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Math Help - Evaluate this integral

  1. #1
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    Evaluate this integral

    Evaluate the integral. (Give your answer as an improper fraction.)


    Ok my work:

    b/c of abs value law x { x when x>/= to 0
    { -x when x<0

    Intergral from -2 to 0 (x-2(-x))dx + Integral from 0 to 3 (x-2(x)) dx


    = integral from -2 to 0 (3x) dx + integral from 0 to 3 -x dx

    Antiderivative= (3x^2)/2 | -2 to 0 intervals

    Antiderivative= -x^2/2 | 0 to 3 intervals


    you plug in each and i got

    = (0 - (3(4))/2) + ( 9/2 - 0) = -1.5 which they said was wrong?! and they wanted as an improper fraction so that is -3/2 but I only have one more try to enter in an answer and I want to be 100% sure I am correct.

    Any help greatly appreciated!!


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  2. #2
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    Re: Evaluate this integral

    Look again at the second of the two integrals, I think that you've lost a negative sign. And yes, if the question asks for the answer to be given as an improper fraction then that is what you should do.
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  3. #3
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    Re: Evaluate this integral

    ok so it should be (0 - (3(4))/2) + (-9/2 - 0) = 6+ (-9/2) which is 1.5 positive sign ==> 3/2. Do you agree?
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  4. #4
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    Re: Evaluate this integral

    No, now you've lost a negative sign from the first integral.
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  5. #5
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    Re: Evaluate this integral

    Quote Originally Posted by BobP View Post
    No, now you've lost a negative sign from the first integral.
    (0 - (3(4))/2) + (-9/2 - 0) = -6+ (-9/2) which is -10.5 or -21/2 ? right?
    Last edited by Steelers72; February 23rd 2013 at 04:15 PM.
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  6. #6
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    Re: Evaluate this integral

    Yes.
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