Question

integrate $\underset{0}{\overset{2}{\int }}\left(\left|x\right|+\left|x-1\right|\right)dx=$

Answer 1

${\int }_0^2\left(\left|x\right|+\left|x-1\right|\right)dx$

Let, $f\left(x\right)=\left|x\right|+\left|x-1\right|$

$\Rightarrow f\left(x\right)=1$, when $0<x<1$

and $f\left(x\right)=2x-1$, when$1<x<2$

${\int }_0^{2}f\left(x\right)dx={\int }_0^{1}dx+{\int }_1^2\left(2x-1\right)dx$

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

$=1+2$

$=3$