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If $f:\mathbb{R}\to \mathbb{R}$ and $f(x)=0$ if $x\in \mathbb{Z}$ and $f(x)=x-\lfloor x\rfloor-\frac12$ if $x\in \mathbb{R}-\mathbb{Z}$. Let $A(x)=\int_0^x f(t)\mathrm dt$.

Show that $A(x)=\dfrac{x^2-x}{2}$ if $0\leq x\leq 1$.

I am supposed to use the definition of Riemann Integral.

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    [x] denotes the greatest integer less than x.2011-10-19

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Note that on $0 < x < 1$ we have that $f(x) = x - \frac12$ (do you see why?). Can you now find $A(x)$?

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    @Ramana: I see no problem using the definition here. Just set up a small partition around the integer points, and treat everything else as normal. Then by the definition, it still converges.2011-10-19
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What does your integrand $f(t) = t - \lfloor t \rfloor - 1/2$ look like when $0 \leq t \leq x \leq 1$? In particular, the expression $\lfloor t \rfloor$ can be simplified greatly in this situation.

Once you understand how to solve your problem, you should next think about how you would compute $A(x)$ when $1 \leq x \leq 2$.

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    If you solve the problem by computing Riemann sums and taking an appropriate limit, then you'll have to modify two terms in your Riemann sums (corresponding to the endpoints), but you'll see that the integral comes out to be the same as if $f(0)$ and $f(1)$ had any other values.2011-10-19