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Find integration of $\int_{-2}^{2}\lfloor x^2-1\rfloor~dx$ where $\lfloor . \rfloor$ is Box function i.e. greatest integer function . More explicitely [x]= greatest integer not greater than x.

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    Oops! Is that the floor function? Then I shall erase my answer...2012-12-19

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$ \begin{align} \int_{-2}^{2}\lfloor x^2-1\rfloor\,\mathrm{d}x &=2\int_0^{2}\lfloor x^2-1\rfloor\,\mathrm{d}x\\ &=2\int_0^1-1\,\mathrm{d}x +2\int_1^{\sqrt{2}}0\,\mathrm{d}x +2\int_{\sqrt2}^{\sqrt3}1\,\mathrm{d}x +2\int_{\sqrt3}^22\,\mathrm{d}x\\[6pt] &=-2+2(\sqrt3-\sqrt2)+4(2-\sqrt3)\\[12pt] &=6-2\sqrt2-2\sqrt3 \end{align} $

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    Thanks for your help. Now I got it.2012-12-19