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Math Help - Help in integration

  1. #1
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    Help in integration

    Hi, I could not figure out two questions in my homework. The first problem deals with a piecewise function.

    9
    g(x)dx, where g(x)={1, x> 0 and x^3, 1 < x < 4 and sqrt x, 4 < x < 9
    0


    also,

    Find the area below the interval [-2, -1], but above the curve y=x^3

    Does this just mean
    -1
    (x^3)dx
    -2

    ??
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  2. #2
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    Quote Originally Posted by kn336a View Post
    Hi, I could not figure out two questions in my homework. The first problem deals with a piecewise function.

    9
    g(x)dx, where g(x)={1, x> 0 and x^3, 1 < x < 4 and sqrt x, 4 < x < 9
    0


    also,

    Find the area below the interval [-2, -1], but above the curve y=x^3

    Does this just mean
    -1
    (x^3)dx
    -2

    ??
    For Q1 try

    I_n=\int_0^1 1 dx + \int_1^4 x^3 dx + \int_4^9 \sqrt{x} dx

    For Q2 you are correct.
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  3. #3
    MHF Contributor Amer's Avatar
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    Ok hi

    I attach my answer the second question you just should multiply the answer by -1




    Attachment 11481

    Attachment 11482
    Last edited by Amer; June 28th 2009 at 12:22 PM.
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  4. #4
    MHF Contributor Amer's Avatar
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    I was late sorry
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  5. #5
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    Smile

    Thanks guys! I got 929/12 for the first one or roughly 77.4167

    the second one I got -15/4

    is this right? I hope so!
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  6. #6
    MHF Contributor Amer's Avatar
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    you answer is correct but the second I think you should take the absolute value cuz it is area and the area positive value

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  7. #7
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    sorry, I'm a little confused. So do I just add an absolute value sign after the integration sign, then calculate?

    I've gotten negative numbers for some of my integration problems and didn't get marked wrong. Is it because it says "but above the curve y=x^3?"

    I thought that just meant the area between the interval and the curve. Thanks!
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