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Question
Evaluate `int_(-1)^2 "f"(x) "d"x`, where f(x) = |x + 1| + |x| + |x – 1|
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Solution
We can redefine f as f(x) = `{{:(2 - x",", "if" - 1 < x ≤ 0),(x + 2",", "if" 0 < ≤ 1),(3x",", "if" 1 < x ≤ 2):}`
Therefore, `int_(-1)^2 "f"(x)"d"x = int_(-1)^0 (2 - x)"d"x + int_0^1 (x + 2)"d"x + int_1^2 3x"d"x` ....(By P2)
= `(2x = x^2/2)_(-1)^0 + (x^2/2 + 2x)_0^1 + ((3x^2)/2)_1^2`
= `0 - (-2 - 1/2) + (1/2 + 2) + 3(4/2 - 1/2)`
= `5/2 + 5/2 + 9/2`
= `19/2`.
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