Answer
$$16$$
Work Step by Step
Given $$f(x)=3 x^{2}+4 x \text { over }[0,2]$$ Since $\Delta x= \dfrac{b-a}{N}=\dfrac{2}{N}$, and $$ x_i =a+i\Delta x= \dfrac{2i}{N}$$ 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{2}{N} \sum_{i=1}^{N} 3\left(\frac{2 i}{N}\right)^2+4\left(\frac{2 i}{N}\right)\\ &=\frac{2}{N} \sum_{i=1}^{N}\left(\frac{12 i^{2}}{N^{2}}+\frac{8 i}{N}\right)\\ &=\frac{8}{N}\left(\frac{3}{N^{2}} \sum_{i=1}^{N} i^{2}+\frac{2}{N} \sum_{i=1}^{N} i\right),\ \text{Use ( 1), (2)}\\ &= \frac{8}{N}\left(\frac{3}{N^{2}} \frac{N(N+1)(2 N+1)}{6} +\frac{2}{N} \frac{N(N+1)}{2}\right)\\ &=\frac{8}{N}\left( \frac{N(N+1)(2 N+1)}{2N} + N(N+1) \right)\\ &= 8+\frac{12}{N}+\frac{4}{N^{2}}+8+\frac{8}{N} \end{align*} Then \begin{align*} \lim_{N\to\infty}R_N&=\lim_{N\to\infty}8+\frac{12}{N}+\frac{4}{N^{2}}+8+\frac{8}{N}\\ &=16 \end{align*}
