Let [.] denote the greatest integer function, then the value of $\int\limits_0^{1.5} x\left[x^2\right] d x$, is

$\frac{3}{4}$

We have,

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

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

$\Rightarrow \int\limits_0^{1.5} x\left[x^2\right] d x=\left[\frac{x^2}{2}\right]_1^{\sqrt{2}}+\left[x^2\right]_{\sqrt{2}}^{1.5}$

$\Rightarrow \int\limits_0^{1.5} x\left[x^2\right] d x=\left(1-\frac{1}{2}\right)+\left(\frac{9}{4}-2\right)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$