Answer
$6$
Work Step by Step
Here, we have $\triangle x=\dfrac{b-a}{n}$ and $c_k=a+\dfrac{k(b-a)}{n}$
Consider the RIEMANN SUM:
$\Sigma_{k=1}^n f(c_k) \triangle x=\Sigma_{k=1}^n (\dfrac{8k}{n^2}+\dfrac{2}{n})=\dfrac{8k}{n^2}\Sigma_{k=1}^n k+\dfrac{2}{n}) \Sigma_{k=1}^n$
or, $\dfrac{8k}{n^2}\Sigma_{k=1}^n k+\dfrac{2}{n}) \Sigma_{k=1}^n=\dfrac{4(n+1)}{n}+2$
Now, $\lim\limits_{n \to \infty} \Sigma_{k=1}^n f(c_k) \triangle x=\lim\limits_{n \to \infty}\dfrac{4(n+1)}{n}+2$
Thus, $\lim\limits_{n \to \infty}6+\dfrac{4}{n}=6$
