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Find out the value of the integral $$\int_{-2}^{2} \lfloor x^2-1\rfloor dx$$ where $[x]$ denotes the floor function (i.e., $[x]$ is the greatest integer $\le x$.)

My attempt ..... $$\int_{-2}^2 \lfloor x^2 – 1\rfloor dx = 2\int_0^2 \lfloor x^2-1\rfloor dx$$ Because $\lfloor x^2 – 1\rfloor$ is even. $$2\int_0^2 \lfloor x^2-1\rfloor dx =\\ 2\int_0^1 \lfloor x^2-1\rfloor dx+2\int_1^{\sqrt{2}} \lfloor x^2-1\rfloor dx +2\int_{\sqrt{2}}^{\sqrt{3}} \lfloor x^2-1\rfloor dx + 2\int_{\sqrt{3}}^2 \lfloor x^2-1\rfloor dx$$

But how to evaluate this or am I wrong in the whole assumption?

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  • $\begingroup$ By square brackets, do you mean entier function? $\endgroup$
    – Jakobian
    Feb 13, 2019 at 14:30
  • $\begingroup$ By [x] I mean greatest integer $\le$ x $\endgroup$
    – Saradamani
    Feb 13, 2019 at 14:30
  • $\begingroup$ Sorry for the bad edits.. $\endgroup$
    – Saradamani
    Feb 13, 2019 at 14:36
  • $\begingroup$ You already have done what is needed...by finding the values of $x$ where the floor function changes it's values. $\endgroup$
    – dmtri
    Feb 13, 2019 at 14:46

1 Answer 1

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Since $f$ is even we have $$ I/2 =\int_0^2[x^2-1]\;dx=$$ $$=\int_0^1-1\;dx+\int_1^\sqrt{2}0\;dx+\int_\sqrt{2}^\sqrt{3}1\;dx+\int_\sqrt{3}^22\;dx=...$$ $$=(-1)+0+(\sqrt{3}-\sqrt{2})+(4-2\sqrt{3})=3-\sqrt{3}-\sqrt{2}$$

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  • $\begingroup$ Can you just elaborate atleast one of your integrations meaning how you obtained -1 or 0 etc.. I know it is naive to ask such a silly question after proceeding upto this level. But I still ask you this. I haven't ever solved such problem .. It's silly and pity. $\endgroup$
    – Saradamani
    Feb 13, 2019 at 14:40
  • $\begingroup$ Oh now I get it get it... I get it no need to elaborate. $\endgroup$
    – Saradamani
    Feb 13, 2019 at 14:42
  • $\begingroup$ Of course, for each $x\in [0,1)$ we have $f(x)=-1$ $\endgroup$
    – nonuser
    Feb 13, 2019 at 14:42
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    $\begingroup$ I have solved it myself.Thanks. I understood what it is for the first case -1 comes because from 0 to 1 f(x) is always -1. You just have to put a value between 0 to 1 $\endgroup$
    – Saradamani
    Feb 13, 2019 at 14:45

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