Question

Evaluate the definite integral: \( \int_{-2}^{2} |x^2 - x - 2| \, dx \)

• A. \( \frac{40}{3} \)
• B. \( \frac{28}{3} \)
• C. \( \frac{36}{5} \)
• D. \( \frac{44}{3} \)

Hint:

For definite integrals involving modulus functions, always split the integral at the points where the expression inside the modulus changes sign.

Answer 1

The Correct Option is A

Analyze the expression within the modulus: \[ f(x) = x^2 - x - 2 = (x - 2)(x + 1) \] The sign of \( f(x) \) changes at \( x = -1 \) and \( x = 2 \). For the interval \([-2, 2]\), we divide it into regions: 1. For \( x \in [-2, -1] \), \( f(x) \ge 0 \). \ 2. For \( x \in [-1, 2] \), \( f(x) \le 0 \). Thus: \[ \int_{-2}^{2} |x^2 - x - 2| dx = \int_{-2}^{-1} (x^2 - x - 2) dx + \int_{-1}^{2} -(x^2 - x - 2) dx \] Compute each integral: \[ \int_{-2}^{-1} (x^2 - x - 2) dx = \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{-2}^{-1} = \left( \frac{-1}{3} - \frac{1}{2} + 2 \right) - \left( \frac{-8}{3} - 2 - (-4) \right) = \frac{13}{6} \] \[ \int_{-1}^{2} -(x^2 - x - 2) dx = \int_{-1}^{2} (-x^2 + x + 2) dx = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^{2} = \frac{27}{6} \] Sum the results: \[ \frac{13}{6} + \frac{27}{6} = \frac{40}{6} = \frac{20}{3} \]