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Find the area (in square units) enclosed by the parabola y = x² - 1, the tangent to it at the point (2, 3), and the y-axis.
- 14/3
- 56/3
- 8/3
- 32/3
Correct answer: 8/3
Solution
Between x=0 and x=2 the parabola lies above the tangent; area = integral₀² [(x² - 1) - (4x - 5)] dx = 8/3.
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