Question

The area between the parabolas \( y^2 = 4 - x \) and \( y^2 = x \) is given by:

• A. \( \frac{3\sqrt{2}}{16} \)
• B. \( \frac{16\sqrt{3}}{5} \)
• C. \( \frac{5\sqrt{3}}{16} \)
• D. \( \frac{16\sqrt{2}}{3} \)

Hint:

For areas enclosed between curves, integrate the difference of the upper and lower functions with respect to \( x \) or \( y \).

Answer 1

The Correct Option is D

Step 1: Find points of intersection. Equating \( y^2 = 4 - x \) and \( y^2 = x \), \[ 4 - x = x \quad \Rightarrow \quad 4 = 2x \quad \Rightarrow \quad x = 2. \] So, the region extends from \( x = 0 \) to \( x = 2 \).
Step 2: Compute area using integration. \[ A = \int_0^2 \left( \sqrt{4-x} - \sqrt{x} \right) dx. \] Solving the integral, we get: \[ A = \frac{16\sqrt{2}}{3}. \]
Step 3: Selecting the correct option. Since \( \frac{16\sqrt{2}}{3} \) matches, the correct answer is (D).