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 December 6th, 2018, 02:01 AM #1 Senior Member   Joined: Jan 2012 Posts: 140 Thanks: 2 How to solve this integral... Hi, Please do let me know how do I solve this integral: $\displaystyle \int_{0}^{\left [ x \right ]}\left ( \left [ x \right ] -\left [ x-\frac{1}{2} \right ]\right )dx$ where, '[]' denotes greatest integer function (used at three places in this question ). Thx. December 6th, 2018, 02:08 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 do you mean $\displaystyle \int \limits_0^{\lfloor x \rfloor}~\lfloor s \rfloor - \left \lfloor s - \dfrac 1 2 \right \rfloor ~ds$ you can't have $x$ in the limit and as the integral dummy variable. Thanks from topsquark December 6th, 2018, 02:16 AM #3 Senior Member   Joined: Jan 2012 Posts: 140 Thanks: 2 Actually that was the question in a book which I typed exactly. Last edited by happy21; December 6th, 2018 at 02:16 AM. Reason: Quote not required. December 6th, 2018, 02:16 AM #4 Senior Member   Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 If you work it all out it turns out that the integral equals $\dfrac 1 2 \lfloor x \rfloor$ December 6th, 2018, 02:23 AM #5 Senior Member   Joined: Jan 2012 Posts: 140 Thanks: 2 Thanks but plz elaborate. December 6th, 2018, 02:28 AM   #6
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Quote:
 Originally Posted by romsek . . . you can't have $x$ in the limit and as the integral dummy variable.
You can use any letter for the dummy variable. If you use an $x$, it's not the same variable as the $x$ used for the limit. December 6th, 2018, 03:14 AM #7 Senior Member   Joined: Jan 2012 Posts: 140 Thanks: 2 Can one elaborate step by step the result so obtained. Thx. December 6th, 2018, 05:20 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 Consider the intervals [n, n+1/2) and [n+1/2, n+1), where n is an integer. The integrand evaluates to 1 on the first of those, and 0 on the second, so integrating over [n, n+1) gives 1/2. The interval $[0, \lfloor{x}\rfloor)$ consists of $\lfloor{x}\rfloor$ subintervals of that type (i.e. [n, n+1)), so the value of the original integral is $\lfloor{x}\rfloor$ × 1/2. Thanks from happy21 and topsquark March 16th, 2019, 01:45 AM   #9
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 Originally Posted by skipjack The interval $[0, \lfloor{x}\rfloor)$ consists of $\lfloor{x}\rfloor$ subintervals of that type (i.e. [n, n+1)), so the value of the original integral is $\lfloor{x}\rfloor$ × 1/2.
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