To find the area of the region bounded by the curves \(4y = |4 - x^2|\) and \(y = 7 - |x|\), we will follow these steps:
### Step 1: Understand the curves
1. **Curve 1: \(4y = |4 - x^2|\)**
This can be split into two cases:
- When \(4 - x^2 \geq 0\) (i.e., \(-2 \leq x \leq 2\)), we have \(4y = 4 - x^2\) or \(y = 1 - \frac{x^2}{4}\).
- When \(4 - x^2 < 0\) (i.e., \(x < -2\) or \(x > 2\)), we have \(4y = -(4 - x^2)\) or \(y = \frac{x^2}{4} - 1\).
2. **Curve 2: \(y = 7 - |x|\)**
This is a V-shaped graph with a vertex at \((0, 7)\) and intersects the x-axis at \((-7, 0)\) and \((7, 0)\).
### Step 2: Find points of intersection
To find the area, we need to find the points where these two curves intersect.
1. **For \(x \in [-2, 2]\)**:
Set \(1 - \frac{x^2}{4} = 7 - |x|\):
\[
1 - \frac{x^2}{4} = 7 - x \quad \text{(for } x \geq 0\text{)}
\]
\[
\Rightarrow -\frac{x^2}{4} + x - 6 = 0 \quad \Rightarrow x^2 - 4x + 24 = 0
\]
The discriminant is negative, so there are no intersections in this range.
2. **For \(x < -2\) or \(x > 2\)**:
Set \(\frac{x^2}{4} - 1 = 7 + x\) (for \(x < 0\)):
\[
\frac{x^2}{4} - x - 8 = 0 \quad \Rightarrow x^2 - 4x - 32 = 0
\]
Solving this using the quadratic formula:
\[
x = \frac{4 \pm \sqrt{16 + 128}}{2} = \frac{4 \pm 12}{2} \Rightarrow x = 8 \text{ or } x = -4
\]
### Step 3: Set up the integral for the area
The area \(A\) between the curves from \(x = -4\) to \(x = 2\) can be calculated as:
\[
A = \int_{-4}^{-2} \left(7 + x - \left(\frac{x^2}{4} - 1\right)\right) \, dx + \int_{-2}^{2} \left(7 - |x| - \left(1 - \frac{x^2}{4}\right)\right) \, dx
\]
### Step 4: Calculate the area
1. **For \(x \in [-4, -2]\)**:
\[
A_1 = \int_{-4}^{-2} \left(7 + x - \left(\frac{x^2}{4} - 1\right)\right) \, dx
\]
Simplifying:
\[
= \int_{-4}^{-2} \left(8 + x - \frac{x^2}{4}\right) \, dx
\]
2. **For \(x \in [-2, 2]\)**:
\[
A_2 = \int_{-2}^{2} \left(7 - |x| - \left(1 - \frac{x^2}{4}\right)\right) \, dx
\]
This simplifies to:
\[
= \int_{-2}^{2} \left(6 - |x| + \frac{x^2}{4}\right) \, dx
\]
### Step 5: Evaluate the integrals
1. **Calculating \(A_1\)**:
\[
A_1 = \left[8x + \frac{x^2}{2} - \frac{x^3}{12}\right]_{-4}^{-2}
\]
2. **Calculating \(A_2\)**:
\[
A_2 = 2 \int_{0}^{2} \left(6 - x + \frac{x^2}{4}\right) \, dx
\]
### Step 6: Combine areas
Finally, add \(A_1\) and \(A_2\) to get the total area.
