Answer
$A=\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{4(1+\frac{4i}{n})\sqrt{(1+\frac{4i}{n})^3+8}}{n}$
Work Step by Step
Using the Definition 2
$A=\lim\limits_{n \to \infty}\sum_{i=1}^nf(x_i)\Delta x$ where $\Delta x=\frac{\text{upper bound - lower bound}}{n}$
Find $\Delta x$ and simplify:
$\Delta x=\frac{5-1}{n}=\frac{4}{n}$
Then,
$A=\lim\limits_{n \to \infty}\sum_{i=1}^nx_i\sqrt{x_i^3+8}\cdot \frac{4}{n}$
$A=\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{4x_i\sqrt{x_i^3+8}}{n}$
where $x_i=1+\frac{4i}{n}$
