Step 1: Understanding the Concept:
We are dealing with two intersecting parabolas. To find the area between them, we need to find their points of intersection and then integrate the difference of their y-functions.
Step 2: Detailed Explanation:
1. Find points of intersection:
Given \( y = x^{2}/2 \) and \( y^{2} = 2x \).
Substitute \( y \) into the second equation:
\[ (x^{2}/2)^{2} = 2x \implies x^{4}/4 = 2x \implies x^{4} = 8x \implies x(x^{3} - 8) = 0 \]
Intersection points are \( x = 0 \) and \( x = 2 \).
For \( x = 0, y = 0 \). For \( x = 2, y = 2 \). Points are \( (0,0) \) and \( (2,2) \).
2. Calculate Area:
Between \( x=0 \) and \( x=2 \), the upper curve is \( y = \sqrt{2x} \) and the lower curve is \( y = x^{2}/2 \).
\[ \text{Area} = \int_{0}^{2} (\sqrt{2x} - x^{2}/2) dx \]
\[ \text{Area} = \left[ \sqrt{2} \cdot \frac{x^{3/2}}{3/2} - \frac{x^{3}}{6} \right]_{0}^{2} \]
\[ \text{Area} = \left[ \frac{2\sqrt{2}}{3} x^{3/2} - \frac{x^{3}}{6} \right]_{0}^{2} \]
Evaluating at boundaries:
\[ \text{Area} = ( \frac{2\sqrt{2}}{3} \cdot 2\sqrt{2} - \frac{8}{6} ) - 0 \]
\[ \text{Area} = \frac{8}{3} - \frac{4}{3} = \frac{4}{3} \text{ sq unit} \].
Step 3: Final Answer:
The included area is 4/3 sq unit.