int_{-2}^2 |x^2-x-2| dx = | vedclass

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(B) First,we factor the quadratic expression inside the absolute value: $x^2 - x - 2 = (x - 2)(x + 1)$.The roots are $x = -1$ and $x = 2$.We examine t

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$\frac{19}{3}$

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(B) First,we factor the quadratic expression inside the absolute value: $x^2 - x - 2 = (x - 2)(x + 1)$.The roots are $x = -1$ and $x = 2$.We examine the sign of $f(x) = x^2 - x - 2$ on the interval $[-2, 2]$:For $x \in [-2, -1]$,$f(x) \ge 0$.For $x \in [-1, 2]$,$f(x) \le 0$,so $|x^2 - x - 2| = -(x^2 - x - 2) = -x^2 + x + 2$.Thus,the integral is $\int_{-2}^{-1} (x^2 - x - 2) dx + \int_{-1}^2 (-x^2 + x + 2) dx$.Evaluating the first part: $[\frac{x^3}{3} - \frac{x^2}{2} - 2x]_{-2}^{-1} = (-\frac{1}{3} - \frac{1}{2} + 2) - (-\frac{8}{3} - 2 + 4) = \frac{7}{6} - (-\frac{2}{3}) = \frac{7}{6} + \frac{4}{6} = \frac{11}{6}$.Evaluating the second part: $[-\frac{x^3}{3} + \frac{x^2}{2} + 2x]_{-1}^2 = (-\frac{8}{3} + 2 + 4) - (\frac{1}{3} + \frac{1}{2} - 2) = \frac{10}{3} - (-\frac{7}{6}) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$.Total sum: $\frac{11}{6} + \frac{27}{6} = \frac{38}{6} = \frac{19}{3}$.

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

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