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Define aₙ = ∫₀^(π/2) l(1-sinⁿ t r)sin 2t dt. Then the value of lim_(n→∞) (1)/(n)∑_(i=1)ⁿ a_i is

1. 1/2
2. 1
3. 4/3
4. 3/2

Correct answer: 1

Solution

a_n = int_0^(pi/2)(1-sin^n t)sin2t dt = 1 - 2/(n+2) using int sin^n t sin2t dt = 2/(n+2). Since a_n -> 1, the average (1/n)sum a_i also -> 1.

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