The area (in square units) of the region described by `{(x, y):y^(2)le 2x and y le 4-x}` is

Options

A. `(7)/(32)`
B. `(5)/(64)`
C. `(15)/(64)`
D. `18`

Answer

To find the area of the region described by the inequalities \( y^2 \leq 2x \) and \( y \leq 4 - x \), we will follow these steps:

### Step 1: Identify the curves
The first inequality \( y^2 \leq 2x \) represents a parabola that opens to the right. The second inequality \( y \leq 4 - x \) represents a straight line.

### Step 2: Find the points of intersection
To find the points where the curves intersect, we set \( y^2 = 2x \) and \( y = 4 - x \) equal to each other.

Substituting \( y = 4 - x \) into \( y^2 = 2x \):
\[ (4 - x)^2 = 2x \]
Expanding the left side:
\[ 16 - 8x + x^2 = 2x \]
Rearranging gives:
\[ x^2 - 10x + 16 = 0 \]
Now, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} \]
Calculating the discriminant:
\[ = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm \sqrt{36}}{2} = \frac{10 \pm 6}{2} \]
This gives us:
\[ x = \frac{16}{2} = 8 \quad \text{and} \quad x = \frac{4}{2} = 2 \]

### Step 3: Find corresponding y-values
Now we find the y-values for these x-values using \( y = 4 - x \):
- For \( x = 8 \):
\[ y = 4 - 8 = -4 \]
- For \( x = 2 \):
\[ y = 4 - 2 = 2 \]

Thus, the points of intersection are \( (8, -4) \) and \( (2, 2) \).

### Step 4: Set up the integral for the area
The area between the curves can be found by integrating the difference of the functions from the leftmost intersection point to the rightmost intersection point.

The area \( A \) is given by:
\[ A = \int_{2}^{8} \left((4 - x) - \sqrt{2x}\right) \, dx \]

### Step 5: Evaluate the integral
We will compute the integral:
\[ A = \int_{2}^{8} (4 - x) \, dx - \int_{2}^{8} \sqrt{2x} \, dx \]

Calculating the first integral:
\[ \int (4 - x) \, dx = 4x - \frac{x^2}{2} \]
Evaluating from 2 to 8:
\[ = \left[4(8) - \frac{8^2}{2}\right] - \left[4(2) - \frac{2^2}{2}\right] = (32 - 32) - (8 - 2) = 0 - 6 = -6 \]

Now for the second integral:
\[ \int \sqrt{2x} \, dx = \frac{2}{3}(2x)^{3/2} \cdot \frac{1}{\sqrt{2}} = \frac{2\sqrt{2}}{3}(x^{3/2}) \]
Evaluating from 2 to 8:
\[ = \left[\frac{2\sqrt{2}}{3}(8^{3/2})\right] - \left[\frac{2\sqrt{2}}{3}(2^{3/2})\right] \]
Calculating \( 8^{3/2} = 8\sqrt{8} = 16\sqrt{2} \) and \( 2^{3/2} = 2\sqrt{2} \):
\[ = \frac{2\sqrt{2}}{3}(16\sqrt{2}) - \frac{2\sqrt{2}}{3}(2\sqrt{2}) = \frac{32}{3} - \frac{4}{3} = \frac{28}{3} \]

### Step 6: Combine the results
Now we combine the results:
\[ A = -6 - \frac{28}{3} = -\frac{18}{3} - \frac{28}{3} = -\frac{46}{3} \]
Since area cannot be negative, we take the absolute value:
\[ A = \frac{46}{3} \]

### Final Answer
The area of the region described by the inequalities is \( \frac{46}{3} \) square units.