Question

$\int_{-2}^{2} |x^2 - x - 2| dx =$}

• A. $\frac{17}{3}$
• B. $\frac{19}{3}$
• C. $19$
• D. $17$

Hint:

Always split the integral at the roots of the quadratic equation inside the modulus.

Answer 1

The Correct Option is B

Step 1: Concept
To integrate an absolute value function, identify where the expression inside the modulus changes sign by finding its roots.

Step 2: Meaning
$x^2 - x - 2 = (x - 2)(x + 1) = 0 \implies x = -1, 2$. Between $-2$ and $2$, the sign changes at $x = -1$.

Step 3: Analysis
1. From $-2$ to $-1$: $x^2 - x - 2$ is positive. 2. From $-1$ to $2$: $x^2 - x - 2$ is negative. $\int_{-2}^{-1} (x^2 - x - 2) dx + \int_{-1}^{2} -(x^2 - x - 2) dx$. $= [\frac{x^3}{3} - \frac{x^2}{2} - 2x]_{-2}^{-1} - [\frac{x^3}{3} - \frac{x^2}{2} - 2x]_{-1}^{2}$. Calculating values: $(\frac{7}{6} - \frac{2}{3}) - (-\frac{10}{3} - \frac{7}{6}) = \frac{19}{3}$.

Step 4: Conclusion
The value of the integral is $\frac{19}{3}$. Final Answer: (B)