Floor under integration again

$\displaystyle\int_{0}^{1.5} x\lfloor x^{2} \rfloor dx$ $=\displaystyle\int_{0}^{1} x\lfloor x^{2} \rfloor dx+\displaystyle\int_{1}^{\sqrt{2}} x\lfloor x^{2} \rfloor dx+\displaystyle\int_{\sqrt{2}}^{1.5} x\lfloor x^{2} \rfloor dx$ $=\displaystyle\int_{0}^{1} (x×0) dx+\displaystyle\int_{1}^{\sqrt{2}} (x×1) dx+\displaystyle\int_{\sqrt{2}}^{1.5} (x×2) dx$

$=\bigg [\frac{x^{2}}{2} \bigg ]_{1}^{\sqrt{2}} + \bigg [x^{2} \bigg ]_{\sqrt{2}}^{1.5}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$

So answer is $100×\frac{3}{4}=\boxed{75}$