Question

Integrate:
${\int }_0^1|3x-1|dx$ ?

Answer 1

$|3x-1|=$ $\{\begin{array}{c}ifx<\frac{1}{3}-(3x-1)\\ ifx>\frac{1}{3}(3x-1)\end{array}$

$={\int }_0^{1/3}(1-3x).dx+{\int }_{1/3}^1(3x-1).dx.$

$={\int }_0^{1/3}dx-{\int }_0^{1/3}3x.dx+{\int }_{1/3}^{1}3x.dx-{\int }_{1/3}^{1}dx$

$={\left[x\right]}_0^{1/3}-\frac{3}{2}{\left[x^2\right]}_0^{1/3}+\frac{3}{2}{\left[x^2\right]}_1^{1/3}-{\left[x\right]}_{1/3}^1$

$=\frac{1}{3}-\frac{3}{2}\times {\left(\frac{1}{3}\right)}^2+\frac{3}{2}[1-{\left(\frac{1}{3}\right)}^2]-[1-\frac{1}{3}]$

$=\frac{1}{3}-\frac{1}{6}+\frac{4}{3}-\frac{2}{3}=\frac{5}{6}$