The area (in sq.units ) of the region enclosed between the parabola `y^(2)=2x` and the line `x+y = 4` is _______.

Answer

To find the area enclosed between the parabola \( y^2 = 2x \) and the line \( x + y = 4 \), we will follow these steps:

### Step 1: Find the points of intersection
To find the points of intersection, we need to solve the equations of the parabola and the line simultaneously.

1. From the line equation \( x + y = 4 \), we can express \( x \) in terms of \( y \):
\[ x = 4 - y \]

2. Substitute this expression for \( x \) into the parabola equation \( y^2 = 2x \):
\[ y^2 = 2(4 - y) \]
\[ y^2 = 8 - 2y \]
Rearranging gives:
\[ y^2 + 2y - 8 = 0 \]

3. Now, we can solve this quadratic equation using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \]
\[ = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2} \]
This gives us the solutions:
\[ y = 2 \quad \text{and} \quad y = -4 \]

4. Now, substitute these \( y \) values back into the line equation to find the corresponding \( x \) values:
- For \( y = 2 \):
\[ x = 4 - 2 = 2 \quad \Rightarrow \quad (2, 2) \]
- For \( y = -4 \):
\[ x = 4 - (-4) = 8 \quad \Rightarrow \quad (8, -4) \]

### Step 2: Set up the integral for the area
The area \( A \) between the curves can be found using the integral:
\[ A = \int_{y_1}^{y_2} (x_{\text{line}} - x_{\text{parabola}}) \, dy \]
Where:
- \( y_1 = -4 \)
- \( y_2 = 2 \)
- \( x_{\text{line}} = 4 - y \)
- \( x_{\text{parabola}} = \frac{y^2}{2} \)

Thus, the area becomes:
\[ A = \int_{-4}^{2} \left( (4 - y) - \frac{y^2}{2} \right) dy \]

### Step 3: Evaluate the integral
Now we compute the integral:
\[ A = \int_{-4}^{2} \left( 4 - y - \frac{y^2}{2} \right) dy \]
Breaking it down:
\[ = \int_{-4}^{2} 4 \, dy - \int_{-4}^{2} y \, dy - \int_{-4}^{2} \frac{y^2}{2} \, dy \]

Calculating each part:
1. \( \int_{-4}^{2} 4 \, dy = 4[y]_{-4}^{2} = 4(2 - (-4)) = 4 \cdot 6 = 24 \)
2. \( \int_{-4}^{2} y \, dy = \left[\frac{y^2}{2}\right]_{-4}^{2} = \frac{2^2}{2} - \frac{(-4)^2}{2} = 2 - 8 = -6 \)
3. \( \int_{-4}^{2} \frac{y^2}{2} \, dy = \frac{1}{2}\left[\frac{y^3}{3}\right]_{-4}^{2} = \frac{1}{2}\left(\frac{2^3}{3} - \frac{(-4)^3}{3}\right) = \frac{1}{2}\left(\frac{8}{3} - \frac{-64}{3}\right) = \frac{1}{2}\left(\frac{72}{3}\right) = \frac{36}{3} = 12 \)

Putting it all together:
\[ A = 24 - (-6) - 12 = 24 + 6 - 12 = 18 \]

### Final Answer
The area enclosed between the parabola \( y^2 = 2x \) and the line \( x + y = 4 \) is \( 18 \) square units.