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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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    Did you notice that $x \in \mathbb Z$ only at $x = 0$ and $x = 1$ in your domain?2011-10-19
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    @JonasTeuwen Yeah I have some other things to prove in my homework assignment so it's useful for that part.2011-10-19
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    Well then, what does $[x]$ mean to you?2011-10-19
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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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    At $x=1$ $[x]=1$ not $0$ That's the problem. You didn't consider the fact that f(x)=0 at $x\in \mathbb{Z}$2011-10-19
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    @RamanaVenkata, that's not a problem, since it's a single point. You can consider the open domain and the integral doesn't change.2011-10-19
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    You're right, I have changed the domain. If you change a single point in your function the integral will not change. So you could just define $f(1) = \frac12$.2011-10-19
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    I am supposed to use only to use Reimann Integral definition and prove it.2011-10-19
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    That would be a silly question.2011-10-19
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    Why would it be a silly question? I was just asked to prove something from its definition.2011-10-19
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    That would unnecessarily handicap yourself. Anyway, you just note what we have written here and then you can compute the Riemann integral from the definition using this. It should not be necessary for us to spell everything out.2011-10-19
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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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    You didn't consider the fact that f(x)=0 at $x\in \mathbb{Z}$2011-10-19
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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