A student calculates ∫(0 to 2π) sin(x) dx = 0 and concludes the area bounded by y = sin(x) and the x-axis from 0 to 2π is 0. What is the correct area?
A student calculates ∫(0 to 2π) sin(x) dx = 0 and concludes the area bounded by y = sin(x) and the x-axis from 0 to 2π is 0. What is the correct area?
- A. 4
- B. 0
- C. 2
- D. 2π
Answer: A) 4
Explanation: Definite integral gives net signed area. True geometric area = ∫(0 to π) sin(x) dx + |∫(π to 2π) sin(x) dx| = 2 + |−2| = 4 sq units.
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