Ta có \({x^2} - x = 0 \Leftrightarrow x = 0\) hoặc \(x = 1\).
Như vậy,
\(\int\limits_0^2 {\left| {{x^2} - x} \right|dx} = \int\limits_0^1 {\left| {{x^2} - x} \right|dx} + \int\limits_1^2 {\left| {{x^2} - x} \right|dx} = \left| {\int\limits_0^1 {\left( {{x^2} - x} \right)dx} } \right| + \left| {\int\limits_1^2 {\left( {{x^2} - x} \right)dx} } \right|\)
\( = \left| {\left. {\left( {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right)} \right|_0^1} \right| + \left| {\left. {\left( {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right)} \right|_1^2} \right| = \left| {\frac{{ - 1}}{6} - 0} \right| + \left| {\frac{2}{3} - \left( { - \frac{1}{6}} \right)} \right| = 1\)
Vậy đáp án đúng là B.