The area (in sq units) of the region described by `{(x,y):y^(2)le2x and yge4x-1}` is

Options

A. `7/32`
B. `5/64`
C. `15/64`
D. `9/32`

Answer

To find the area of the region described by the inequalities \( y^2 \leq 2x \) and \( y \geq 4x - 1 \), we will follow these steps:

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

### Step 2: Find the points of intersection
To find the area of the bounded region, we need to find the points where the parabola intersects the line. We set \( y = 4x - 1 \) into the equation of the parabola \( y^2 = 2x \).

Substituting:
\[ (4x - 1)^2 = 2x \]
Expanding and rearranging gives:
\[ 16x^2 - 8x + 1 = 2x \implies 16x^2 - 10x + 1 = 0 \]

### Step 3: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 16 \cdot 1}}{2 \cdot 16} \]
\[ x = \frac{10 \pm \sqrt{100 - 64}}{32} = \frac{10 \pm \sqrt{36}}{32} = \frac{10 \pm 6}{32} \]
This gives us two values:
\[ x_1 = \frac{16}{32} = \frac{1}{2}, \quad x_2 = \frac{4}{32} = \frac{1}{8} \]

### Step 4: Find corresponding y-values
For \( x = \frac{1}{2} \):
\[ y = 4 \left(\frac{1}{2}\right) - 1 = 2 - 1 = 1 \quad \Rightarrow \quad \text{Point: } \left(\frac{1}{2}, 1\right) \]

For \( x = \frac{1}{8} \):
\[ y = 4 \left(\frac{1}{8}\right) - 1 = \frac{1}{2} - 1 = -\frac{1}{2} \quad \Rightarrow \quad \text{Point: } \left(\frac{1}{8}, -\frac{1}{2}\right) \]

### Step 5: Set up the integral for area
The area \( A \) can be calculated by integrating the difference between the upper curve (line) and the lower curve (parabola) with respect to \( y \):
\[ A = \int_{-\frac{1}{2}}^{1} \left( x_{\text{line}} - x_{\text{parabola}} \right) dy \]
Where:
- \( x_{\text{line}} = \frac{y + 1}{4} \)
- \( x_{\text{parabola}} = \frac{y^2}{2} \)

Thus, the area becomes:
\[ A = \int_{-\frac{1}{2}}^{1} \left( \frac{y + 1}{4} - \frac{y^2}{2} \right) dy \]

### Step 6: Evaluate the integral
Calculating the integral:
\[ A = \int_{-\frac{1}{2}}^{1} \left( \frac{y + 1}{4} - \frac{y^2}{2} \right) dy \]
\[ = \int_{-\frac{1}{2}}^{1} \left( \frac{y}{4} + \frac{1}{4} - \frac{y^2}{2} \right) dy \]
\[ = \left[ \frac{y^2}{8} + \frac{y}{4} - \frac{y^3}{6} \right]_{-\frac{1}{2}}^{1} \]

### Step 7: Calculate the definite integral
Calculating the limits:
1. For \( y = 1 \):
\[ \frac{1^2}{8} + \frac{1}{4} - \frac{1^3}{6} = \frac{1}{8} + \frac{2}{8} - \frac{4}{24} = \frac{3}{8} - \frac{1}{6} = \frac{9}{24} - \frac{4}{24} = \frac{5}{24} \]

2. For \( y = -\frac{1}{2} \):
\[ \frac{(-\frac{1}{2})^2}{8} + \frac{-\frac{1}{2}}{4} - \frac{(-\frac{1}{2})^3}{6} = \frac{1/4}{8} - \frac{1/8} + \frac{1/48} = \frac{1}{32} - \frac{1}{8} + \frac{1}{48} \]
Finding a common denominator (48):
\[ \frac{1}{32} = \frac{3}{96}, \quad -\frac{1}{8} = -\frac{6}{48} = -\frac{12}{96}, \quad \frac{1}{48} = \frac{2}{96} \]
Combining gives:
\[ \frac{3 - 12 + 2}{96} = \frac{-7}{96} \]

### Step 8: Final area calculation
Thus, the area \( A \) is:
\[ A = \left( \frac{5}{24} - \left(-\frac{7}{96}\right) \right) \]
Finding a common denominator (96):
\[ A = \frac{20}{96} + \frac{7}{96} = \frac{27}{96} = \frac{9}{32} \]

### Final Answer
The area of the region described is \( \frac{9}{32} \) square units.

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