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Math Help - Continuous Function on an Interval (Cal 2)

  1. #1
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    Continuous Function on an Interval (Cal 2)

    Let f(x) be a continuous function defined on the interval [2, infinity) such that
    f(4) = 10, |f(x)| < x^{9} + 6
    and


    INT (f(x) e^(-x/2))dx = 6.
    4

    (integral from 4 to infinity of f(x) times e to the negative x over 2)

    Determine the value of


    INT (f '(x) e^(-x/2))dx
    4


    So, it gives a value for the definite integral, as well as an x,y coordinate of the function itself and an equation that f(x) is less than...and it asks for the same definite integral with the derivative of f(x) replacing f(x).

    Not quite sure where to start (or end for that matter!), if anyone could point me in the right direction (quickly) I would appreciate it!
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  2. #2
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    Quote Originally Posted by habsfan View Post
    Let f(x) be a continuous function defined on the interval [2, infinity) such that
    f(4) = 10, |f(x)| < x^{9} + 6
    and


    INT (f(x) e^(-x/2))dx = 6.
    4

    (integral from 4 to infinity of f(x) times e to the negative x over 2)

    Determine the value of


    INT (f '(x) e^(-x/2))dx
    4


    So, it gives a value for the definite integral, as well as an x,y coordinate of the function itself and an equation that f(x) is less than...and it asks for the same definite integral with the derivative of f(x) replacing f(x).

    Not quite sure where to start (or end for that matter!), if anyone could point me in the right direction (quickly) I would appreciate it!
    \int f(x)e^{-x/2}dx

    if we integrate by parts with u=f(x) \mbox{ then } du=f'(x)dx
    and dv=e^{-x/2}dx \mbox{ then} v=-2e^{-x/2}


    \int_{4}^{\infty} f(x)e^{-x/2}dx=-2e^{-x/2}f(x)  |_{4}^{\infty}+2\int_{4}^{\infty} f'(x)e^{-x/2}dx

    use what is given to finish from here.

    hint: \lim_{x \to \infty} f(x)e^{-x/2}=0 Why? think about it and good luck.
    Last edited by TheEmptySet; March 28th 2008 at 10:27 PM. Reason: Latex error
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