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 will follow these steps:

### Step 1: Understand the curves
The parabola given by the equation \( y^2 = 8x \) opens to the right. The line \( x = 2 \) is a vertical line. We need to find the points where these two curves intersect.

### Step 2: Find the points of intersection
To find the points of intersection, we substitute \( x = 2 \) into the parabola's equation:

\[ y^2 = 8(2) = 16 \]

Taking the square root gives us:

\[ y = 4 \quad \text{and} \quad y = -4 \]

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 curves from \( x = 0 \) to \( x = 2 \) can be found by integrating the upper curve minus the lower curve. The upper curve is given by \( y = \sqrt{8x} \) and the lower curve is \( y = -\sqrt{8x} \).

The area can be expressed as:

\[ A = \int_{0}^{2} \left( \sqrt{8x} - (-\sqrt{8x}) \right) \, dx \]

This simplifies to:

\[ A = \int_{0}^{2} 2\sqrt{8x} \, dx \]

### Step 4: Simplify the integral
We can factor out the constant:

\[ A = 2 \int_{0}^{2} \sqrt{8} \sqrt{x} \, dx = 2 \sqrt{8} \int_{0}^{2} x^{1/2} \, dx \]

Since \( \sqrt{8} = 2\sqrt{2} \), we have:

\[ A = 2 \cdot 2\sqrt{2} \int_{0}^{2} x^{1/2} \, dx = 4\sqrt{2} \int_{0}^{2} x^{1/2} \, dx \]

### Step 5: Evaluate the integral
The integral \( \int x^{1/2} \, dx \) is evaluated as follows:

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

Now we evaluate this from 0 to 2:

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

Calculating the limits:

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

### Step 6: Simplify the expression
Calculating further:

\[ = 4\sqrt{2} \cdot \frac{4\sqrt{2}}{3} = \frac{16 \cdot 2}{3} = \frac{32}{3} \]

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

\[ \boxed{\frac{32}{3}} \text{ square units} \]

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