Question

If $I={\int }_0^{1.7}\left[x^2\right]dx$ then $I$ equals:

• A. $2.4+\sqrt{2}$
• B. $2.4-\sqrt{2}$
• C. $2.4+\frac{1}{\sqrt{2}}$
• D. $2.4-\frac{1}{\sqrt{2}}$

Answer 1

The correct option is B $2.4-\sqrt{2}$

Put $x^2=t$
or $x=\sqrt{t}$ or $dx=\frac{1}{2\sqrt{t}}dt$
$I={\int }_0^{2.89}\frac{\left[t\right]}{2\sqrt{t}}dt$
$={\int }_0^1\frac{\left[t\right]}{2\sqrt{t}}dt+{\int }_1^2\frac{\left[t\right]}{2\sqrt{t}}dt+{\int }_2^{2.89}\frac{\left[t\right]}{2\sqrt{t}}dt$
$=0+{\int }_1^2\frac{1}{2\sqrt{t}}dt+{\int }_2^{2.89}\frac{2}{2\sqrt{t}}dt$
$=\sqrt{t}{]}_1^2+2\sqrt{t}{]}_2^{2.89}$
$=(\sqrt{2}-1)+2(1.7-\sqrt{2})$
$=2.4-\sqrt{2}$