Question

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

Answer 1

$$\begin{gathered} {\int }_{-1}^1\left|2x+1\right|\mathit{}\mathit{d}x \\ \mathrm{We}\mathrm{know}\mathrm{that},\left|2x+1\right|=\left\{\begin{array}{l}-\left(2x+1\right),-1\le x\le -\frac{1}{2}\\ \left(2x+1\right),-\frac{1}{2}<x\le 1\end{array}\right. \\ \therefore I={\int }_{-1}^{\frac{-1}{2}}-\left(2x+1\right)\mathit{d}x+{\int }_{-\frac{1}{2}}^1\left(2x+1\right)\mathit{}\mathit{d}x \\ \Rightarrow I=-{\left[x^2+x\right]}_{-1}^{\frac{-1}{2}}+{\left[x^2+x\right]}_{-\frac{1}{2}}^1 \\ \Rightarrow I=-\frac{1}{4}+\frac{1}{2}+1-1+1+1-\frac{1}{4}+\frac{1}{2} \\ \Rightarrow I=\frac{5}{2} \end{gathered}$$