Question

Evaluate the following integrals:
$\underset{0}{\overset{4}{\int }}\left(\left|x\right|+\left|x-2\right|+\left|x-4\right|\right)dx$

Answer 1

$$\begin{gathered} I={\int }_0^4\left\{\left|x\right|+\left|x-2\right|+\left|x-4\right|\right\}dx \\ \Rightarrow I={\int }_0^4\left|x\right|\mathit{}\mathit{d}x+{\int }_0^4\left|x-2\right|\mathit{}\mathit{d}x+{\int }_0^4\left|x-4\right|\mathit{}\mathit{d}x \\ \mathrm{We}\mathrm{know}\mathrm{that},\left|x\right|=\left\{\begin{array}{l}-x,-5\le x\le 0\\ x,x>0\end{array}\right. \\ \left|x-2\right|=\left\{\begin{array}{l}-\left(x-2\right),0\le x\le 2\\ x-2,2<x\le 4\end{array}\right. \\ \left|x-4\right|=\left\{\begin{array}{l}-\left(x-4\right),0\le x\le 4\\ x-4,x>4\end{array}\right. \\ \therefore I={\int }_0^{4}x\mathit{}\mathit{d}x-{\int }_0^2\left(x-2\right)\mathit{}\mathit{d}x+{\int }_2^4\left(x-2\right)\mathit{}\mathit{d}x-{\int }_0^4\left(x-4\right)\mathit{}\mathit{d}x \\ \Rightarrow I={\left[\frac{x^2}{2}\right]}_0^4-{\left[\frac{x^2}{2}-2x\right]}_0^2+{\left[\frac{x^2}{2}-2x\right]}_2^4-{\left[\frac{x^2}{2}-4x\right]}_0^4 \\ \Rightarrow I=8-\left(2-4\right)+8-8-2+4-\left(8-16\right) \\ \Rightarrow I=20 \end{gathered}$$