Averigua el valor de la integral$\int_{-2}^{2} \lfloor x^2-1\rfloor dx$

Question

Averigüe el valor de la 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 $ .)

Mi intento ..... $$\int_{-2}^2 \lfloor x^2 – 1\rfloor dx = 2\int_0^2 \lfloor x^2-1\rfloor dx$$ 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$ $

Pero, ¿cómo evaluar esto o me equivoco en el supuesto general?

Answer 1

Ya que $f$ es igual tenemos $$ I/2 =\int_0^2[x^2-1]\;dx=$$ $$=\int_0^1-1\;dx+\int_1^\sqrt{2}0\;dx+\int_\sqrt{2}^\sqrt{3}1\;dx+\int_\sqrt{3}^22\;dx=...$ $ $$=(-1)+0+(\sqrt{3}-\sqrt{2})+(4-2\sqrt{3})=3-\sqrt{3}-\sqrt{2}$ $