

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