Answer
$R_N=\frac{1}{N^{3}}\cdot\displaystyle \sum_{j=1}^{N}i^{2}$
$A=\frac{1}{3}$
Work Step by Step
$f(x)=x^{2}$, $[0,1]$
So using the definition of $R_N$ it follows:
$$R_N=\displaystyle \sum_{j=1}^{N}\left(0+i\frac{1-0}{N}\right)^{2}\frac{1-0}{N}$$
$$R_N=\displaystyle \sum_{j=1}^{N}\left(i\frac{1}{N}\right)^{2}\frac{1}{N}$$
$$R_N=\displaystyle \sum_{j=1}^{N}i^{2}\frac{1}{N^{2}}\cdot \frac{1}{N}$$
$$R_N=\displaystyle \sum_{j=1}^{N}i^{2}\cdot\frac{1}{N^{3}}$$
$$R_N=\frac{1}{N^{3}}\cdot\displaystyle \sum_{j=1}^{N}i^{2}$$
So the area is:
$$A=\lim\limits_{N \to \infty}R_N$$
$$A=\lim\limits_{N \to \infty}\displaystyle \frac{1}{N^{3}} \cdot \left(\frac{N^{3}}{3}+\frac{N^{2}}{2}+\frac{N}{6}\right)$$
$$A=\lim\limits_{N \to \infty}\displaystyle \left(\frac{1}{3}+\frac{1}{2N}+\frac{1}{6N^{2}}\right)$$
$$A=\displaystyle \left(\frac{1}{3}+0+0\right)$$
$$A=\frac{1}{3}$$
