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Here's a problem from my probability textbook:

If two points be taken at random on a finite straight line their average distance apart will be one third of the line.

What I did: I got the integral$${{\int_0^1 \int_0^1 |x-y|\,\text{d}x\,\text{d}y}\over{1^2}},$$which ends up evaluating to ${1\over3}$ according to Wolfram Alpha. However, I am not sure how to evaluate the integral by hand, given that there's the absolute value sign. Any help would be appreciated.

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    $\begingroup$ Split the integral into two parts, one where $y<x$ and one where $y>x$. Then you can rewrite the integral without the absolute value bars because you'll know the sign of $x-y$. $\endgroup$
    – John_Krampf
    Sep 21, 2021 at 6:16
  • $\begingroup$ math.stackexchange.com/q/3004029/321264 $\endgroup$
    – StubbornAtom
    Sep 21, 2021 at 6:56

1 Answer 1

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Hint $:$ $$\begin{align*}\int_{0}^{1}\int_{0}^{1} |x - y|\ dx\ dy & = \int_{0}^{1}\int_{0}^{y} (y - x) \ dx\ dy + \int_{0}^{1}\int_{y}^{1} (x - y)\ dx\ dy. \end{align*}$$

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  • $\begingroup$ Thanks @RabinKumarChakraborty. Can you elaborate in detail how you arrived at the two integrals? $\endgroup$
    – Emperor Concerto
    Sep 21, 2021 at 6:24
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    $\begingroup$ @Emperor Concerto$:$ I just break the interval in which $x$ varies. First I let $x$ to vary from $0$ to $y$ and then from $y$ to $1.$ Draw pictures to convince yourself. $\endgroup$
    – RKC
    Sep 21, 2021 at 6:28
  • $\begingroup$ Second integral on the right should be $ \displaystyle \int_{0}^{1}\int_{y}^{1} (x - y)\ dx\ dy$ $\endgroup$
    – Math Lover
    Sep 21, 2021 at 6:28
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    $\begingroup$ @Math Lover$:$ That's exactly what I did. Please have a look at it. $\endgroup$
    – RKC
    Sep 21, 2021 at 6:29
  • $\begingroup$ it shows correct now. It showed $ \displaystyle \int_{0}^{1}\int_{y}^{x} (x - y)\ dx\ dy$ earlier. $\endgroup$
    – Math Lover
    Sep 21, 2021 at 6:31

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