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Math Help - integration and the greatest function help exam in 7 hrs

  1. #1
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    integration and the greatest function help exam in 7 hrs

    integrate f(x) from 0 to n. where n is a positive integer and f(x) is the greatest integer function.
    I Know that i have to use the definiton of greatest integer function and break it into sum of the integrals how do i do this.
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  2. #2
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    Quote Originally Posted by myoplex11 View Post
    integrate f(x) from 0 to n. where n is a positive integer and f(x) is the greatest integer function.
    I Know that i have to use the definiton of greatest integer function and break it into sum of the integrals how do i do this.
    F(b) - F(a) is the signed area, where a and b are the limits of integration (from a to b).

    In this case, your b is n and 0 is a. Do you have a specific problem?
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    this is the problem we had on our practice exam
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    Quote Originally Posted by myoplex11 View Post
    integrate f(x) from 0 to n. where n is a positive integer and f(x) is the greatest integer function.
    I Know that i have to use the definiton of greatest integer function and break it into sum of the integrals how do i do this.
    Okay.

    A defined function on a closed interval is Riemann integrable if and only if it is continous almost everywhere.

    We we need to find,
    \int_0^n [ x ] dx
    Since it is continous almost everywhere we can use the subdivision rule,
    \sum_{k=1}^n \int_{k-1}^k [x] dx
    But,
    \int_{k-1}^k [x] dx =k-1
    Because in that [x]=k-1 on this interval.

    Thus,
    \sum_{k=1}^n k-1=\frac{n(n-1)}{2}
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    Quote Originally Posted by AfterShock View Post
    F(b) - F(a) is the signed area, where a and b are the limits of integration (from a to b).

    In this case, your b is n and 0 is a. Do you have a specific problem?
    That theorem does not work. The fundamental theorem of calculus only applies when a function is countinous on the closed interval.
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    If you draw a graph for say n=8, you will see a series of seven ‘stair steps’.
    The area under the ‘stairs’ is the sum of the areas of seven rectangles.
    So 1+2+3…+7. To do it for n we get a sum of \frac{{\left( {n - 1} \right)n}}{2}.
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