# The integral ∫02 ||x − 1| − x | dx is equal to :

more_vert

The integral $\int\limits_0^2{||x-1|-x|dx}$ is equal to :

Numerical Value Type

more_vert

verified

Ans. (01.50)

Sol. $\int\limits_0^2{||x-1|-x|dx}=\int\limits_0^1{|1-x-x|dx}+\int\limits_1^2{|x-1-x|dx}$

$=\int\limits_0^{1/2}{(1-2x)dx}+\int\limits_{1/2}^1{(2x-1)dx}+\int\limits_1^2{dx}$

$=\Big[x-x^2\Big]_0^{1\over2}+\Big[x^2-x\Big]_{1\over2}^1+\big[x\big]_1^2$

$={1\over2}-{1\over4}+(1-1)-\left({1\over4}-{1\over2}\right)+2-1$

$={1\over4}+{1\over4}+1={3\over2}$

more_vert
Can we solve this by graph
more_vert
Yes you can