The area (in sq. units) of the region described by `{(x, y) : y^2 le 2x`, and `y ge 4x – 1}` is

Options

A. `frac{9}{32}`
B. `frac{11}{12}`
C. `frac{11}{32}`
D. `frac{8}{9}`

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: Graph the equations
First, we need to graph the equations \( y^2 = 2x \) and \( y = 4x - 1 \).

1. **For \( y^2 = 2x \)**:
- This is a parabola that opens to the right.
- To find the vertex, set \( y = 0 \): \( 0^2 = 2x \) gives \( x = 0 \). So, the vertex is at (0, 0).
- When \( y = 2 \), \( 2^2 = 2x \) gives \( x = 2 \). Thus, the point (2, 2) is on the parabola.
- When \( y = -2 \), \( (-2)^2 = 2x \) gives \( x = 2 \). Thus, the point (2, -2) is also on the parabola.

2. **For \( y = 4x - 1 \)**:
- This is a straight line with a slope of 4 and a y-intercept of -1.
- To find the x-intercept, set \( y = 0 \): \( 0 = 4x - 1 \) gives \( x = \frac{1}{4} \).
- To find another point, set \( x = 0 \): \( y = 4(0) - 1 = -1 \). So, the point (0, -1) is on the line.

### Step 2: Find the points of intersection
To find the area of the region, we need to determine where the two curves intersect.

Set \( y^2 = 2x \) equal to \( y = 4x - 1 \):
1. Substitute \( y \) from the line equation into the parabola equation:
\[ (4x - 1)^2 = 2x \]
Expanding this gives:
\[ 16x^2 - 8x + 1 = 2x \]
Rearranging gives:
\[ 16x^2 - 10x + 1 = 0 \]

2. Use the quadratic formula to find \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 16 \cdot 1}}{2 \cdot 16} \]
\[ x = \frac{10 \pm \sqrt{100 - 64}}{32} = \frac{10 \pm \sqrt{36}}{32} = \frac{10 \pm 6}{32} \]
This gives us:
\[ x_1 = \frac{16}{32} = \frac{1}{2}, \quad x_2 = \frac{4}{32} = \frac{1}{8} \]

### Step 3: Determine the corresponding y-values
Now, we find the y-values for these x-values:
- For \( x = \frac{1}{2} \):
\[ y = 4\left(\frac{1}{2}\right) - 1 = 2 - 1 = 1 \]
- For \( x = \frac{1}{8} \):
\[ y = 4\left(\frac{1}{8}\right) - 1 = \frac{1}{2} - 1 = -\frac{1}{2} \]

### Step 4: Set up the integral for the area
The area \( A \) can be found by integrating the difference between the line and the parabola from \( y = -\frac{1}{2} \) to \( y = 1 \):
\[ A = \int_{-\frac{1}{2}}^{1} \left( \frac{y + 1}{4} - \frac{y^2}{2} \right) dy \]

### Step 5: Calculate the integral
1. Calculate the integral:
\[ A = \int_{-\frac{1}{2}}^{1} \left( \frac{y + 1}{4} - \frac{y^2}{2} \right) dy \]
\[ = \int_{-\frac{1}{2}}^{1} \left( \frac{y}{4} + \frac{1}{4} - \frac{y^2}{2} \right) dy \]

2. Integrate term by term:
\[ = \left[ \frac{y^2}{8} + \frac{y}{4} - \frac{y^3}{6} \right]_{-\frac{1}{2}}^{1} \]

3. Evaluate at the limits:
- At \( y = 1 \):
\[ = \frac{1^2}{8} + \frac{1}{4} - \frac{1^3}{6} = \frac{1}{8} + \frac{2}{8} - \frac{4}{24} = \frac{3}{8} - \frac{1}{6} = \frac{9 - 4}{24} = \frac{5}{24} \]
- At \( y = -\frac{1}{2} \):
\[ = \frac{(-\frac{1}{2})^2}{8} + \frac{-\frac{1}{2}}{4} - \frac{(-\frac{1}{2})^3}{6} = \frac{1/4}{8} - \frac{1/8} + \frac{1/48} = \frac{1}{32} - \frac{1}{8} + \frac{1}{48} \]
Convert to a common denominator (96):
\[ = \frac{3}{96} - \frac{12}{96} + \frac{2}{96} = \frac{-7}{96} \]

4. Combine:
\[ A = \left( \frac{5}{24} + \frac{7}{96} \right) = \frac{20}{96} + \frac{7}{96} = \frac{27}{96} = \frac{9}{32} \]

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

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