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Math Help - Interpretation of an Integral

  1. #1
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    Interpretation of an Integral

    Hello MathHelp World,

    I have a survival function that estimates the amount of time that a message remains unopened once dispatched. (See attachment) The data is shown on the left, for instance, after one day, 87% of messages remain unopened, and after 3 days 83% remain unopened.

    After 30 days the messages expire and they are pulled. At that point in time about 47% of the messages were not opened.

    I fit a polynomial function to the curve and figured a definite integral over the 30 day time frame. The result was about 19.

    My question:
    What are the units and how can I use this information? This is 19 'Days'? 19 'Message Days' Can I say that the messages are in the system for an average of 19 days? If I have a system capacity of say 100 messages, and I dispatch 50 messages that I expect will remain in the system for an average of 19 days, can I say that it is using an estimated (19/365) * (50/100) = 2.6% of the annual message capacity?

    Any help appreciated,

    Will
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  2. #2
    A Plied Mathematician
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    Re: Interpretation of an Integral

    Given an integral of the form

    I=\int_{a}^{b}f(x)\,dx,

    the units of I are given as the product of units of f(x) and units of x.

    So what are the units of the function you're integrating? And what are the units of the independent variable?
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  3. #3
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    Re: Interpretation of an Integral

    From 'a' to 'b' is in 'days'; the y-axis is in '% messages remaining', so 'I' must be '% messages remaining * days' (and I'm not sure that means anything). So I have to change the y-axis units, which I could do by starting with 100, so that 84 messages remain after 3 days. That would allow me to have 'message * days' (or number of days that a message exists).

    I can use this information to estimate the share of annual message capacity that will be used. Dispatch 100 messages, they use 19 'message * days' on average (an expected 19 * 100 = 1900 'message * days' total. If capacity is 100, there are 100 * 365 = 36500 'message * days' in a year. 100 messages could be expected to use 5.2% of the annual capacity.

    Please point out if I have misinterpreted anything. Thanks for the help!
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  4. #4
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    Re: Interpretation of an Integral

    Message-days are an appropriate unit to use IMO since the OP implies the further question of "How long would a typical message remain unread?" and if you know that each message is read in 19 days then you can, as you have done, estimate how much space is required.

    To make a comparison: in SI units Planck's Constant has units of Js (energy * time) yet nobody disputes it's validity
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