Question

Evaluate the following integrals:
$\underset{1}{\overset{4}{\int }}\left\{\left|x-1\right|+\left|x-2\right|+\left|x-4\right|\right\}dx$

Answer 1

$$\begin{gathered} I={\int }_1^4\left\{\left|x-1\right|+\left|x-2\right|+\left|x-4\right|\right\}\mathit{}\mathit{d}x \\ \Rightarrow I={\int }_1^4\left|x-1\right|\mathit{}\mathit{d}x+{\int }_1^4\left|x-2\right|\mathit{}\mathit{d}x+{\int }_1^4\left|x-4\right|\mathit{}\mathit{d}x \\ \mathrm{We}\mathrm{know}\mathrm{that},\left|x-1\right|=\left\{\begin{array}{l}-\left(x-1\right),x\le 1\\ x-1,1<x\le 4\end{array}\right. \\ \left|x-2\right|=\left\{\begin{array}{l}-\left(x-2\right),1\le x\le 2\\ x-2,2<x\le 4\end{array}\right. \\ \left|x-4\right|=\left\{\begin{array}{l}-\left(x-4\right),1\le x\le 4\\ x-4,x>4\end{array}\right. \\ \therefore I={\int }_1^4\left(x-1\right)\mathit{}\mathit{d}x-{\int }_1^2\left(x-2\right)\mathit{}\mathit{d}x+{\int }_2^4\left(x-2\right)\mathit{}\mathit{d}x-{\int }_1^4\left(x-4\right)\mathit{}\mathit{d}x \\ \Rightarrow I={\left[\frac{x^2}{2}-x\right]}_1^4-{\left[\frac{x^2}{2}-2x\right]}_1^2+{\left[\frac{x^2}{2}-2x\right]}_2^4-{\left[\frac{x^2}{2}-4x\right]}_1^4 \\ \Rightarrow I=8-4-\frac{1}{2}+1-\left(2-4-\frac{1}{2}+2\right)+8-8-2+4-\left(8-16-\frac{1}{2}+4\right) \\ \Rightarrow I=\frac{23}{2} \end{gathered}$$