What is the area of the region enclosed by` y = 2 |x|` and `y = 4` ?

Options

A. 2 square unit
B. `4` square unit
C. 8 square unit
D. 16 squre unit

Explanation

To find the area of the region enclosed by the curves \( y = 2|x| \) and \( y = 4 \), we will follow these steps:

### Step 1: Find the points of intersection
We need to find the points where the two curves intersect. This occurs when:

\[ 2|x| = 4 \]

Dividing both sides by 2 gives:

\[ |x| = 2 \]

This results in two solutions:

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

### Step 2: Sketch the curves
The graph of \( y = 2|x| \) is a V-shaped graph that opens upwards, with its vertex at the origin (0,0). The line \( y = 4 \) is a horizontal line. The two curves intersect at the points \( (-2, 4) \) and \( (2, 4) \).

### Step 3: Set up the integral for the area
The area between the curves from \( x = -2 \) to \( x = 2 \) can be calculated using the integral:

\[ \text{Area} = \int_{-2}^{2} (y_{\text{top}} - y_{\text{bottom}}) \, dx \]

Here, \( y_{\text{top}} = 4 \) and \( y_{\text{bottom}} = 2|x| \).

### Step 4: Calculate the area
We can split the integral into two parts due to symmetry:

\[ \text{Area} = 2 \int_{0}^{2} (4 - 2x) \, dx \]

Now, we will compute the integral:

\[ \int_{0}^{2} (4 - 2x) \, dx \]

Calculating the integral:

\[ = \left[ 4x - x^2 \right]_{0}^{2} \]

Evaluating at the limits:

\[ = \left( 4(2) - (2)^2 \right) - \left( 4(0) - (0)^2 \right) \]
\[ = (8 - 4) - 0 = 4 \]

### Step 5: Multiply by 2 for symmetry
Since we only calculated the area from \( 0 \) to \( 2 \), we multiply by 2 to account for the area from \( -2 \) to \( 0 \):

\[ \text{Total Area} = 2 \times 4 = 8 \]

### Final Answer
The area of the region enclosed by the curves \( y = 2|x| \) and \( y = 4 \) is:

\[ \boxed{8} \text{ square units} \]