Answer
This limit is equal to the area under the graph: $~~~f(x) = \sqrt{1+x},~~~$
on the interval: $~~~0 \leq x \leq 3$
Work Step by Step
For each $x_i$, such that $1 \leq i \leq n$, note that $x_i = (i\frac{3}{n}) = \frac{3i}{n}$
$\Delta x = \frac{3}{n}$
We can express the area under the curve as a limit:
$A = \lim\limits_{n \to \infty} [f(x_1)\Delta x+f(x_2)\Delta x+...+f(x_n)\Delta x]$
$A = \lim\limits_{n \to \infty} \sum_{i=1}^{n}~\frac{3}{n}~\sqrt{1+\frac{3i}{n}}$
This limit is equal to the area under the graph: $~~~f(x) = \sqrt{1+x},~~~$
on the interval: $~~~0 \leq x \leq 3$
