∫₀^π sin x dx =
A2 (= [-cos x]₀^π = 1 + 1)
Bπ
C0
D-2
Answer & Solution
Correct answer: A. 2 (= [-cos x]₀^π = 1 + 1)
∫₀^π sin x dx = -cos x | from 0 to π = -cos π - (-cos 0) = -(-1) - (-1) = 1 + 1 = 2. Geometric meaning: area under sin curve from 0 to π = 2.
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