Answer
$R_N=\frac{216}{3}+\frac{216}{2N}+\frac{216}{6N^{2}}-\frac{72}{2}-\frac{72}{2N}+6$
$A=42$
Work Step by Step
$f(x)=x^{2}$, $[-1,5]$
So using the definition of $R_N$ it follows:
$$R_N=\displaystyle \sum_{j=1}^{N}\left(-1+j\frac{5-(-1)}{N}\right)^{2}\frac{5-(-1)}{N}$$
$$R_N=\displaystyle \sum_{j=1}^{N}\left(-1+j\frac{6}{N}\right)^{2}\frac{6}{N}$$
$$R_N=\displaystyle \sum_{j=1}^{N}\left[\left(j\frac{6}{N}\right)^{2}-\frac{12}{N}j+1\right]\frac{6}{N}$$
$$R_N=\displaystyle \sum_{j=1}^{N}\left[j^{2}\frac{216}{N^{3}}-\frac{72}{N^{2}}j+\frac{6}{N}\right]$$
$$R_N=\displaystyle \sum_{j=1}^{N}j^{2}\frac{216}{N^{3}}-\displaystyle \sum_{j=1}^{N}\frac{72}{N^{2}}j+\displaystyle \sum_{j=1}^{N} \frac{6}{N}$$
$$R_N=\frac{216}{N^{3}}\displaystyle \sum_{j=1}^{N}j^{2}-\frac{72}{N^{2}}\displaystyle \sum_{j=1}^{N}j+\frac{6}{N}\displaystyle \sum_{j=1}^{N} 1$$
$$R_N=\frac{216}{N^{3}} (\frac{N^{3}}{3}+\frac{N^{2}}{2}+\frac{N}{6})-\frac{72}{N^{2}}(\frac{N^{2}}{2}+\frac{N}{2})+\frac{6}{N}\cdot N $$
$$R_N=\frac{216}{3}+\frac{216}{2N}+\frac{216}{6N^{2}}-\frac{72}{2}-\frac{72}{2N}+6$$
So the area is:
$$A=\lim\limits_{N \to \infty}R_N$$
$$A=\lim\limits_{N \to \infty}\left(\frac{216}{3}+\frac{216}{2N}+\frac{216}{6N^{2}}-\frac{72}{2}-\frac{72}{2N}+6\right)$$
$$A=\displaystyle (\frac{216}{3}+0+0-\frac{72}{2}-0+6)$$
$$A=42$$
