The value of int_{-2}^{2}left|3 x^{2}-3 x-6ri | vedclass

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(A) We need to evaluate the integral $I = \int_{-2}^{2} |3x^2 - 3x - 6| dx$.First,factor the expression inside the absolute value: $3x^2 - 3x - 6 = 3(

Vedclass · Accepted Answer

$19$

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(A) We need to evaluate the integral $I = \int_{-2}^{2} |3x^2 - 3x - 6| dx$.First,factor the expression inside the absolute value: $3x^2 - 3x - 6 = 3(x^2 - x - 2) = 3(x - 2)(x + 1)$.The roots of the quadratic are $x = -1$ and $x = 2$.In the interval $[-2, -1]$,$3(x^2 - x - 2) \ge 0$.In the interval $[-1, 2]$,$3(x^2 - x - 2) \le 0$.Thus,$I = 3 \left[ \int_{-2}^{-1} (x^2 - x - 2) dx + \int_{-1}^{2} -(x^2 - x - 2) dx ight]$.Evaluating the first integral: $\int_{-2}^{-1} (x^2 - x - 2) dx = [\frac{x^3}{3} - \frac{x^2}{2} - 2x]_{-2}^{-1} = (-\frac{1}{3} - \frac{1}{2} + 2) - (-\frac{8}{3} - 2 + 4) = \frac{7}{6} - (-\frac{2}{3}) = \frac{7}{6} + \frac{4}{6} = \frac{11}{6}$.Evaluating the second integral: $-\int_{-1}^{2} (x^2 - x - 2) dx = -[\frac{x^3}{3} - \frac{x^2}{2} - 2x]_{-1}^{2} = -[(\frac{8}{3} - 2 - 4) - (-\frac{1}{3} - \frac{1}{2} + 2)] = -[-\frac{10}{3} - \frac{7}{6}] = -[-\frac{27}{6}] = \frac{9}{2} = \frac{27}{6}$.Total value $I = 3 	imes (\frac{11}{6} + \frac{27}{6}) = 3 	imes \frac{38}{6} = 3 	imes \frac{19}{3} = 19$.

The value of $\int_{-2}^{2}\left|3 x^{2}-3 x-6\right| d x$ is ...... .

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