∫03 x2−3x+2dx =∫01 x2−3x+2dx+∫12 −x2−3x+2dx+∫23 x2−3x+2dx=x33−3x22+2x01−x33−3x22+2x12+x33−3x22+2x23=13−32+2−83−6+4+13−32+2 +9−272+6−83−6+4=56−23+56+32−23=5−4+5+9−46=116

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$\int_{0}^{3} |x^{2} - 3 x + 2| d x =$

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3/2

1/6

11/6

11/2

answer is C.

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Detailed Solution

$\begin{array}{l} \int_{0}^{3} |x^{2} - 3 x + 2| d x = \int_{0}^{1} (x^{2} - 3 x + 2) d x + \int_{1}^{2} - (x^{2} - 3 x + 2) d x + \int_{2}^{3} (x^{2} - 3 x + 2) d x \\ = {[\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x]}_{0}^{1} - {[\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x]}_{1}^{2} + {[\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x]}_{2}^{3} \\ = (\frac{1}{3} - \frac{3}{2} + 2) - (\frac{8}{3} - 6 + 4) + (\frac{1}{3} - \frac{3}{2} + 2) + (9 - \frac{27}{2} + 6) - (\frac{8}{3} - 6 + 4) \\ = \frac{5}{6} - \frac{2}{3} + \frac{5}{6} + \frac{3}{2} - \frac{2}{3} = \frac{5 - 4 + 5 + 9 - 4}{6} = \frac{11}{6} \end{array}$