Evaluate ∫[0 to π] x sin x dx.
Evaluate ∫[0 to π] x sin x dx.
- A. π/2
- B. 0
- C. 2π
- D. π
Answer: D) π
Explanation: Using parts: let u = x, dv = sin x dx. Then du = dx, v = -cos x. Integral is [-x cos x] + ∫cos x dx = [-x cos x + sin x]. Evaluating from 0 to π gives (-π(-1) + 0) - (0 + 0) = π.
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