Question

If $\left[x\right]$ denotes the greatest integer function, the value of ${\int }_0^{1.5}x\left[x^2\right]dx$ is:

• A. $\frac{5}{4}$
• B. $0$
• C. $\frac{3}{2}$
• D. $\frac{3}{4}$

Answer 1

The correct option is D

$\frac{3}{4}$

Explanation for the correct option.

$\begin{array}{rcl}{\int }_0^{1.5}x\left[x^2\right]dx& =& {\int }_0^{1}x\left[x^2\right]dx+{\int }_1^{\sqrt{2}}x\left[x^2\right]dx+{\int }_{\sqrt{2}}^{1.5}x\left[x^2\right]dx\\ & =& {\int }_0^1\left(x\times 0\right)dx+{\int }_1^{\frac{3}{2}}x\left[x^2\right]dx\\ & =& {\int }_1^{\frac{3}{2}}x\left[x^2\right]dx\end{array}$

Let $x^2=t$ then $2xdx=dt$. So,

$\begin{array}{rcl}{\int }_1^{\frac{3}{2}}x\left[x^2\right]dx& =& \frac{1}{2}{\int }_1^{\frac{9}{4}}\left[t\right]dt\\ & =& \frac{1}{2}{\int }_1^2\left[t\right]dt+\frac{1}{2}{\int }_2^{\frac{9}{4}}\left[t\right]dt\\ & =& \frac{1}{2}{\int }_1^{2}1dt+\frac{1}{2}{\int }_2^{\frac{9}{4}}2dt\\ & =& \frac{1}{2}{\left[t\right]}_1^2+\frac{1}{2}{\left[2t\right]}_2^{\frac{9}{4}}\\ & =& \frac{1}{2}\left(2-1\right)+\frac{1}{2}\left[2\left(\frac{9}{4}\right)-2\left(2\right)\right]\\ & =& \frac{1}{2}+\frac{1}{2}\left(\frac{9-8}{2}\right)\\ & =& \frac{3}{4}\end{array}$

Hence, option D is correct.