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Let lambda be the area (in square units) of the region {(x, y) in R²: 4x² <= y <= 8x + 12}. Find the value of 3*lambda.
- 64
- 96
- 128
- 256
Correct answer: 128
Solution
The line and parabola meet at x = -1 and x = 3; integrating their difference gives lambda = 128/3, so 3*lambda = 128.
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