Question

Area bounded by the curves \(y = x^2\) and \(y = 2 - x^2\) is

• A. \(\frac{8}{3}\) sq units
• B. \(\frac{3}{8}\) sq units
• C. \(\frac{3}{2}\) sq units
• D. None of these

Hint:

For symmetric curves, integrate from 0 to intersection and double.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Find intersection points and integrate the difference of the functions.

Step 2: Detailed Explanation:
Intersection: \(x^2 = 2 - x^2 \Rightarrow 2x^2 = 2 \Rightarrow x = \pm 1\). Area = \(\int_{-1}^1 [(2 - x^2) - x^2] dx = \int_{-1}^1 (2 - 2x^2) dx = 2 \int_0^1 (2 - 2x^2) dx\) (even function) = \(4 \int_0^1 (1 - x^2) dx = 4 [x - \frac{x^3}{3}]_0^1 = 4(1 - \frac{1}{3}) = \frac{8}{3}\).

Step 3: Final Answer:
\(\frac{8}{3}\) sq units, which corresponds to option (A).