Answer
$$\frac{81}{2}$$
Work Step by Step
Given $$f(x)=3 x+6, \quad[1,4]$$ Since $\Delta x= \dfrac{b-a}{N}=\dfrac{3}{N}$, and $$ x_i =a+i\Delta x= 1+\dfrac{3i}{N}$$ Since \begin{align*} R_N& =\Delta x \sum_{i=1}^{N} f\left(x_{i}\right)\\ &=\frac{3}{N} \sum_{i=1}^{N} 3\left(1+\dfrac{3i}{N}\right)+6\\ &=\frac{3 }{N } \sum_{i=1}^{N} 9+\dfrac{9i}{N}\\ &=\frac{3}{N}\left(\frac{9N(N+1)}{2N}+9N \right)\\ &=27\left(1+\frac{N}{2 N}+\frac{1}{2 N}\right) \end{align*} Then \begin{align*} \lim_{N\to\infty}R_N&=\lim_{N\to\infty}27\left(1+\frac{N}{2 N}+\frac{1}{2 N}\right)\\ &=27+\frac{27}{2}=\frac{81}{2} \end{align*}
