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 is

• A. $2\sqrt{3}-\sqrt{2}-1$
• B. $0$
• C. $\sqrt{2}-\sqrt{3}+1$
• D. $\sqrt{3}-\sqrt{2}+1$

Answer 1

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

Let $I={\int }_0^{\sqrt{3}}f\left(x^2\right)dx={\int }_0^{\sqrt{3}}\left\{x^2\right\}dx$

$={\int }_0^{\sqrt{3}}(x^2-[x^2\left]\right)dx$

$={\int }_0^{\sqrt{3}}{x}^{2}dx-{\int }_0^{\sqrt{3}}\left[x^2\right]dx$

$={\left[\frac{x^3}{3}\right]}_0^{\sqrt{3}}-\left[{\int }_0^{\sqrt{3}}\right[x^2]dx+{\int }_1^{\sqrt{2}}[x^2]dx+{\int }_{\sqrt{2}}^{\sqrt{3}}[x^2\left]dx\right]$

$=\sqrt{3}-[{\int }_0^{1}0dx+{\int }_1^{\sqrt{2}}1dx+{\int }_{\sqrt{2}}^{\sqrt{3}}2dx]$

$=\sqrt{3}-\{0+[x{]}_1^{\sqrt{2}}+\left[2x{]}_{\sqrt{2}}^{\sqrt{3}}\right\}$

$=\sqrt{3}-\sqrt{2}+1-2\sqrt{3}+2\sqrt{2}$

$=\sqrt{2}-\sqrt{3}+1$