The area (in square units) of the region bounded by `y^(2)=2x and y=4x-1` , is

Options

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

Explanation

To find the area of the region bounded by the curves \( y^2 = 2x \) (a parabola) and \( y = 4x - 1 \) (a straight line), we will follow these steps:

### Step 1: Find Points of Intersection
We need to find the points where the two curves intersect. To do this, we will set \( y^2 = 2x \) equal to \( (4x - 1)^2 \).

1. Substitute \( y = 4x - 1 \) into \( y^2 = 2x \):
\[ (4x - 1)^2 = 2x \]
Expanding the left side:
\[ 16x^2 - 8x + 1 = 2x \]
Rearranging gives:
\[ 16x^2 - 10x + 1 = 0 \]

### Step 2: Solve the Quadratic Equation
We will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( x \):
- Here, \( a = 16 \), \( b = -10 \), and \( c = 1 \).
- Calculate the discriminant:
\[ b^2 - 4ac = (-10)^2 - 4 \cdot 16 \cdot 1 = 100 - 64 = 36 \]
- Now, applying the quadratic formula:
\[ x = \frac{10 \pm \sqrt{36}}{32} = \frac{10 \pm 6}{32} \]
This gives us:
\[ x_1 = \frac{16}{32} = \frac{1}{2}, \quad x_2 = \frac{4}{32} = \frac{1}{8} \]

### Step 3: Set Up the Area Integral
The area \( A \) between the curves from \( x = \frac{1}{8} \) to \( x = \frac{1}{2} \) can be expressed as:
\[ A = \int_{x_1}^{x_2} (y_{\text{top}} - y_{\text{bottom}}) \, dx \]
Where:
- \( y_{\text{top}} = 4x - 1 \) (the line)
- \( y_{\text{bottom}} = \sqrt{2x} \) (the parabola)

Thus, the area becomes:
\[ A = \int_{\frac{1}{8}}^{\frac{1}{2}} \left( (4x - 1) - \sqrt{2x} \right) \, dx \]

### Step 4: Evaluate the Integral
We will evaluate the integral:
\[ A = \int_{\frac{1}{8}}^{\frac{1}{2}} (4x - 1) \, dx - \int_{\frac{1}{8}}^{\frac{1}{2}} \sqrt{2x} \, dx \]

1. **First Integral:**
\[ \int (4x - 1) \, dx = 2x^2 - x \]
Evaluating from \( \frac{1}{8} \) to \( \frac{1}{2} \):
\[ \left[ 2\left(\frac{1}{2}\right)^2 - \frac{1}{2} \right] - \left[ 2\left(\frac{1}{8}\right)^2 - \frac{1}{8} \right] \]
\[ = \left[ 2 \cdot \frac{1}{4} - \frac{1}{2} \right] - \left[ 2 \cdot \frac{1}{64} - \frac{1}{8} \right] \]
\[ = \left[ \frac{1}{2} - \frac{1}{2} \right] - \left[ \frac{1}{32} - \frac{1}{8} \right] \]
\[ = 0 - \left[ \frac{1}{32} - \frac{4}{32} \right] = \frac{3}{32} \]

2. **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 \( \frac{1}{8} \) to \( \frac{1}{2} \):
\[ \left[ \frac{2\sqrt{2}}{3} \left(\frac{1}{2}\right)^{3/2} \right] - \left[ \frac{2\sqrt{2}}{3} \left(\frac{1}{8}\right)^{3/2} \right] \]
\[ = \frac{2\sqrt{2}}{3} \cdot \frac{1}{2\sqrt{2}} - \frac{2\sqrt{2}}{3} \cdot \frac{1}{8\sqrt{2}} \]
\[ = \frac{1}{3} - \frac{1}{12} = \frac{4}{12} - \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \]

### Step 5: Combine the Results
Now, we combine the results of the two integrals:
\[ A = \frac{3}{32} - \frac{1}{4} = \frac{3}{32} - \frac{8}{32} = -\frac{5}{32} \]
Since area cannot be negative, we take the absolute value:
\[ A = \frac{5}{32} \]

### Final Answer
The area of the region bounded by the curves \( y^2 = 2x \) and \( y = 4x - 1 \) is \( \frac{5}{32} \) square units.

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