Answer
$A = \lim\limits_{n \to \infty} \sum_{i=1}^{n}~[~(4+\frac{3i}{n})^2+\sqrt{1+2(4+\frac{3i}{n})}~~]\cdot \frac{3}{n}$
Work Step by Step
For each $x_i$, such that $1 \leq i \leq n$, note that $x_i = (4+i\frac{3}{n}) = (4+\frac{3i}{n})$
$\Delta x = \frac{3}{n}$
We can express the area under the curve as a limit:
$A = \lim\limits_{n \to \infty} [f(x_1)\Delta x+f(x_2)\Delta x+...+f(x_n)\Delta x]$
$A = \lim\limits_{n \to \infty} [f(4+\frac{3}{n})(\frac{3}{n})+f(4+\frac{6}{n})(\frac{3}{n})+...+f(4+\frac{3n}{n})(\frac{3}{n})]$
$A = \lim\limits_{n \to \infty} [(4+\frac{3}{n})^2+\sqrt{1+2(4+\frac{3}{n})}]\cdot \frac{3}{n} + [(4+\frac{6}{n})^2+\sqrt{1+2(4+\frac{6}{n})}]\cdot \frac{3}{n} +...+[(4+\frac{3n}{n})^2+\sqrt{1+2(4+\frac{3n}{n})}]\cdot \frac{3}{n} $
$A = \lim\limits_{n \to \infty} \sum_{i=1}^{n}~[~(4+\frac{3i}{n})^2+\sqrt{1+2(4+\frac{3i}{n})}~~]\cdot \frac{3}{n}$
