Find limit of sequence using integral Given sequence $$a_m = \frac{1}{m^2} \sum \limits_{k = 1}^m \sqrt[3]{(mx + k + 1)\cdot (mx + k)^2}$$
Find its limit using integral.
I thought it may be solved using either Euler-Mclaurin formula or Riemann sum. Unfortunately, the function under the sum sign is pretty uncomfy to operate with.

Hint. One may write, as $n \to \infty$,
$$
\begin{align}
a_n = \frac{1}{n^2} \sum \limits_{k = 1}^n \sqrt[3]{(nx + k + 1)\cdot (nx + k)^2}&= \frac{1}{n} \sum \limits_{k = 1}^n \sqrt[3]{\left(x + \frac{k+1}{n}\right)\cdot \left(x + \frac{k}{n}\right)^2}
\end{align}
$$ giving
$$
\frac{1}{n} \sum \limits_{k = 1}^n\sqrt[3]{\left(x + \frac{k}{n}\right)^3}\le a_n \le \frac{1}{n} \sum \limits_{k = 1}^n\sqrt[3]{\left(x + \frac{k+1}{n}\right)^3}
$$ or
$$
\frac{1}{n} \sum \limits_{k = 1}^n\left(x + \frac{k}{n}\right)\le a_n \le \frac{1}{n} \sum \limits_{k = 1}^n\left(x + \frac{k+1}{n}\right)
$$ that is $$
\color{blue}{\frac{1}{n} \sum \limits_{k = 1}^n\left(x + \frac{k}{n}\right)}\le a_n \le \color{blue}{\frac{1}{n} \sum \limits_{k = 1}^n\left(x + \frac{k}{n}\right)}+\frac{1}{n} \sum \limits_{k = 1}^n\frac{1}{n}
$$ Can you finish it?
