Question

Find the value of ${\int }_0^2|1-x|dx$.

Answer 1

Since, $1-x\ge 0$ in $[0,1]$ and $1-x\le 0$ in $[1,2]$

$\Rightarrow {\int }_0^2|1-x|dx$

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

$={\int }_0^{1}xdx-{\int }_0^{1}1dx+{\int }_1^{2}1dx-{\int }_1^{2}xdx$

$={\left[\frac{x^2}{2}\right]}_0^1-{\left[x\right]}_0^1+{\left[x\right]}_1^2-{\left[\frac{x^2}{2}\right]}_1^2$ [$\because \int x^{n}dx=\frac{x^{n+1}}{n+1},n\ne -1$]

$=[\frac{1^2}{2}-0]-[1-0]+[2-1]-[\frac{2^2}{2}-\frac{1^2}{2}]$ [$\because {\left[f\right(x\left)\right]}_a^b=f\left(b\right)-f\left(a\right)$]

$=\frac{1}{2}-1+1-2+\frac{1}{2}$

$=1-2$

$=-1$

$\therefore {\int }_0^2|1-x|dx=-1$