Question

Evaluate the following integrals:
$\underset{0}{\overset{3}{\int }}\left|3x-1\right|dx$

Answer 1

$$\begin{gathered} {\int }_0^3\left|3x-1\right|\mathit{}\mathit{d}x \\ \mathrm{We}\mathrm{know}\mathrm{that},\left|3x-1\right|=\left\{\begin{array}{l}-\left(3x-1\right),0\le x\le \frac{1}{3}\\ \left(3x-1\right),\frac{1}{3}<x\le 3\end{array}\right. \\ \therefore I=={\int }_0^{\frac{1}{3}}-\left(3x+1\right)dx+{\int }_{\frac{1}{3}}^0\left(3x+1\right)dx \\ \Rightarrow I={\left[\frac{-3x^2}{2}-x\right]}_0^{\frac{1}{3}}+{\left[\frac{3x^2}{2}+x\right]}_{\frac{1}{3}}^3 \\ \Rightarrow I=-\frac{1}{6}+\frac{1}{3}-0+\frac{27}{2}+3-\frac{1}{6}-\frac{1}{3} \\ \Rightarrow I=\frac{65}{6} \end{gathered}$$