Answer
$\displaystyle \int_{0}^{1}x^{4}dx$
Work Step by Step
$\lim\limits_{n \to \infty} \displaystyle \sum_{i=1}^{n} \frac{i^{4}}{n^{5}}$
$\lim\limits_{n \to \infty} \displaystyle \sum_{i=1}^{n} \frac{i^{4}}{n^{4+1}}$
$\lim\limits_{n \to \infty} \displaystyle \sum_{i=1}^{n} \frac{i^{4}}{n^{1}\cdot n^{4}}$
$\lim\limits_{n \to \infty} \displaystyle \sum_{i=1}^{n} \frac{i^{4}}{n\cdot n^{4}}$
$\lim\limits_{n \to \infty} \displaystyle \sum_{i=1}^{n} \frac{i^{4}}{ n^{4}}\cdot \frac{1}{n}$
$\lim\limits_{n \to \infty} \displaystyle \sum_{i=1}^{n} \left(\frac{i}{ n}\right)^{4}\cdot \frac{1}{n}$
$\lim\limits_{n \to \infty} \displaystyle \sum_{i=1}^{n} \left(0+i\frac{1}{ n}\right)^{4}\cdot \frac{1}{n}$
$\lim\limits_{n \to \infty} \displaystyle \sum_{i=1}^{n} \left(0+i\frac{1-0}{ n}\right)^{4}\cdot \frac{1-0}{n}$
By the theorem $4$ it follows:
$\lim\limits_{n \to \infty} \displaystyle \sum_{i=1}^{n} (0+i\frac{1-0}{ n})^{4}\cdot \frac{1-0}{n}$=$\displaystyle \int_{0}^{1}x^{4}dx$
