To find the area of the region enclosed by the graphs of the functions, we need to carefully set up an integral that represents the area.

Problem:

The graphs are given as:

1. $y = \sqrt{x} + 2$
2. $x = -2$
3. $y = 3$

The process involves finding the area between the curve $y = \sqrt{x} + 2$ and the horizontal line $y = 3$, bounded by $x = -2$.

Steps:

1. Find the intersection points:

   The area is bound vertically between $y = 3$ and $y = \sqrt{x} + 2$, so we need to determine the limits of integration by solving for where these two curves intersect: $y = \sqrt{x} + 2 = 3$ $\sqrt{x} = 1 \quad \Rightarrow \quad x = 1$ So the curves intersect at $x = 1$.

2. Set up the integral:

   The integral will be set up from $x = -2$ to $x = 1$, and we are calculating the area between the line $y = 3$ and the curve $y = \sqrt{x} + 2$.

   The area is given by the integral of the difference between $y = 3$ and $y = \sqrt{x} + 2$: $A = \int_{-2}^{1} \left( 3 - (\sqrt{x} + 2) \right) dx$ Simplifying the expression inside the integral: $A = \int_{-2}^{1} (3 - \sqrt{x} - 2) dx = \int_{-2}^{1} (1 - \sqrt{x}) dx$

3. Evaluate the integral:

   Now, we can compute the integral term by term: $A = \int_{-2}^{1} 1 \, dx - \int_{-2}^{1} \sqrt{x} \, dx$

Let's compute the value of the area.