Câu hỏi 4

\(\begin{array}{l}I = \int\limits_0^2 {\left| {1 - x} \right|dx}  = \int\limits_0^1 {\left| {1 - x} \right|dx}  + \int\limits_1^2 {\left| {1 - x} \right|dx} \\\,\,\,\, = \int\limits_0^1 {\left( {1 - x} \right)dx}  - \int\limits_1^2 {\left( {1 - x} \right)dx} \\\,\,\,\, = \left. {\left( {x - \dfrac{{{x^2}}}{2}} \right)} \right|_0^1 - \left. {\left( {x - \dfrac{{{x^2}}}{2}} \right)} \right|_1^2\\\,\,\,\, = \dfrac{1}{2} - \left( {0 - \dfrac{1}{2}} \right) = 1\end{array}\).

Chọn B.