$\int\limits_0^{1.5}\left[x^2\right] d x$ is equal to

$2-\sqrt{2}$

We have,

$\int\limits_0^{1.5}\left[x^2\right] d x=\int\limits_0^1\left[x^2\right] d x+\int\limits_1^{\sqrt{2}}\left[x^2\right] d x+\int\limits_{\sqrt{2}}^{1.5}\left[x^2\right] d x$

$\Rightarrow \int\limits_0^{1.5}\left[x^2\right] d x=\int\limits_0^1 0 d x+\int\limits_1^{\sqrt{2}} 1 \cdot d x+\int\limits_{\sqrt{2}}^{1.5} 2 d x$

$\Rightarrow \int\limits_0^{1.5}\left[x^2\right] d x=0+(\sqrt{2}-1)+2(1.5-\sqrt{2})=2-\sqrt{2}$