I am interested in the following formula: $$f_{2m+1}(x)=\sum_{i=0}^m {2m+1\choose i}x^{2m+1-i}(1-x)^i$$ This is half of the binomial representation of $1=(x+1-x)^k$ for $k=2m+1$.*
I treat the term above as a function. Intuitively: $\int_0^1 f_{k}(x)=\frac{1}{2}\text{ for any }k=2m+1.$
Now, what is $\int_0^1 x f_k(x)?$
My hypothesis (based on results for small $k$) is that: $$\int_0^1 x f_k(x)=\frac{\frac{3}{2}(k+1)+1}{4(k+2)}=\frac{3m+4}{8m+12.}$$
However, I don't know how to prove it in a clean manner.
*I chose $k$ odd specifically to have symmetric halves and not worry about the middle element. Also, the functions $x^k(1-x)^j$ are associated with some beta distributions, if it is of any help.
I think now I figured out all the details. The suggestion in the comment was most helpful, only the sign was opposite. In fact:$$f_{2m+1}(x)=P(U_{m+1}^{2m+1}< x)=P(U_{m+1}^{2m+1}\leq x),$$ where $U_{m+1}^{2m+1}$ is an $m+1$-st order statistic of a uniform sample of size $2m+1.$ Since the distribution of $U_{m+1}^{2m+1}$ is $Beta(m+1,m+1)$, then we can use known results. Integrating by parts (with $F_{Beta}, f_{Beta}$ denoting the cdf and pdf of beta distribution), we get: $$\int_0^1xF_{Beta}(x)dx=\left .\left(\frac{x^2}{2}F_{Beta}(x)\right)\right|_0^1-\int_0^1\frac{x^2}{2}f_{Beta}(x)dx=\frac{1}{2}-\frac{1}{2}E(u_{m+1}^{2m+1})^2$$
Since the second moment of beta distribution is: $$E(U_{m+1}^{2m+1})^2=\frac{(m+1)(m+2)}{(2m+3)(2m+2)}=\frac{m+2}{2(2m+3)}.$$ Then: $$\int_0^1 xf_{2m+1}(x)dx=\frac{1}{2}-\frac{m+2}{4(2m+3)}=\frac{3m+4}{8m+12}.$$