https://r2.erweima.ai/mathbot/image/172320298986258x6w0dv.jpg

To solve this problem, we'll find the area of the shaded region bounded below by the function \( p(x) \) and above by the function \( q(x) \).

### Step 1: Identify the functions

The functions provided are:
\[
p(x) = -8\sqrt{\frac{x + 2}{4}} + 2
\]
\[
q(x) = \frac{-8x - 8}{4}
\]

### Step 2: Simplify \( q(x) \)

Simplify \( q(x) \) to make it easier to work with:
\[
q(x) = -2x - 2
\]

### Step 3: Find the points of intersection

To find the area, we first need the points of intersection, which are where \( p(x) = q(x) \).

Set the equations equal:
\[
-8\sqrt{\frac{x + 2}{4}} + 2 = -2x - 2
\]

First, isolate the square root term:
\[
-8\sqrt{\frac{x + 2}{4}} = -2x - 4
\]
\[
8\sqrt{\frac{x + 2}{4}} = 2x + 4
\]
Divide both sides by 8:
\[
\sqrt{\frac{x + 2}{4}} = \frac{x}{4} + \frac{1}{2}
\]
Square both sides to eliminate the square root:
\[
\frac{x + 2}{4} = \left(\frac{x}{4} + \frac{1}{2}\right)^2
\]
Expand the square on the right side:
\[
\frac{x + 2}{4} = \frac{x^2}{16} + \frac{x}{4} + \frac{1}{4}
\]
Multiply through by 16 to clear the fraction:
\[
4(x + 2) = x^2 + 4x + 4
\]
Simplify and collect like terms:
\[
4x + 8 = x^2 + 4x + 4
\]
\[
0 = x^2 - 4
\]
Factor the quadratic:
\[
(x - 2)(x + 2) = 0
\]

So, the points of intersection are \( x = 2 \) and \( x = -2 \).

### Step 4: Set up the integral for the area

The area \( A \) of the shaded region is given by:
\[
A = \int_{-2}^{2} \left[q(x) - p(x)\right] \, dx
\]

Substitute the functions:
\[
A = \int_{-2}^{2} \left[(-2x - 2) - \left(-8\sqrt{\frac{x + 2}{4}} + 2\right)\right] dx
\]
Simplify the integrand:
\[
A = \int_{-2}^{2} \left[-2x - 2 + 8\sqrt{\frac{x + 2}{4}} - 2\right] dx
\]
\[
A = \int_{-2}^{2} \left[-2x - 4 + 8\sqrt{\frac{x + 2}{4}}\right] dx
\]

### Step 5: Evaluate the integral

Split the integral into two parts:
\[
A = \int_{-2}^{2} (-2x - 4) \, dx + \int_{-2}^{2} 8\sqrt{\frac{x + 2}{4}} \, dx
\]

1. **First Integral**:
   \[
   \int_{-2}^{2} (-2x - 4) \, dx
   \]
   Since the integrand is an odd function, its integral over a symmetric interval \([-2, 2]\) is 0.

2. **Second Integral**:
   \[
   \int_{-2}^{2} 8\sqrt{\frac{x + 2}{4}} \, dx
   \]
   Simplify the square root:
   \[
   \int_{-2}^{2} 8\sqrt{\frac{x + 2}{4}} \, dx = \int_{-2}^{2} 4\sqrt{x + 2} \, dx
   \]
   Let \( u = x + 2 \), then \( du = dx \) and the limits change from 0 to 4:
   \[
   A = \int_{0}^{4} 4\sqrt{u} \, du
   \]
   Integrate:
   \[
   \int 4u^{1/2} \, du = \frac{4}{3} u^{3/2}
   \]
   Evaluate from 0 to 4:
   \[
   A = \frac{4}{3} \left[(4)^{3/2} - 0^{3/2}\right]
   \]
   \[
   A = \frac{4}{3} \times 8 = \frac{32}{3}
   \]

Thus, the area of the shaded region is approximately **10.67 square units**.

Would you like a further breakdown of these steps or any clarification? 

Here are some related questions to explore:

1. How would the area change if the bounds of integration were different?
2. What happens if the functions intersect at different points?
3. How do you handle absolute values in the context of area calculation?
4. Can you find the area using numerical methods?
5. How do you find the area when the integrand contains a more complex function?
6. How would this problem change if one of the functions was a trigonometric function?
7. What role does symmetry play in the simplification of integrals?
8. How would you calculate the volume of a solid formed by revolving this region around the x-axis?

**Tip:** When integrating square roots, substitution often simplifies the integral.

Learn how to calculate the area of the shaded region bounded by two functions using integral calculus. This problem involves finding points of intersection, setting up and evaluating integrals, and understanding the concept of area under a curve. Suitable for college-level students and those interested in advanced calculus problems.

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