Find the area of the region bounded by line x = 2 and parabola `y^(2)=8x`.

Explanation

To find the area of the region bounded by the line \( x = 2 \) and the parabola \( y^2 = 8x \), we can follow these steps:

### Step 1: Understand the curves
The equation of the parabola \( y^2 = 8x \) can be rewritten as \( y = \pm \sqrt{8x} \). This means the parabola opens to the right and has two branches, one above the x-axis and one below.

### Step 2: Identify the points of intersection
We need to find the points where the line \( x = 2 \) intersects the parabola. Substituting \( x = 2 \) into the parabola's equation:

\[ y^2 = 8(2) = 16 \implies y = \pm 4 \]

Thus, the points of intersection are \( (2, 4) \) and \( (2, -4) \).

### Step 3: Set up the integral for the area
The area bounded by the line and the parabola can be calculated by integrating the function that represents the upper part of the region minus the lower part. Since the area above the x-axis and below the x-axis are symmetric, we can calculate the area above the x-axis and then double it.

The area \( A \) can be expressed as:

\[ A = 2 \int_{0}^{2} y \, dx \]

### Step 4: Express \( y \) in terms of \( x \)
From the parabola \( y^2 = 8x \), we have:

\[ y = \sqrt{8x} = 2\sqrt{2x} \]

### Step 5: Set up the integral
Now, substituting \( y \) into the area formula:

\[ A = 2 \int_{0}^{2} 2\sqrt{2x} \, dx = 4 \int_{0}^{2} \sqrt{2x} \, dx \]

### Step 6: Evaluate the integral
To evaluate the integral \( \int \sqrt{2x} \, dx \):

\[ \int \sqrt{2x} \, dx = \int \sqrt{2} \cdot x^{1/2} \, dx = \sqrt{2} \cdot \frac{x^{3/2}}{3/2} = \frac{2\sqrt{2}}{3} x^{3/2} \]

Now we can evaluate it from 0 to 2:

\[ A = 4 \left[ \frac{2\sqrt{2}}{3} x^{3/2} \right]_{0}^{2} \]

Calculating this:

\[ = 4 \cdot \frac{2\sqrt{2}}{3} \left[ (2)^{3/2} - (0)^{3/2} \right] = 4 \cdot \frac{2\sqrt{2}}{3} \cdot (2\sqrt{2}) = 4 \cdot \frac{2\sqrt{2}}{3} \cdot 2\sqrt{2} \]

\[ = 4 \cdot \frac{2 \cdot 2 \cdot 2}{3} = \frac{32}{3} \]

### Step 7: Final area calculation
Thus, the area of the region bounded by the line \( x = 2 \) and the parabola \( y^2 = 8x \) is:

\[ A = \frac{32}{3} \text{ square units} \]