I'm having problems trying to compute the limit of the sum
$\lim_{n\to\infty}\sum_{k=1}^{n}(\frac{k}{n})^{4}(\frac{2k-1}{n^{2}})$
As you can see you can't take $\xi_{k}$ as the extreme of some partition, for example $k/n$. because we obtain $\frac{1}{n}\sum_{k=1}^{n}(\frac{k}{n})^{4}(\frac{2k-1}{n})=\frac{1}{n}\sum_{k=1}^{n}\frac{2k^{5}-k^{4}}{n^{5}}=\frac{1}{n}\sum_{k=1}^{n}(2(\frac{k}{n})^{5}-\frac{1}{n}(\frac{k}{n})^{4})$ but what do I do with the second $\frac{1}{n}$?
I'm thinking on taking another values for $\xi_{k}$ and another interval distinct from $[0,1]$ but I don't get which one could be useful.
Another idea I had is that as we can take inside the sum $(\frac{k}{n})^{4}\frac{2k-1}{n}=2(\frac{k}{n})^{4}(\frac{\frac{k-1}{n}+\frac{k}{n}}{2})$ with $x_{k-1}=\frac{k-1}{n}$ and $x_{k}=\frac{k}{n}$ the second parentheses would be the mid point of $[x_{k-1},x_{k}]$ but in the first parentheses I have one extrema $x_{k}=\frac{k}{n}$ so this would let me to take two distinct points in each subinterval