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Question
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Sum
Solution
`int_(-1)^1 "f"(x) "d"x = int_(-1)^0 "f"(x) "d"x + int_0^1 "f"(x) "d"x`
= `int_(-1)^0 (-x) "d"x + int_0^1 x "d"x`
= `- [x^2/2]_(-1)^0 + [x^2/2]_0^1`
= `- [0 - (-1)^2/2] + [(1)^2/2 - ((0))/2]`
= `- [-1/2] + [1/2]`
= `1/2 + 1/2`
= 1
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Definite Integrals
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