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Let g(x) = cos(x²), f(x) = sqrt(x), and let alpha, beta (alpha < beta) be the roots of 18x² - 9pi*x + pi² = 0. Find the area (in sq. units) bounded by y = (g of f)(x), the lines x = alpha, x = beta and y = 0.

1. 1/2 (sqrt3 + 1)
2. 1/2 (sqrt3 - sqrt2)
3. 1/2 (sqrt2 - 1)
4. 1/2 (sqrt3 - 1)

Correct answer: 1/2 (sqrt3 - 1)

Solution

(g of f)(x) = cos x. Roots are pi/6 and pi/3, and integral of cos x over [pi/6, pi/3] = sin(pi/3) - sin(pi/6) = (sqrt3 - 1)/2.

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