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December 6th, 2018, 02:01 AM  #1 
Senior Member Joined: Jan 2012 Posts: 133 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: 2,404 Thanks: 1306 
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. 
December 6th, 2018, 02:16 AM  #3 
Senior Member Joined: Jan 2012 Posts: 133 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: 2,404 Thanks: 1306 
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: 133 Thanks: 2 
Thanks but plz elaborate.

December 6th, 2018, 02:28 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,472 Thanks: 2039  
December 6th, 2018, 03:14 AM  #7 
Senior Member Joined: Jan 2012 Posts: 133 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: 20,472 Thanks: 2039 
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. 
March 16th, 2019, 01:45 AM  #9 
Senior Member Joined: Jan 2012 Posts: 133 Thanks: 2  

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