Question

The value of \( \int_{-2}^{2} x |x| \,dx \) is:

• A. \( \frac{1}{8} \)
• B. \( \frac{1}{4} \)
• C. \( -\frac{1}{4} \)
• D. \( -\frac{1}{8} \)
• E. \( 0 \)

Hint:

For absolute value integrals, break the integral at the points where the function inside changes behavior.

Answer 1

Answer: (E)

Step 1: Break the integral into two parts The function \( x|x| \) behaves differently for positive and negative values: 
- For \( x \geq 0 \), \( |x| = x \), so \( x|x| = x^2 \). 
- For \( x<0 \), \( |x| = -x \), so \( x|x| = -x^2 \). 
Thus, we split the integral at \( x = 0 \): \[ \int_{-2}^{2} x |x| \,dx = \int_{-2}^{0} (-x^2) \,dx + \int_{0}^{2} x^2 \,dx. \] Step 2: Evaluate the integrals Evaluating the first integral: \[ \int_{-2}^{0} -x^2 \,dx = - \left[ \frac{x^3}{3} \right]_{-2}^{0} \] \[ = - \left[ \frac{0^3}{3} - \frac{(-2)^3}{3} \right] \] \[ = - \left[ 0 - \left(-\frac{8}{3} \right) \right] = -\left(\frac{8}{3}\right) = -\frac{8}{3}. \] Evaluating the second integral: \[ \int_{0}^{2} x^2 \,dx = \left[ \frac{x^3}{3} \right]_{0}^{2} \] \[ = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}. \] Step 3: Compute the final result Summing both integrals: \[ -\frac{8}{3} + \frac{8}{3} = 0. \] Thus, the integral evaluates to \( 0 \), confirming option (E).