$\int\limits_0^{1.5}\left[x^2\right] d x$ (where [.] denotes greatest integer function) =

$2-\sqrt{2}$

$\int\limits_0^{3 / 2}\left[x^2\right] d x$

$=\int\limits_0^1 0 d x+\int\limits_1^{\sqrt{2}} 1 d x+\int\limits_{\sqrt{2}}^{3 / 2} 2 d x$

$=(\sqrt{2}-1)+2\left(\frac{3}{2}-\sqrt{2}\right)=\sqrt{2}-1+3-2 \sqrt{2}=2+\sqrt{2}-2 \sqrt{2}=2-\sqrt{2}$