Step 1: Understanding the Question:
The goal is to calculate the area enclosed by a rightward-opening parabola and a vertical line.
The symmetry of the parabola suggests that the area is distributed equally above and below the x-axis.
Step 2: Key Formula or Approach:
The area for a curve \(y^2 = 4ax\) from \(x=0\) to \(x=h\) is:
\[ \text{Area} = 2 \times \int_{0}^{h} y \, dx \]
For \(y^2 = 8x\), we have \(y = \sqrt{8x} = 2\sqrt{2}\sqrt{x}\).
Step 3: Detailed Explanation:
The required area is the integral from \(x=0\) to \(x=2\).
Total Area = \(2 \times \int_{0}^{2} 2\sqrt{2}x^{1/2} \, dx\)
\[ \text{Area} = 4\sqrt{2} \int_{0}^{2} x^{1/2} \, dx \]
Applying the power rule for integration \(\int x^n \, dx = \frac{x^{n+1}}{n+1}\):
\[ \text{Area} = 4\sqrt{2} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{2} \]
\[ \text{Area} = 4\sqrt{2} \times \frac{2}{3} \times [x^{3/2}]_{0}^{2} \]
\[ \text{Area} = \frac{8\sqrt{2}}{3} \times (2^{3/2} - 0) \]
Note that \(2^{3/2} = 2\sqrt{2}\).
\[ \text{Area} = \frac{8\sqrt{2}}{3} \times 2\sqrt{2} \]
\[ \text{Area} = \frac{16 \times (\sqrt{2} \times \sqrt{2})}{3} \]
\[ \text{Area} = \frac{16 \times 2}{3} = \frac{32}{3} \]
Step 4: Final Answer:
The area of the bounded region is \(\frac{32}{3}\) square units.