Question

Evaluate the following integrals:
$\underset{-2}{\overset{2}{\int }}\left|2x+3\right|dx$

Answer 1

$$\begin{gathered} {\int }_{-2}^2\left|2x+3\right|\mathit{d}x \\ \mathrm{We}\mathrm{know}\mathrm{that},\left|2x+3\right|=\left\{\begin{array}{l}-\left(2x+3\right),-2\le x\le -\frac{3}{2}\\ \left(2x+3\right),-\frac{3}{2}<x\le 2\end{array}\right. \\ \therefore I={\int }_{-2}^{\frac{-3}{2}}-\left(2x+3\right)\mathit{d}x+{\int }_{-\frac{3}{2}}^2\left(2x+3\right)\mathit{d}x \\ \Rightarrow I=-{\left[x^2+3x\right]}_{-2}^{\frac{-3}{2}}+{\left[x^2+3x\right]}_{-\frac{3}{2}}^2 \\ \Rightarrow I=-\frac{9}{4}+\frac{9}{2}+4-6+4+6-\frac{9}{4}+\frac{9}{2} \\ \Rightarrow I=\frac{25}{2} \end{gathered}$$