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Math Help - Area under curve

  1. #1
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    Area under curve

    Hi can you please explain this to me,

    For
    \int^k_0 x(x-4)\ dx = 13, why must k be from 0 to 4 such that the area is 13units^2?
    I integrated x(x-4) from 0 to 4, I only got the area as 10.something...it's not 13?

    If that is so, then what should the value for k be such that
    \int^k_0 x^3(x-9)\ dx = 13
    Made this one up myself, I just wanted to know how to get k=4 in the previous question..

    Please help, thank you so much!
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  2. #2
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    Re: Area under curve

    Yep you're correct if you integrate that you'll get F(x)=(1/3)x^3-2x^2+c and when you integrate that from 0 to 4 you get 10.67. Might be a typo.
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  3. #3
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    Re: Area under curve

    Plot the graph of f(x)=x(x-4). It's a parabola with roots 0 and 4.

    Area under curve-pix.png

    If you use limits from 0 to 4, you'll get -32/3 which indicates an area BELOW the x-axis. So k has to be greater than 4, if you want the value of the definite integral to be 13 for your answer.
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  4. #4
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    Re: Area under curve

    Thank you I understand!
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