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We note that s i n x has a period or cycle of 2 π . As an odd function, its area under the curve over one cycle is 0 and half a cycle has positive and the other half has equal area but negative. The absolute value of area of half a cycle is ∫ 0 π sin x d x = − cos x ∣ ∣ ∣ ∣ 0 π = 2 . Then we have I s ( n ) = ∫ 0 n π sin x d x = { 2 0 if n is odd if n is even ∀ n ∈ N
I 1 = ∫ − 2 0 1 7 π 2 0 1 8 π sin x d x = ∫ − 2 0 1 7 π 0 sin x d x + ∫ 0 2 0 1 8 π sin x d x = − ∫ 0 2 0 1 7 π sin x d x + I s ( 2 0 1 8 ) = − I s ( 2 0 1 7 ) + 0 = − 2 Since sin x is odd
I 2 = ∫ − 2 0 1 7 π 2 0 1 8 π ∣ sin x ∣ d x = ( 2 0 1 8 − ( − 2 0 1 7 ) ) I s ( 1 ) = 8 0 7 0 All half-cycles are positive.
I 3 = ∫ − 2 0 1 7 π 2 0 1 8 π sin ∣ x ∣ d x = ∫ − 2 0 1 7 π 0 sin ∣ x ∣ d x + ∫ 0 2 0 1 8 π sin ∣ x ∣ d x = ∫ 0 2 0 1 7 π sin x d x + ∫ 0 2 0 1 8 π sin x d x = I s ( 2 0 1 7 ) + I s ( 2 0 1 8 ) = 2 Since sin ∣ x ∣ is even
Therefore, I = I 1 + I 2 + I 3 = − 2 + 8 0 7 0 + 2 = 8 0 7 0 .