Question

Evaluate the integral

${\int }_0^{1.5}\left[x^2\right]dx$, where $[.]$ denotes the greatest integer function.

• A. $\sqrt{2}-2$
• B. $2-\sqrt{2}$
• C. $2+\sqrt{2}$
• D. None of these

Answer 1

The correct option is B $2-\sqrt{2}$

${\int }_0^{1.5}\left[x^2\right]dx={\int }_0^1\left[x^2\right]dx+{\int }_1^{\sqrt{2}}\left[x^2\right]dx+{\int }_{\sqrt{2}}^{1.5}\left[x^2\right]dx$

$[\because \left[x^2\right]=\{\begin{array}{c}\begin{array}{cc}0& 0\le x<1\end{array}\\ \begin{array}{cc}1& 1\le x<\sqrt{2}\end{array}\\ \begin{array}{cc}2& \sqrt{2}\le x<1.5\end{array}\end{array}]$

$={\int }_0^{1}0dx+{\int }_1^{\sqrt{2}}1.dx+{\int }_{\sqrt{2}}^{1.5}2dx={\left[x\right]}_1^{\sqrt{2}}+{\left[2x\right]}_{\sqrt{2}}^{1.5}$

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