The integral $\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx}$
is equal to______.

$\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx}$ <br><br>= $\int\limits_0^1 {\left| { - \left( {x - 1} \right) - x} \right|} dx$ + $+ \int\limits_1^2 {\left| {\left( {x - 1} \right) - x} \right|} dx$ <br><br>= $\int\limits_0^1 {\left| {1 - x - x} \right|} dx + \int\limits_1^2 {dx}$ <br><br>= $\int\limits_0^1 {\left| {1 - 2x} \right|} dx + \int\limits_1^2 {dx}$ <br><br>= $$\int\limits_0^{{1 \over 2}} {\left( {1 - 2x} \right)} dx + \int\limits_{{1 \over 2}}^1 {\left( {2x - 1} \right)dx} + \int\limits_1^2 {dx} $$ <br><br>= $$\left[ {x - {x^2}} \right]_0^{{1 \over 2}} + \left[ {x - {x^2}} \right]_{{1 \over 2}}^1 + \left( {2 - 1} \right)$$ <br><br>= $$\left( {{1 \over 2} - {1 \over 4}} \right) + \left[ {\left( {1 - 1} \right) - \left( {{1 \over 4} - {1 \over 2}} \right)} \right] + 1$$ <br><br>= ${3 \over 2}$ = 1.5