What is the area of the region enclosed by the curve y = 4 - x^2 and the x-axis?
Answer and explanation
Correct answer: A. 32/3
- Step 1: Find x-intercepts: 4 - x^2 = 0 => x = -2 and x = 2. Step 2. Set up integral: int_{-2}^{2} (4 - x^2) dx = 2 int_{0}^{2} (4 - x^2) dx by symmetry. Step 3. Antiderivative: [4x - x^3 / 3]_{0}^{2} = 4(2) - 8/3 = 8 - 8/3 = 16/3. Step 4. Double for total area: 2 (16/3) = 32/3. Answer: 32/3. Why not the others: 16/3 computes only the right half (from 0 to 2); 8 assumes a triangular area; 28/3 makes an integration error.
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