Evaluate ∫[-π/4 to π/4] x² sin x dx.

Evaluate ∫[-π/4 to π/4] x² sin x dx.

• A. 1
• B. π/4
• C. 0
• D. π²/16

Answer: C) 0

Explanation: Let f(x) = x² sin x. Then f(-x) = (-x)² sin(-x) = -x² sin x = -f(x). Since f(x) is an odd function, its integral over the symmetric interval [-π/4, π/4] is 0.