QuestionReportQuestion: \(\int _ { - 1 } ^ { 1 } | 1 - x | d x =\)...∫−11∣1−x∣dx=\int _ { - 1 } ^ { 1 } | 1 - x | d x =∫−11∣1−x∣dx=A– 2B0C2D4Answer2ExplanationSolution∫−11∣1−x∣dx=∫−11(1−x)dx=[x−x22]−11=2\int _ { - 1 } ^ { 1 } | 1 - x | d x = \int _ { - 1 } ^ { 1 } ( 1 - x ) d x = \left[ x - \frac { x ^ { 2 } } { 2 } \right] _ { - 1 } ^ { 1 } = 2∫−11∣1−x∣dx=∫−11(1−x)dx=[x−2x2]−11=2.
