Question

Evaluate the integral: $\int_{-\pi}^{\pi} x^2 \sin(x) \, dx$

• A. \( \pi^2 \)
• B. \( \frac{\pi^2}{2} \)
• C. 0
• D. \( 2\pi^2 \)

Hint:

The integral of an odd function over a symmetric interval is always 0.

Answer 1

The Correct Option is C

The function \( x^2 \sin(x) \) is an odd function because \( x^2 \) is even and \( \sin(x) \) is odd. Therefore, the integral of an odd function over a symmetric interval such as \( [-\pi, \pi] \) is 0.