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Find the area of the curvilinear triangle in the first quadrant bounded by the y-axis, the curve y = tan x, and the curve y = (2/3)cos x.

1. (1/3) - (1/2)ln(4/3)
2. (1/3) - (1/2)ln(3/2)
3. (1/3) + (1/2)ln(4/3)
4. (1/2)ln(4/3)

Correct answer: (1/3) - (1/2)ln(4/3)

Solution

The curves meet at x = pi/6 (sin x = 1/2); integrating ((2/3)cos x - tan x) from 0 to pi/6 gives 1/3 - (1/2)ln(4/3).

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