Mathematics | VITEEE 2014 | Discussion Page

The value of $\int_{0}^{\sqrt{2}} [x^{2}]dx$ , where [.] is the greatest integer function is

$2- \sqrt{2}$

$2+\sqrt{2}$

$\sqrt{2}-1$

$\sqrt{2}-2$

Option C

$\int_{0}^{\sqrt{2}} \left[x^{2}\right]dx=\int_{0}^{1} \left[x^{2}\right]dx+\int_{1}^{\sqrt{2}}\left[x^{2}\right]dx$

$=\int_{0}^{1} 0dx+\int_{1}^{\sqrt{2}}1dx$

$= \left[x\right]_{1}^{\sqrt{2}}=\sqrt{2}-1$

Discussion Forum