question_answer

Evaluate \[\int_{-1}^{2}{(|x+1|+|x|+|x-1|)}\,dx\]

Answer:

Let\[l=\int_{-\,1}^{2}{(|x+1|+|x|+|x-1|)\,dx}\] \[l=\int_{-\,1}^{0}{(|x+1|+|x|+|x-1|)\,dx}\] \[+\int_{0}^{1}{(|x+1|+|x|+|x-1|)\,dx}\] \[+\int_{1}^{2}{(|x+1|+|x|+|x-1|)\,dx}\] \[+\int_{1}^{2}{[(x+1)+(x)+(x-1)]\,dx}\]\[+\int_{1}^{2}{[(x+1)+(x)+(x-1)]\,dx}\] \[=\int_{-\,1}^{0}{(-x+2)\,dx+\int_{0}^{1}{(x+2)\,dx+\int_{1}^{2}{(3x)}}\,dx}\] \[=\left[ \frac{-\,\,{{x}^{2}}}{2}+2x \right]_{-\,1}^{0}+\left[ \frac{{{x}^{2}}}{2}+2x \right]_{0}^{1}+\left[ \frac{3{{x}^{2}}}{2} \right]_{1}^{2}\] \[g'(c)=\frac{3-1}{b-a}=\frac{2}{b-a}\]   \[=\frac{5}{2}+\frac{5}{2}+\frac{9}{2}=\frac{19}{2}\]