The area bounded by the curve `y^(2)=8x` and the line x=2 is

Options

A. `32/3` sq. units
B. `23/3` sq. units
C. `16/3` sq. units
D. `13/2` sq. units

Answer

To find the area bounded by the curve \(y^2 = 8x\) and the line \(x = 2\), we will follow these steps:

### Step 1: Understand the curve
The equation \(y^2 = 8x\) represents a parabola that opens to the right. We can express \(y\) in terms of \(x\):
\[ y = \sqrt{8x} \quad \text{and} \quad y = -\sqrt{8x} \]

### Step 2: Identify the points of intersection
To find the area bounded by the curve and the line \(x = 2\), we first need to find the points where the curve intersects the line. We substitute \(x = 2\) into the equation of the curve:
\[ y^2 = 8(2) = 16 \implies y = 4 \quad \text{and} \quad y = -4 \]
Thus, the points of intersection are \((2, 4)\) and \((2, -4)\).

### Step 3: Set up the integral for the area
The area \(A\) between the curve and the line can be calculated using the integral:
\[ A = \int_{0}^{2} (y_{\text{upper}} - y_{\text{lower}}) \, dx \]
In this case, \(y_{\text{upper}} = \sqrt{8x}\) and \(y_{\text{lower}} = -\sqrt{8x}\).

### Step 4: Calculate the area
The area can be expressed as:
\[ A = \int_{0}^{2} \left(\sqrt{8x} - (-\sqrt{8x})\right) \, dx = \int_{0}^{2} 2\sqrt{8x} \, dx \]
This simplifies to:
\[ A = 2\int_{0}^{2} \sqrt{8x} \, dx = 2\sqrt{8} \int_{0}^{2} \sqrt{x} \, dx = 4\sqrt{2} \int_{0}^{2} \sqrt{x} \, dx \]

### Step 5: Integrate \(\sqrt{x}\)
Now we compute the integral:
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]
Evaluating this from 0 to 2:
\[ \int_{0}^{2} \sqrt{x} \, dx = \left[\frac{2}{3} x^{3/2}\right]_{0}^{2} = \frac{2}{3} (2^{3/2}) - 0 = \frac{2}{3} (2\sqrt{2}) = \frac{4\sqrt{2}}{3} \]

### Step 6: Substitute back into the area formula
Now substituting back into the area formula:
\[ A = 4\sqrt{2} \cdot \frac{4\sqrt{2}}{3} = \frac{16 \cdot 2}{3} = \frac{32}{3} \]

### Final Answer
The area bounded by the curve \(y^2 = 8x\) and the line \(x = 2\) is:
\[ \boxed{\frac{32}{3}} \]