Q92 / 160 · WBJEE 2015 · Mathematics
Let $f(x)$ denotes the fractional part of a real number $x$. Then, the value of $\int_{0}^{\sqrt{3}} f\left(x^{2}\right) d x$ is
- A$2 \sqrt{3}-\sqrt{2}-1$
- B0
- Correct answerC$\sqrt{2}-\sqrt{3}+1$
- D$\sqrt{3}-\sqrt{2}+1$
Solution
Let $I=\int_{0}^{\sqrt{3}} f\left(x^{2}\right) d x=\int_{0}^{\sqrt{3}}\left\{x^{2}\right\} d x$
$\left.=\int_{0}^{\sqrt{3}}\left(x^{2}-\mid x^{2}\right)\right) d x$
$=\int_{0}^{\sqrt{3}} x^{2} d x-\int_{0}^{\sqrt{3}}\left|x^{2}\right| d x$
$=\left[\frac{x^{3}}{3}\right]_{0}^{\sqrt{3}}-1 \int_{0}^{1}\left|x^{2}\right| d x+\int_{1}^{\sqrt{2}}\left|x^{2}\right| d x+\int_{\sqrt{2}}\left|x^{2}\right| d x \mid$
$=\sqrt{3}-\left[\int_{0}^{1} 0 d x+\int_{1}^{\sqrt{2}} 1 d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x\right]$
$=\sqrt{3}-\left\{0+|x|_{i}^{\sqrt{2}}+|2 x| \sqrt{2}\right\}$
$=\sqrt{3}-\sqrt{2}+1-2 \sqrt{3}+2 \sqrt{2}$
$=\sqrt{2}-\sqrt{3}+1$
Hence, the correct answer is option: 'C'.