$\int_{- 3}^{2} {| x + 1 | + | x + 2 | + | x - 1 |} d x is$

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Detailed Solution

$\begin{array}{l} Let I = \int_{- 3}^{2} {| x + 1 | + | x + 2 | + | x - 1 |} dx \\ Breaking points are \\ x + 1 = 0 \Rightarrow x = - 1 \\ x + 2 = 0 \Rightarrow x = - 2 \\ x - 1 = 0 \Rightarrow x = 1 \\ ∴ I = \int_{- 3}^{- 2} f (x) dx + \int_{- 2}^{- 1} f (x) dx + \int_{- 1}^{1} f (x) dx + \int_{1}^{2} f (x) dx \\ where f (x) = | x + 1 | + | x + 2 | + | x - 1 | \end{array}$

$\begin{array}{l} I_{1} = \int_{- 3}^{- 2} [-(x+1)-(x+2)-(x-1)]dx \\ = - {[\frac{x^{2}}{2} + x + \frac{x^{2}}{2} + 2 x + \frac{x^{2}}{2} + x]}_{- 3}^{- 2} = \frac{7}{2} \\ I_{2} = \int_{- 2}^{- 1} [-(x+1)+(x+2)-(x-1)] d x \\ = \frac{- x^{2}}{2} - x + \frac{x^{2}}{2} + 2 x - \frac{x^{2}}{2} + {x|}_{- 2}^{- 1} \\ = \frac{- x^{2}}{2} + {2 x|}_{- 2}^{- 1} = (\frac{- 1}{2} - 2) - (- 2 - 4) \\ = - \frac{5}{2} + 6 = \frac{7}{2} \\ I_{3} = \int_{- 1}^{1} [(x + 1) + (x + 2) - (x - 1)] d x \\ = \int_{- 1}^{1} (x + 4) d x = \frac{x^{2}}{2} + {4 x|}_{- 1}^{1} = 8 \\ I_{4} = \int_{1}^{2} [(x+1)+(x+2)+(x-1) d x] \\ = \int_{1}^{2} (3 x + 2) d x = \frac{3 x^{2}}{2} + {2 x|}_{1}^{2} = \frac{13}{2} \\ ∴ I = I_{1} + I_{2} + I_{3} + I_{4} \\ = \frac{7}{2} + \frac{7}{2} + 8 + \frac{13}{2} = \frac{43}{2} \\ = 21.5 \end{array}$