A) \[\frac{31}{2}\]
B) \[\frac{35}{2}\]
C) \[\frac{37}{2}\]
D) \[\frac{39}{2}\]
Correct Answer: D
Solution :
Option is not correct Let \[I=\int_{-3}^{2}{(|x+1|+|x+2|+|x-1|)dx}\] Again let \[f(x)=|x+1|+|x+2|+|x-1|\] \[=\left\{ \begin{matrix} -(x+1)-(x+2)-(x-1), & -3<x\le -2 \\ -(x+1)+x+2-(x-1), & -2<x\le -1 \\ 1+x+x+2-(x-1) & -1<x\le 0 \\ 1+x+x+2-(x-1) & 0\le x<1 \\ 1+x+x+2+x-1 & 1\le x<2 \\ \end{matrix} \right.\] \[=\left\{ \begin{matrix} -3x-2, & -3<x\le -2 \\ -x+2, & -2<x\le -1 \\ x+4, & -1\le x<1 \\ 3x+2, & 1\le x<2 \\ \end{matrix} \right.\] \[\therefore \]\[I=\int_{-3}^{-2}{(-3x-2)}dx+\int_{-2}^{-1}{(-x+2)}dx\] \[+\int_{-1}^{1}{(x+4)dx+\int_{1}^{2}{(3x+2)dx}}\] \[=\left[ -\frac{3{{x}^{2}}}{2}-2x \right]_{-3}^{-2}+\left[ -\frac{{{x}^{2}}}{2}+2x \right]_{-2}^{-1}\] \[+\left[ \frac{{{x}^{2}}}{2}+4x \right]_{-1}^{1}+\left[ \frac{3{{x}^{2}}}{2}+2x \right]_{1}^{2}\] \[=\left[ -6+4-\left( -\frac{27}{2} \right)+6 \right]+\left[ -\frac{1}{2}-2-(-2-4) \right]\] \[+\left[ \frac{1}{2}+4-\left( \frac{1}{2}-4 \right) \right]+\left[ 6+4-\,\left( \frac{3}{2}+2 \right) \right]\] \[=\frac{11}{2}+\frac{7}{2}+8+\frac{13}{2}\] \[=\frac{31}{2}+8=\frac{47}{2}\]
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