The value of the integral, $$\int_1^3 [x^2 - 2x - 2] dx$$, where $$[x]$$ denotes the greatest integer less than or equal to $$x$$, is

Rewrite:

$$x^2-2x-2=(x-1)^2-3$$

Hence,

$$[x^2-2x-2]=[(x-1)^2-3]$$

Since

$$3$$ is an integer,

$$[(x-1)^2-3]= [(x-1)^2]-3$$

Therefore,

$$I= \int_1^3[(x-1)^2]\,dx - \int_1^33\,dx$$

Put

$$u=x-1,$$

then

$$du=dx$$

and limits become

$$x=1\to u=0,\qquad x=3\to u=2$$

Thus,

$$I=\int_0^2[u^2]\,du-6$$

Now split according to integer values of $$u^2$$ on $$[0,2].$$

For

$$0\le u<1,$$

$$[u^2]=0$$

For

$$1\le u<\sqrt2,$$

$$[u^2]=1$$

For

$$\sqrt2\le u<\sqrt3,$$

$$[u^2]=2$$

For

$$\sqrt3\le u<2,$$

$$[u^2]=3$$

Hence,

$$\int_0^2[u^2]\,du = 0+\int_1^{\sqrt2}1\,du +\int_{\sqrt2}^{\sqrt3}2\,du +\int_{\sqrt3}^23\,du$$

$$= (\sqrt2-1)+2(\sqrt3-\sqrt2)+3(2-\sqrt3)$$

$$=\sqrt2-1+2\sqrt3-2\sqrt2+6-3\sqrt3$$

$$=5-\sqrt2-\sqrt3$$

Therefore,

$$I=(5-\sqrt2-\sqrt3)-6$$

$$=-1-\sqrt2-\sqrt3$$