Question

The integral $\underset{0}{\overset{2}{\int }}||x-1|-x|dx$ is equal to

Answer 1

$\underset{0}{\overset{2}{\int }}||x-1|-x|dx$
$=\underset{0}{\overset{1}{\int }}|1-x-x|dx+{\int }_1^2|x-1-x|dx$
$=\underset{0}{\overset{1}{\int }}|2x-1|dx+{\int }_1^{2}1dx$
$=\underset{0}{\overset{\frac{1}{2}}{\int }}(1-2x)dx+{\int }_{\frac{1}{2}}^1(2x-1)dx+{\int }_1^{2}1dx$
$=[(\frac{1}{2}-0)-(\frac{1}{4}-0)]+(1-\frac{1}{4})-(1-\frac{1}{2})+1$
$=\frac{1}{2}-\frac{1}{4}+\frac{3}{4}-\frac{1}{2}+1=\frac{3}{2}$