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 \( y^2 = 2x \) and \( y = 4x - 1 \), we will follow these steps:

### Step 1: Find the points of intersection
To find the points of intersection, we need to solve the equations \( y^2 = 2x \) and \( y = 4x - 1 \) simultaneously.

Substituting \( y = 4x - 1 \) 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
Now we will solve the quadratic equation \( 16x^2 - 10x + 1 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 16, b = -10, c = 1 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-10)^2 - 4 \cdot 16 \cdot 1 = 100 - 64 = 36 \]

Now applying the quadratic formula:

\[ x = \frac{10 \pm \sqrt{36}}{2 \cdot 16} = \frac{10 \pm 6}{32} \]

This gives us:

\[ x_1 = \frac{16}{32} = \frac{1}{2}, \quad x_2 = \frac{4}{32} = \frac{1}{8} \]

### Step 3: Find corresponding y-values
Now we will find the corresponding y-values for both x-values:

For \( x = \frac{1}{2} \):

\[ y = 4 \left(\frac{1}{2}\right) - 1 = 2 - 1 = 1 \]

For \( x = \frac{1}{8} \):

\[ y = 4 \left(\frac{1}{8}\right) - 1 = \frac{4}{8} - 1 = \frac{1}{2} - 1 = -\frac{1}{2} \]

Thus, the points of intersection are \( \left(\frac{1}{2}, 1\right) \) and \( \left(\frac{1}{8}, -\frac{1}{2}\right) \).

### Step 4: Set up the integral for area
The area \( A \) between the curves from \( x = \frac{1}{8} \) to \( x = \frac{1}{2} \) can be found using the formula:

\[ A = \int_{x_1}^{x_2} (y_{\text{top}} - y_{\text{bottom}}) \, dx \]

Here, \( y_{\text{top}} = 4x - 1 \) and \( y_{\text{bottom}} = \sqrt{2x} \) (since \( y^2 = 2x \) gives two values for y, we take the positive root for the area calculation).

### Step 5: Calculate the area
We will integrate:

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

Calculating the integral:

1. **Integrate \( 4x - 1 \)**:
\[ \int (4x - 1) \, dx = 2x^2 - x \]

2. **Integrate \( \sqrt{2x} \)**:
\[ \int \sqrt{2x} \, dx = \frac{2}{3}(2x)^{3/2} = \frac{2}{3} \cdot 2^{3/2} x^{3/2} = \frac{4\sqrt{2}}{3} x^{3/2} \]

Now we combine the results:

\[ A = \left[2x^2 - x - \frac{4\sqrt{2}}{3} x^{3/2}\right]_{\frac{1}{8}}^{\frac{1}{2}} \]

Calculating the limits:

For \( x = \frac{1}{2} \):

\[ = 2\left(\frac{1}{2}\right)^2 - \frac{1}{2} - \frac{4\sqrt{2}}{3} \left(\frac{1}{2}\right)^{3/2} \]

For \( x = \frac{1}{8} \):

\[ = 2\left(\frac{1}{8}\right)^2 - \frac{1}{8} - \frac{4\sqrt{2}}{3} \left(\frac{1}{8}\right)^{3/2} \]

### Step 6: Final calculation
After calculating both limits and subtracting, we will arrive at the area of the bounded region.

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

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