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December 6th, 2018, 03:01 AM   #1
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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.
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December 6th, 2018, 03:08 AM   #2
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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.
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December 6th, 2018, 03:16 AM   #3
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Actually that was the question in a book which I typed exactly.

Last edited by happy21; December 6th, 2018 at 03:16 AM. Reason: Quote not required.
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December 6th, 2018, 03:16 AM   #4
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If you work it all out it turns out that the integral equals

$\dfrac 1 2 \lfloor x \rfloor$
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December 6th, 2018, 03:23 AM   #5
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Thanks but plz elaborate.
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December 6th, 2018, 03:28 AM   #6
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Quote:
Originally Posted by romsek View Post
. . . 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.
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December 6th, 2018, 04:14 AM   #7
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Can one elaborate step by step the result so obtained. Thx.
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December 6th, 2018, 06:20 AM   #8
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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.
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