Answer
The area under the graph of $f(x)=x^3$ for $0\leq x\leq 1$
Work Step by Step
Consider $x_i=\frac{i}{n}$.
The lower bound: $x_0=\frac{0}{n}=0$
The upper bound: $x_n=\frac{n}{n}=1$
Then,
$\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{1}{n}(\frac{i}{n})^3=\lim\limits_{n \to \infty}\sum_{i=1}^nf(x_i)\Delta x$
$\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{1}{n}(\frac{i}{n})^3=\lim\limits_{n \to \infty}\sum_{i=1}^nf(x_i)\cdot \frac{\text{upper bound - lower bound}}{n}$
$\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{1}{n}x_i^3=\lim\limits_{n \to \infty}\sum_{i=1}^nf(x_i)\cdot \frac{1-0}{n}$
$\lim\limits_{n \to \infty}\sum_{i=1}^nx_i^3\frac{1}{n}=\lim\limits_{n \to \infty}\sum_{i=1}^nf(x_i)\cdot \frac{1}{n}$
Comparing both sides, we get the function $f(x)=x^3$.
Thus, the limit represents the area under the graph of $f(x)=x^3$ for $0\leq x\leq 1$.
