${\int }_0^{1.5} \left[x^2\right]dx$ is equal to

Solution:

We have,

$\begin{array}{l}{\int }_0^{1.5} \left[x^2\right]dx={\int }_0^1 \left[x^2\right]dx+{\int }_1^{\sqrt{2}} \left[x^2\right]dx+{\int }_{\sqrt{2}}^{1.5} \left[x^2\right]dx\\ \Rightarrow {\int }_0^{1.5} \left[x^2\right]dx={\int }_0^1 0dx+{\int }_1^{1.5} 1\cdot dx+{\int }_{\sqrt{2}}^1 2dx\end{array}$

$\Rightarrow {\int }_0^{1.5} \left[x^2\right]dx=0+(\sqrt{2}-1)+2(1.5-\sqrt{2})=2-\sqrt{2}$