CUET Preparation Today
If $I=\int_{0}^{1.7}[x^2]dx$, where [x] denotes the greatest integer ≤ x, then I equals: |
$2.4+\sqrt{2}$ $2.4-\sqrt{2}$ $2.4+\frac{1}{\sqrt{2}}$ $2.4-\frac{1}{\sqrt{2}}$ |
$2.4-\sqrt{2}$ |
$I=\int\limits_{0}^{1}[x^2]dx+\int\limits_{0}^{\sqrt{2}}[x^2]dx+\int\limits_{\sqrt{2}}^{1.7}[x^2]dx=(\sqrt{2}-1)+2(1.7-\sqrt{2})=2.4-\sqrt{2}$ |
