1) The value of $ \int_{0}^{\sqrt{2}} [x^{2}]dx$ , where [.] is the greatest integer function is A) $2- \sqrt{2}$ B) $2+\sqrt{2}$ C) $\sqrt{2}-1$ D) $\sqrt{2}-2$ Answer: Option CExplanation:$\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 $ $= \left[x\right]_{1}^{\sqrt{2}}=\sqrt{2}-1$
