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Let f, g: R -> R with f(x) = x² + 5/12 and g(x) = 2(1 - 4|x|/3) for |x| <= 3/4, and g(x) = 0 for |x| > 3/4. If alpha is the area of the region {(x,y): |x| <= 3/4, 0 <= y <= min(f(x), g(x))}, find the value of 9*alpha.
- 5
- 6
- 4
- 7
Correct answer: 6
Solution
The two graphs cross at x = 1/4; min is the parabola on [0,1/4] and the line on [1/4,3/4]. Integrating and doubling gives alpha = 2/3, so 9*alpha = 6.
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