Draw a rough sketch of the curve and find the area of the region bounded by curve `y^2=8x` and the line x=2.

Answer

To solve the problem of finding the area of the region 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 rewrite it in terms of \( y \):
\[ y = \pm \sqrt{8x} \]
This means for each \( x \), there are two values of \( y \) (one positive and one negative).

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

### Step 3: Sketch the Curve
We can plot the curve \( y^2 = 8x \) and the line \( x = 2 \):
- The vertex of the parabola is at the origin (0,0).
- The points we calculated give us the bounds on the right side of the parabola.

### Step 4: Set Up the Area Integral
The area \( A \) between the curve and the line can be calculated by integrating the upper curve minus the lower curve from \( x = 0 \) to \( x = 2 \):
\[ A = 2 \int_{0}^{2} \sqrt{8x} \, dx \]
We multiply by 2 because the area above the x-axis is equal to the area below the x-axis.

### Step 5: Simplify the Integral
The integral can be simplified:
\[ A = 2 \int_{0}^{2} \sqrt{8} \sqrt{x} \, dx = 2 \cdot \sqrt{8} \int_{0}^{2} x^{1/2} \, dx \]
Since \( \sqrt{8} = 2\sqrt{2} \), we have:
\[ A = 4\sqrt{2} \int_{0}^{2} x^{1/2} \, dx \]

### Step 6: Calculate the Integral
Now we calculate the integral:
\[ \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]
Evaluating from 0 to 2:
\[ \left[ \frac{2}{3} x^{3/2} \right]_{0}^{2} = \frac{2}{3} (2^{3/2} - 0) = \frac{2}{3} (2\sqrt{2}) = \frac{4\sqrt{2}}{3} \]

### Step 7: Final Area Calculation
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 of the region bounded by the curve \( y^2 = 8x \) and the line \( x = 2 \) is:
\[ \boxed{\frac{32}{3}} \]