Question

Evaluate the following integrals:
$\underset{1}{\overset{2}{\int }}\left|x-3\right|dx$

Answer 1

$$\begin{gathered} {\int }_1^2\left|x-3\right|\mathit{d}x \\ \mathrm{We}\mathrm{know}\mathrm{that},\left|x+1\right|=\left\{\begin{array}{l}-\left(x+1\right),1\le x\le 3\\ \left(x+1\right),x>3\end{array}\right. \\ \therefore I={\int }_1^2\left|x-3\right|\mathit{d}x \\ \Rightarrow I={\int }_1^2-\left(x-3\right)dx \\ \Rightarrow I={\left[\frac{-x^2}{2}-3x\right]}_1^2 \\ \Rightarrow I=-2-6+\frac{1}{2}+3 \\ \Rightarrow I=\frac{3}{2} \end{gathered}$$