If $I=\int_{0}^{1.7}[x^2]dx$, where [x] denotes the greatest integer ≤ x, then I equals:
Answer & explanation
Correct answer: option 2
$I=\int\limits_{0}^{1}[x^2]dx+\int\limits_{0}^{\sqrt{2}}[x^2]dx+\int\limits_{\sqrt{2}}^{1.7}[x^2]dx=(\sqrt{2}-1)+2(1.7-\sqrt{2})=2.4-\sqrt{2}$