Question

Let $f\left(x\right)$ denotes the fractional part of a real number $x$. Then the value of ${\int }_0^{\sqrt{3}}f\left(x^2\right)dx$

• A. $2\sqrt{3}-\sqrt{2}-1$
• B. $0\left(zero\right)$
• C. $\sqrt{2}-\sqrt{3}+1$
• D. $\sqrt{3}-\sqrt{2}+1$

Answer 1

The correct option is C $\sqrt{2}-\sqrt{3}+1$

${\int }_0^{\sqrt{3}}f\left(x^2\right)dx={\int }_0^{\sqrt{3}}\left\{x^2\right\}dx={\int }_0^{\sqrt{3}}(x^2-[x^2\left]\right)dx={\int }_0^{\sqrt{3}}{x}^{2}dx-\left({\int }_0^1\right[x^2]dx+{\int }_1^{\sqrt{2}}[x^2]dx+{\int }_{\sqrt{2}}^{\sqrt{3}}[x^2\left]dx\right)$
$=\sqrt{3}-({\int }_0^{1}0.dx+{\int }_1^{\sqrt{2}}1.dx+{\int }_{\sqrt{2}}^{\sqrt{3}}2.dx)=\sqrt{3}-(\sqrt{2}-1+2\sqrt{3}-2\sqrt{2})=\sqrt{2}-\sqrt{3}+1$