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Thread: Average Income Integral

  1. #1
    Oct 2010

    Average Income Integral

    I'm not sure if this is the right forum, but it's from the course on DE...
    I'm stuck on understanding the first part of a question:

    "In a large population, the proportion with income between x and x+dx is f(x)dx. Express the mean (average) income *mu* as an integral, assumiing that any positive income is possible."

    Can someone point me in the right starting direction?
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  2. #2
    MHF Contributor

    Apr 2005
    The average value of any function, f(x), between x= a and x= b is
    $\displaystyle \frac{\int_a^b f(x)dx}{b- a}$.

    Here, a= 0 while b is "infinity"- there is no upper bound.

    So the average is a limit- $\displaystyle \lim_{t\to\infty}\frac{\int_0^t f(x)dx}{t}$
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  3. #3
    Oct 2010
    Ok, thanks that helps. I see why a=0 and b="infinity" and the limit part at the end, but would you mind explaining why the average income is the average value of f(x)? This is the point I get stuck at...
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  4. #4
    Oct 2010
    For anyone else who is interested, I asked my supervisor for help, so here's the answer;
    If the population has N people, the number with income in the range x to x+dx is Nf(x)dx
    We multiply this by x to get the sum of their incomes.
    So the total income is
    $\displaystyle {\int_0^{\infty} x.Nf(x)dx}$

    Hence the mean income,
    $\displaystyle {\mu}=\frac{\int_0^{\infty} x.Nf(x)dx}{N}={\int_0^{\infty} xf(x)dx}$

    Not simply the integral of f(x), which is perhaps why HallsofIvy's post is a little unclear.
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