Let $f(x) := x - \lfloor x \rfloor - \frac{1}{2}$ for all real numbers $x$ not an integer and $f(x) := 0 $ for all integers $x$. If $$P(x) := \int_{0}^{x} f(t) dt$$ for all real numbers $x$, express $P(x)$ in terms of $\lfloor x \rfloor$.
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HINT: Let $n$ be an integer. If $n\le t<n+1$, then $\lfloor t\rfloor=n$, and $f(t)=t-n-\frac12$. In particular, if $n\le x\le n+1$, then
$$\int_n^xf(t)\,dt=\int_n^x\left(t-n-\frac12\right)dt=\int_{-1/2}^{x-n-1/2}u\,du\;,$$
where $u=t-n-\frac12$.
Brian M. Scott
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Thank you for taking time to write:) – Yes Feb 24 '15 at 03:51
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@YLChou: You’re welcome. If you follow up the hint, you’ll find that $P(x)$ actually has a pretty simple value in terms of $\lfloor x\rfloor$. – Brian M. Scott Feb 24 '15 at 03:53