Find the area of the region enclosed by the parabola y = 4 - x^2 and the x-axis (y = 0).
Answer and explanation
Correct answer: B. 32/3
- Step 1: Find the x-intercepts by setting y = 0: 4 - x^2 = 0 => x = -2 and x = 2.
- Step 2: Since 4 - x^2 >= 0 on [-2, 2], the area is: A = integral from -2 to 2 of (4 - x^2) dx.
- Step 3: By symmetry, A = 2 integral from 0 to 2 of (4 - x^2) dx = 2 [4x - x^3/3] from 0 to 2 = 2 [8 - 8/3] = 2 (16/3) = 32/3.
Answer
32/3.
Why not the others: '16/3' computes the area over only half the interval [0, 2]; '8' integrates incorrectly as 4(2) - 0; '64/3' doubles the complete area twice.
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