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If $ f(x)={\left\lfloor x^2\right\rfloor -\left\lfloor x\right\rfloor ^2}$,where ${\left\lfloor x\right\rfloor }$ denotes the greatest integer $\le x$ then $\int_1^2 f(x)dx?$ please give some hint. thank you.

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Break the interval into regions where $1<x^2<2$, $2<x^2<3$ and $3<x^2<4$. –  Eckhard Mar 19 '13 at 17:07

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up vote 12 down vote accepted

On the interval $(1,2)$ you have $[x] = 1$. On the interval $(1,\sqrt2)$ you have $[x^2] = 1$ and on the interval $(\sqrt{2}, \sqrt{3})$ you have $[x^2] = 3$ and on the interval $(\sqrt{3}, 4)$ you have $[x^2] = 3$. So you can break on the integral into: $$\begin{align} \int_1^2 f(x) \; dx &= \int_1^\sqrt{2} f(x)\; dx + \int_\sqrt{2}^\sqrt{3} f(x)\; dx + \int_\sqrt{3}^2 f(x)\; dx\\ &= \int_1^\sqrt{2} 1 - 1 \; dx + \int_\sqrt{2}^\sqrt{3} 2-1\; dx + \int_\sqrt{3}^2 3- 1\; dx \end{align} $$

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Our resident serial downvoter seems to be visiting us, @Thomas...+1 to balance and for a nice effort to answer. –  DonAntonio Mar 19 '13 at 17:12
    
@DonAntonio: Thanks. –  Thomas Mar 19 '13 at 17:13
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@DonAntonio and Thomas - never mind this downvote. I just notice that my reputation has decreased by 1, and when I checked what happened I saw this answer downvoted by myself (by no means intentionally). Have no idea, how it happened, but I apologize anyway :) now I shall take a look on all my previous downvotes. Btw, +1 for the answer –  Ilya Mar 19 '13 at 17:17
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(+1) Nice answer –  Mhenni Benghorbal Mar 19 '13 at 17:18
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So in fact it was the unconscious downvoter...? :) –  DonAntonio Mar 19 '13 at 17:24

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