∫−22 |[x]|dx is equal to - Infinity Learn by Sri Chaitanya

Solution:

$\begin{array}{l} \int_{- 2}^{2} | [x] | dx \\ = \int_{- 2}^{- 1} | [x] | dx + \int_{- 1}^{0} | [x] | dx + \int_{0}^{1} | [x] | dx + \int_{1}^{2} | [x] | dx \\ = \int_{- 2}^{- 1} 2 dx + \int_{- 1}^{0} 1 dx + \int_{0}^{1} 0 dx + \int_{1}^{2} 1 dx \\ = 2 [x]_{- 2}^{- 1} + [x]_{- 1}^{0} + 0 + [x]_{1}^{2} \\ = 2 (- 1 + 2) + (0 + 1) + (2 - 1) \\ = 2 + 1 + 1 = 4 \end{array}$

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