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The value of $\int_{- 2}^{2} | 1 - x | d x$ is

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Correct option is D

|1−x|=1−x,x≤1x−1,x>1 so ∫−22 |1−x|dx=∫−21 |1−x|dx+∫12 |1−x|dx =∫−21 (1−x)dx+∫12 (x−1)dx =5.

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The value of ∫−22 |1−x|dx is

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The value of $\int_{- 2}^{2} | 1 - x | d x$ is