Calculus

Use geometry to evaluate the integral from negative 3 to 3 of f of x, dx for f of x equals the square root of the quantity 4 minus the square of the quantity x plus 1 for x is between negative 3 and 1 including negative 3 and 1, and equals the absolute value of the quantity (x minus 2) minus 1 for x is greater than 1 and less than or equal to 3.

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  1. If I read all those words correctly, you mean

    ∫[-3,3] f(x) dx
    where
    f(x) =
    f1: √(4-√(x+1)) for -3 <= x <= 1
    f2: |x-2|-1 for 1<x<3

    Unfortunately, for x < -1 the first piece is complex, not real.

    Anyway, fix the presumed typo and just plug in the pieces

    ∫[-3,1] f1(x) dx + ∫[1,3] f2(x) dx

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  2. 2pi-1

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