I=∫−23|x−2|dx=∫−22|x−2|dx+∫23|x−2|dx(make x-2=0⇒x=2 now split the integral into two parts at 2)I=∫−22(x−2)dx+∫23(x−2)dx∵−2 0⇒|x−2|=(x−2)I=2x−x22−22+x22−2x23=(4−2)−(−4−2)+(92−6)−(2−4)=172

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$\int_{- 2}^{3} | x - 2 | d x =$

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$\frac{5}{2}$

$\frac{7}{2}$

$\frac{17}{2}$

answer is D.

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Detailed Solution

$I = \int_{- 2}^{3} | x - 2 | d x$

$= \int_{- 2}^{2} | x - 2 | d x + \int_{2}^{3} | x - 2 | d x$

(make $(x - 2) = 0 \Rightarrow x = 2$ now split the integral into two parts at 2)

$I = \int_{- 2}^{2} (x - 2) d x + \int_{2}^{3} (x - 2) d x$

$(\begin{array}{l} ∵ - 2 < x < 2 \\ \Rightarrow (x - 2) < 0 \\ \Rightarrow | x - 2 | = - (x - 2) \end{array}) (\begin{array}{l} ∵ 2 < x < 3 \\ \Rightarrow (x - 2) > 0 \\ \Rightarrow | x - 2 | = (x - 2) \end{array})$