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Question
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
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Solution
`int_0^2 "f"(x) "d"x = int_0^1 "f"(x) "d"x + int_1^2 "f"(x) "d"x`
= `int_0^1 (3 - 2x - x^2) "d"x + int_1^2 (x^2 + 2x - 3) "d"x`
= `[3x - (2x^2)/2 - x^3/3]_0^1 + [x^3/3 + 2(x^2/2) - 3x]_1^2`
= `[3x - x^2 - x^3/2]_0^1 + [x^3/3 + x^2 - 3x]_1^2`
= `[3(1) - (1)^2 - (1)^3/3 - {0}] + {((2)^3/3 + (2)^2 - 3(2)) - (1/3 + 1 - 3)}`
= `(3 - 1 - 1/3) + {(8/3 + 4 - 6) - (1/3 + 1 - 3)}`
= `(2 - 1/3) + {(8/3 - 2) - (1/3 - 2)}`
= `((6 - 1)/3) + [((8 - 6)/3) - ((1 - 6)/3)]`
= `(5/3) + [2/3 - ((-5)/3)]`
= `5/3 + [2/3 + 5/3]`
= `5/3 + [7/3]`
= `5/3 + 7/3`
= `12/3`
= 4
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