The area (in sq units) of the region
`A={(x,y):(y^(2))/(2) le x le y+4}` is

Options

A. 18
B. `(53)/(3)`
C. 16
D. 30

Explanation

To find the area of the region defined by the inequalities \( A = \{(x,y) : \frac{y^2}{2} \leq x \leq y + 4\} \), we will follow these steps:

### Step 1: Identify the curves
The inequalities represent two curves:
1. The parabola given by \( y^2 = 2x \).
2. The line given by \( x = y + 4 \).

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

From the parabola:
\[ y^2 = 2x \]
From the line:
\[ x = y + 4 \]

Substituting the expression for \( x \) from the line into the parabola:
\[ y^2 = 2(y + 4) \]
\[ y^2 = 2y + 8 \]
Rearranging gives:
\[ y^2 - 2y - 8 = 0 \]

### Step 3: Solve the quadratic equation
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -2, c = -8 \):
\[ y = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \]
\[ y = \frac{2 \pm \sqrt{4 + 32}}{2} \]
\[ y = \frac{2 \pm \sqrt{36}}{2} \]
\[ y = \frac{2 \pm 6}{2} \]
Thus, we have:
\[ y = 4 \quad \text{and} \quad y = -2 \]

### Step 4: Find corresponding x-values
Now, we find the corresponding x-values for these y-values using \( x = y + 4 \):
1. For \( y = 4 \):
\[ x = 4 + 4 = 8 \]
2. For \( y = -2 \):
\[ x = -2 + 4 = 2 \]

The intersection points are \( (8, 4) \) and \( (2, -2) \).

### Step 5: Set up the integral for the area
The area \( A \) between the curves from \( y = -2 \) to \( y = 4 \) can be expressed as:
\[ A = \int_{-2}^{4} [(y + 4) - \left(\frac{y^2}{2}\right)] \, dy \]

### Step 6: Evaluate the integral
Now we calculate the integral:
\[ A = \int_{-2}^{4} \left( y + 4 - \frac{y^2}{2} \right) \, dy \]

Calculating the integral:
\[ = \int_{-2}^{4} \left( y + 4 - \frac{y^2}{2} \right) \, dy = \left[ \frac{y^2}{2} + 4y - \frac{y^3}{6} \right]_{-2}^{4} \]

### Step 7: Substitute the limits
Calculating at the upper limit \( y = 4 \):
\[ = \frac{4^2}{2} + 4(4) - \frac{4^3}{6} = \frac{16}{2} + 16 - \frac{64}{6} = 8 + 16 - \frac{32}{3} = 24 - \frac{32}{3} = \frac{72}{3} - \frac{32}{3} = \frac{40}{3} \]

Calculating at the lower limit \( y = -2 \):
\[ = \frac{(-2)^2}{2} + 4(-2) - \frac{(-2)^3}{6} = \frac{4}{2} - 8 + \frac{8}{6} = 2 - 8 + \frac{4}{3} = -6 + \frac{4}{3} = -\frac{18}{3} + \frac{4}{3} = -\frac{14}{3} \]

### Step 8: Final area calculation
Now, we find the area:
\[ A = \left( \frac{40}{3} - \left(-\frac{14}{3}\right) \right) = \frac{40}{3} + \frac{14}{3} = \frac{54}{3} = 18 \]

### Conclusion
The area of the region \( A \) is \( 18 \) square units.

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