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Math Help - [SOLVED] integral in terms of pi

  1. #1
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    [SOLVED] integral in terms of pi

    Hello,

    Having established that \pi = 2\int^1_{-1} \! \sqrt{1 - x^2} \, dx,
    I need to express \int^2_{-2} \! (x - 3)\sqrt{4 - x^2} \, dx in terms of pi by using the properties of intergrals (linearity with respect to the integrand, additivity with respect to the interval of integration, invariance under translation, and expansion/ contraction of the interval of integration).

    I can get as far as \int^1_{-1} \! 4x \sqrt{1 - x^2} \, dx - 6\pi, and according to the book, the answer is -6pi, so if I'm right so far, the first integral evaluates to 0. I just don't know how to show this.
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  2. #2
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    Lightbulb

    Never mind, I got it
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