Find the area of the region that is bounded above by the curve f(x) = sqrt(x+2) and the line g(x) = -x-2 and bounded below the x axis


To find the area of the region bounded by the curve \( f(x) = \sqrt{x+2} \), the line \( g(x) = -x - 2 \), and the x-axis, we'll follow these steps:

### 1. Determine the Points of Intersection
First, find the points of intersection between the curve \( f(x) \) and the line \( g(x) \). These points occur when \( f(x) = g(x) \).

\[
\sqrt{x+2} = -x - 2
\]

Square both sides to eliminate the square root:

\[
x + 2 = (-x - 2)^2
\]

Expand and simplify:

\[
x + 2 = x^2 + 4x + 4
\]

Move all terms to one side of the equation:

\[
0 = x^2 + 3x + 2
\]

Factor the quadratic equation:

\[
0 = (x + 1)(x + 2)
\]

This gives us the solutions:

\[
x = -1 \quad \text{and} \quad x = -2
\]

### 2. Evaluate the Area
The area of the region is determined by integrating the difference between the functions from \( x = -2 \) to \( x = -1 \) and considering the area under the x-axis as well.

The functions are:
- \( f(x) = \sqrt{x+2} \)
- \( g(x) = -x - 2 \)

The area bounded above the x-axis but below the curve and line is:

\[
\text{Area} = \int_{-2}^{-1} \left( \sqrt{x+2} - (-x-2) \right) \, dx
\]

Simplify the integrand:

\[
\text{Area} = \int_{-2}^{-1} \left( \sqrt{x+2} + x + 2 \right) \, dx
\]

Now, let's compute the integral.

### 3. Solve the Integral
We'll break the integral into two parts:

\[
\text{Area} = \int_{-2}^{-1} \sqrt{x+2} \, dx + \int_{-2}^{-1} (x + 2) \, dx
\]

#### First Integral:

Let \( u = x + 2 \), so \( du = dx \).

\[
\int \sqrt{x+2} \, dx = \int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C = \frac{2}{3} (x+2)^{3/2}
\]

Evaluate from \( x = -2 \) to \( x = -1 \):

\[
\left[ \frac{2}{3} (x+2)^{3/2} \right]_{-2}^{-1} = \frac{2}{3} \left[ (1)^{3/2} - (0)^{3/2} \right] = \frac{2}{3}
\]

#### Second Integral:

\[
\int_{-2}^{-1} (x + 2) \, dx = \left[ \frac{x^2}{2} + 2x \right]_{-2}^{-1} = \left[ \left( \frac{1}{2} - 2 \right) - \left( 0 \right) \right] = -\frac{3}{2}
\]

### 4. Combine the Results

The area is the sum of these two integrals:

\[
\text{Area} = \frac{2}{3} - \frac{3}{2} = \frac{4 - 9}{6} = -\frac{5}{6}
\]

Since the area cannot be negative, we take the absolute value:

\[
\text{Area} = \frac{5}{6} \text{ square units}
\]

Thus, the area of the region is \( \boxed{\frac{5}{6} \text{ square units}} \).

Would you like more details or have any questions?

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Here are 5 related questions for further exploration:

1. How do you find the area between two curves when both functions are above the x-axis?
2. What is the significance of taking the absolute value of the area result?
3. How does the process change if one of the bounding functions intersects the x-axis?
4. Can the integration limits be determined graphically for more complex regions?
5. How does the use of substitution simplify the integration process for more complex integrals?

**Tip:** Always sketch the region first to visualize the boundaries and the intersections clearly.

Learn how to calculate the area of the region bounded above by f(x) = sqrt(x+2), below by g(x) = -x - 2, and bounded below by the x-axis. Detailed step-by-step solution with integration and absolute value considerations. Suitable for advanced high school students.

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