Question

Evaluate the following integrals:
$\underset{-5}{\overset{0}{\int }}f\left(x\right)dx,\mathrm{where}f\left(x\right)=\left|x\right|+\left|x+2\right|+\left|x+5\right|$

Answer 1

$$\begin{gathered} I={\int }_{-5}^0\left\{\left|x\right|+\left|x+2\right|+\left|x+5\right|\right\}dx \\ \Rightarrow I={\int }_{-5}^0\left|x\right|\mathit{}\mathit{d}x+{\int }_{-5}^0\left|x+2\right|\mathit{}\mathit{d}x+{\int }_{-5}^0\left|x+5\right|\mathit{}\mathit{d}x \\ \mathrm{We}\mathrm{know}\mathrm{that},\left|x\right|=\left\{\begin{array}{l}-x,-5\le x\le 0\\ x,x>0\end{array}\right. \\ \left|x+2\right|=\left\{\begin{array}{l}-\left(x+2\right),-5\le x\le -2\\ x+2,-2<x\le 0\end{array}\right. \\ \left|x+5\right|=\left\{\begin{array}{l}-\left(x+5\right),-5\le x\le 0\\ x+5,x>-5\end{array}\right. \\ \therefore I=-{\int }_{-5}^{0}x\mathit{}\mathit{d}x-{\int }_{-5}^{-2}\left(x+2\right)\mathit{}\mathit{d}x+{\int }_{-2}^0\left(x+2\right)\mathit{}\mathit{d}x+{\int }_{-5}^0\left(x+5\right)\mathit{}\mathit{d}x \\ \Rightarrow I=-{\left[\frac{x^2}{2}\right]}_{-5}^0-{\left[\frac{x^2}{2}+2x\right]}_{-5}^{-2}+{\left[\frac{x^2}{2}+2x\right]}_{-2}^0+{\left[\frac{x^2}{2}+5x\right]}_{-5}^0 \\ \Rightarrow I=\frac{25}{2}-\left(2-4-\frac{25}{2}+10\right)-2+4+\left(-\frac{25}{2}+25\right) \\ \Rightarrow I=\frac{63}{2} \end{gathered}$$