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Question
Evaluate: `int_-1^2 (|"x"|)/"x"d"x"`.
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Solution
`|"x"| = "x" "when" "x" ≥0`
= `-"x" "when" "x" < 0`
Therefore, `|"x"|/"x"` = 1 when x ≥ 0
= -1 when x < 0
Thus, `int_-1^2 |"x"|/"x"d"x" = int_-1^0 (-1)d"x" + int_0^2 (1)d"x"`
= `-1 xx ["x"]_1^0 + ["x"]_0^2`
= `(-1) [0 + 1] + [2 - 0] = -1 + 2 = 1`.
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