Is this what you mean?:
\(\displaystyle \lim_{n\to \infty}\sum_{i=1}^{n}\frac{7}{n\left(1+\frac{7i}{n}\right)^{\frac{1}{3}}}\)
If so, this is a Riemann sum. Yes, you are correct. Integration is in order.
It is a Riemann sum with subinterval length \(\displaystyle \frac{7}{n}\)
\(\displaystyle x_{i}=\frac{7i}{n}\)
\(\displaystyle \int_{0}^{7}\frac{1}{(1+x)^{\frac{1}{3}}}dx\)
