Answer
$$426.66$$
Work Step by Step
Given $$f(x)=3 x^{2}- x^2 \text { over }[1,5]$$ Since $\Delta x= \dfrac{b-a}{N}=\dfrac{4}{N}$, and $$ x_i =a+i\Delta x= 1+\dfrac{4i}{N}$$ Since \begin{align} \sum_{j=1}^{N} j&=1+2+\cdots+N=\frac{N(N+1)}{2} \tag{1}\\ \sum_{j=1}^{N} j^{2}&=1^{2}+2^{2}+\cdots+N^{2}=\frac{N(N+1)(2 N+1)}{6} \tag{2}\\ \sum_{j=1}^{N} j^{3}&=1^{3}+2^{3}+\cdots+N^{3}=\frac{N^{2}(N+1)^{2}}{4} \tag{3} \end{align} Then \begin{align*} R_N& =\Delta x \sum_{i=1}^{N} f\left(x_{i}\right)\\ &=\frac{4}{N} \sum_{i=1}^{N} 3\left(1+\dfrac{4i}{N}\right)^3-\left(1+\dfrac{4i}{N}\right)^2\\ &=\frac{4}{N} \sum_{i=1}^{N}\left(3\left(1+\frac{64 i^{3}}{N^{3}}+\frac{12 j}{N}+\frac{48 i^{2}}{N^{2}}\right)-\left(1+\frac{16 i^{2}}{N^{2}}+\frac{8 i}{N}\right)\right.\\ &=\frac{4}{N} \sum_{i=1}^{N}\left(3+\frac{192 i^{3}}{N^{3}}+\frac{36 i}{N}+\frac{144 i^{2}}{N^{2}}-1-\frac{16 i^{2}}{N^{2}}-\frac{8 i}{N}\right)\\ &=\frac{4}{N} \sum_{i=1}^{N}\left(\frac{192 i^{3}}{N^{3}}+\frac{128 i^{2}}{N^{2}}+\frac{28 i}{N}+2\right) \end{align*} Then \begin{align*} \lim_{N\to\infty}R_N&=\lim_{N\to\infty}\frac{4}{N} \sum_{i=1}^{N}\left(\frac{192 i^{3}}{N^{3}}+\frac{128 i^{2}}{N^{2}}+\frac{28 i}{N}+2\right) ,\\ &=\lim_{N\to\infty} \frac{4}{N}\left(\frac{192}{N^{3}} \sum_{i=1}^{N} i^{3}+\frac{128}{N^{2}} \sum_{i=1}^{N} i^{2}+\frac{28}{N} \sum_{i=1}^{N} i+\sum_{i=1}^{N} 2\right) ,\ \ \text{use (1),(2),(3)}\\ &=\lim_{N\to\infty} \frac{4}{N}\left(\frac{192}{N^{3}} \frac{N^{2}(N+1)^{2}}{4}+\frac{128}{N^{2}} \frac{N(N+1)(2 N+1)}{6} +\frac{28}{N} \frac{N(N+1)}{2}+ 2N\right)\\ &=\lim_{N\to\infty} \frac{4}{N}\left(48 N+96+\frac{48}{N}+\frac{128 N}{3}+64+\frac{64}{3 N}+14 N+14+2 N\right)\\ &=\lim_{N\to\infty} \frac{4}{N}\left(64 N+174+\frac{208}{3 N}+\frac{128 N}{3}\right)\\ &=256+\frac{512}{3}\\ &= 426.66
\end{align*}
