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