Find the area bounded by the curves `4 y = |4-x^(2)|, y = 7 - |x|`

Answer

To find the area bounded by the curves \(4y = |4 - x^2|\) and \(y = 7 - |x|\), we will follow these steps:

### Step 1: Rewrite the equations
We start by rewriting the equations of the curves in a more manageable form.

1. For the first curve:
\[ 4y = |4 - x^2| \implies y = \frac{1}{4}|4 - x^2| \]

2. For the second curve:
\[ y = 7 - |x| \]

### Step 2: Find the intersection points
To find the area between the curves, we first need to determine the points where they intersect. We will set the two equations equal to each other:

\[ \frac{1}{4}|4 - x^2| = 7 - |x| \]

This equation can be solved by considering the cases for the absolute values.

#### Case 1: \(x \geq 0\)
In this case, \(|x| = x\) and \(|4 - x^2| = 4 - x^2\) if \(x^2 \leq 4\) (i.e., \(x \leq 2\)) or \(x^2 - 4\) if \(x > 2\).

1. For \(0 \leq x \leq 2\):
\[ \frac{1}{4}(4 - x^2) = 7 - x \]
Multiplying through by 4 gives:
\[ 4 - x^2 = 28 - 4x \implies x^2 - 4x + 24 = 0 \]
The discriminant of this quadratic is negative, indicating no real solutions.

2. For \(x > 2\):
\[ \frac{1}{4}(x^2 - 4) = 7 - x \]
Multiplying through by 4 gives:
\[ x^2 - 4 = 28 - 4x \implies x^2 + 4x - 32 = 0 \]
Solving this using the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{16 + 128}}{2} = \frac{-4 \pm 12}{2} \]
This gives \(x = 4\) (valid) and \(x = -8\) (not valid).

#### Case 2: \(x < 0\)
In this case, \(|x| = -x\) and \(|4 - x^2| = 4 - x^2\) if \(x^2 \leq 4\) (i.e., \(x \geq -2\)) or \(x^2 - 4\) if \(x < -2\).

1. For \(-2 < x < 0\):
\[ \frac{1}{4}(4 - x^2) = 7 + x \]
This leads to the same equation as before, yielding no valid solutions.

2. For \(x < -2\):
\[ \frac{1}{4}(x^2 - 4) = 7 + x \]
This leads to:
\[ x^2 - 4 = 28 + 4x \implies x^2 - 4x - 32 = 0 \]
Solving this gives:
\[ x = \frac{4 \pm \sqrt{16 + 128}}{2} = \frac{4 \pm 12}{2} \]
This gives \(x = 8\) (not valid) and \(x = -4\) (valid).

### Step 3: Determine the area
The intersection points are \(x = -4\) and \(x = 4\). We will find the area between the curves from \(x = -4\) to \(x = 4\).

1. For \(x \in [-4, 0]\):
\[ \text{Upper curve: } y = 7 + x \]
\[ \text{Lower curve: } y = \frac{1}{4}(4 - x^2) \]

2. For \(x \in [0, 4]\):
\[ \text{Upper curve: } y = 7 - x \]
\[ \text{Lower curve: } y = \frac{1}{4}(4 - x^2) \]

### Step 4: Calculate the area
The area \(A\) can be computed as:
\[ A = \int_{-4}^{0} \left( (7 + x) - \frac{1}{4}(4 - x^2) \right) dx + \int_{0}^{4} \left( (7 - x) - \frac{1}{4}(4 - x^2) \right) dx \]

Calculating the first integral:
\[ \int_{-4}^{0} \left( 7 + x - 1 + \frac{x^2}{4} \right) dx = \int_{-4}^{0} \left( 6 + x + \frac{x^2}{4} \right) dx \]

Calculating the second integral:
\[ \int_{0}^{4} \left( 7 - x - 1 + \frac{x^2}{4} \right) dx = \int_{0}^{4} \left( 6 - x + \frac{x^2}{4} \right) dx \]

### Step 5: Evaluate the integrals
1. For the first integral:
\[ \int (6 + x + \frac{x^2}{4}) dx = 6x + \frac{x^2}{2} + \frac{x^3}{12} \Big|_{-4}^{0} \]
Evaluating gives:
\[ = 0 - \left( 6(-4) + \frac{(-4)^2}{2} + \frac{(-4)^3}{12} \right) = 24 - 8 + \frac{64}{12} = 24 - 8 + \frac{16}{3} = 16 + \frac{16}{3} = \frac{48 + 16}{3} = \frac{64}{3} \]

2. For the second integral:
\[ \int (6 - x + \frac{x^2}{4}) dx = 6x - \frac{x^2}{2} + \frac{x^3}{12} \Big|_{0}^{4} \]
Evaluating gives:
\[ = \left( 6(4) - \frac{(4)^2}{2} + \frac{(4)^3}{12} \right) - 0 = 24 - 8 + \frac{64}{12} = 24 - 8 + \frac{16}{3} = 16 + \frac{16}{3} = \frac{64}{3} \]

### Final Area Calculation
Thus, the total area \(A\) is:
\[ A = \frac{64}{3} + \frac{64}{3} = \frac{128}{3} \]

### Final Answer
The area bounded by the curves is:
\[ \boxed{\frac{128}{3}} \text{ square units} \]