Question

The value of \( \int_{-1}^{1} x^2 \sin x \, dx \) is equal to:

• A. \( 2\sin 1 \)
• B. \( 2 \)
• C. \( 4 \)
• D. \( -2\sin 1 \)
• E. \( 0 \)

Hint:

If an integrand is an odd function over a symmetric interval, the integral evaluates to zero without computation.

Answer 1

Answer: (E)

We evaluate the integral: \[ I = \int_{-1}^{1} x^2 \sin x \, dx. \] Step 1: Checking Function Symmetry The given function is: \[ f(x) = x^2 \sin x. \] - \( x^2 \) is an even function because \( x^2 = (-x)^2 \). - \( \sin x \) is an odd function because \( \sin(-x) = -\sin x \). 
- The product of an even and an odd function is an odd function: \[ f(-x) = (-x)^2 \sin(-x) = x^2 (-\sin x) = -f(x). \] 
Step 2: Evaluating the Integral Since \( f(x) \) is an odd function and the integration limits are symmetric about zero \([-a, a]\), we apply the property: \[ \int_{-a}^{a} {odd function} \, dx = 0. \] Thus, \[ I = 0. \]