Question
$\int_{\, - \,2}^{\,2} {\,\left| {\,[x]\,} \right|\,dx = } $

Answer

d
(d) $\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} {2dx\,\,} + \int_{ - 1}^0 {1dx + \int_0^1 {0\,dx + } } \int_1^2 {1dx} $

$ = 2[x]_{ - 2}^{ - 1} + [x]_{ - 1}^0 + 0 + [x]_1^2$

$ = 2( - 1 + 2) + (0 + 1) + (2 - 1) = 2 + 1 + 1 = 4.$

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