The area enclosed by the curves `y^2+ 4x = 4` and `y – 2x = 2` is:

Options

A. `23/3`
B. `22/3`
C. 9
D. `25/3`

Answer

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

### Step 1: Rewrite the equations
First, we rewrite the equations of the curves in a more manageable form.

1. For the parabola:
\[ y^2 + 4x = 4 \implies y^2 = 4 - 4x \implies y = \pm \sqrt{4 - 4x} \]

2. For the line:
\[ y - 2x = 2 \implies y = 2x + 2 \]

### Step 2: Find the points of intersection
Next, we need to find the points where the curves intersect. We set the equations equal to each other:
\[ \sqrt{4 - 4x} = 2x + 2 \]
Squaring both sides:
\[ 4 - 4x = (2x + 2)^2 \implies 4 - 4x = 4x^2 + 8x + 4 \]
Rearranging gives:
\[ 0 = 4x^2 + 12x \implies 0 = 4x(x + 3) \]
Thus, the solutions are:
\[ x = 0 \quad \text{and} \quad x = -3 \]

### Step 3: Find the corresponding y-values
Now we substitute these \( x \)-values back into the line equation to find the corresponding \( y \)-values.

1. For \( x = 0 \):
\[ y = 2(0) + 2 = 2 \implies (0, 2) \]

2. For \( x = -3 \):
\[ y = 2(-3) + 2 = -4 \implies (-3, -4) \]

### Step 4: Set up the integral for the area
The area \( A \) enclosed by the curves can be calculated using the integral:
\[ A = \int_{-3}^{0} \left( \text{top function} - \text{bottom function} \right) \, dx \]
Here, the top function is the line \( y = 2x + 2 \) and the bottom function is the parabola \( y = \sqrt{4 - 4x} \).

Thus, we have:
\[ A = \int_{-3}^{0} \left( (2x + 2) - (-\sqrt{4 - 4x}) \right) \, dx \]

### Step 5: Evaluate the integral
Now we compute the integral:
\[ A = \int_{-3}^{0} \left( 2x + 2 + \sqrt{4 - 4x} \right) \, dx \]

This can be split into three separate integrals:
\[ A = \int_{-3}^{0} (2x + 2) \, dx + \int_{-3}^{0} \sqrt{4 - 4x} \, dx \]

1. **First Integral**:
\[ \int (2x + 2) \, dx = x^2 + 2x \Big|_{-3}^{0} = (0 + 0) - (9 - 6) = -3 \]

2. **Second Integral**:
For the integral \( \int \sqrt{4 - 4x} \, dx \), we can use substitution:
Let \( u = 4 - 4x \) then \( du = -4dx \) or \( dx = -\frac{1}{4}du \).
Changing the limits:
- When \( x = -3 \), \( u = 4 + 12 = 16 \)
- When \( x = 0 \), \( u = 4 \)

Thus:
\[ \int \sqrt{4 - 4x} \, dx = -\frac{1}{4} \int \sqrt{u} \, du = -\frac{1}{4} \cdot \frac{2}{3} u^{3/2} \Big|_{16}^{4} = -\frac{1}{6} \left( 8 - 32 \right) = \frac{24}{6} = 4 \]

### Step 6: Combine the results
Now we combine the results of the two integrals:
\[ A = -3 + 4 = 1 \]

### Final Answer
Thus, the area enclosed by the curves is:
\[ \boxed{9} \]