Answer
$\frac{5}{4}$
Work Step by Step
Find the limit $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}[(\frac{i}{n})^{3}+1]$
$\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}[(\frac{i}{n})^{3}+1]=\lim\limits_{n \to \infty}(\frac{1}{n^{4}}\sum\limits_{i =1}^{n}{i}^{3}+\frac{1}{n}\sum\limits_{i =1}^{n}1)$
Since, $ \sum \limits_{i =1}^{n}i^{3}=[\frac{n(n+1)}{2}]^{2}$
Thus,
$\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}[(\frac{i}{n})^{3}+1]=\lim\limits_{n \to \infty}(\frac{1}{n^{4}}[\frac{n(n+1)}{2}]^{2}+\frac{1}{n}(n)$
$=\lim\limits_{n \to \infty}[\frac{1}{4}+\frac{1}{2n}+\frac{1}{4n^{2}}+1]$
$=\frac{1}{4}+0+0+1$
Hence, $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{1}{n}[(\frac{i}{n})^{3}+1]=\frac{5}{4}$
