CUET Preparation Today
The value of $I=\int_{0}^{1.5} [x^2]dx$, where [] denotes the greatest integer function, is: |
$2-\sqrt{2}$ $\sqrt{2}$ $5\sqrt{2}$ $3-2\sqrt{2}$ |
$2-\sqrt{2}$ |
The correct answer is Option (1) → $2-\sqrt{2}$ $I=\int_{0}^{1.5} [x^2]dx$ $=\int_{0}^10\,dx+\int_{1}^{\sqrt{2}}1\,dx+\int_{\sqrt{2}}^{1.5}2\,dx$ $=0+(\sqrt{2}-1)+2(1.5-\sqrt{2})$ $=\sqrt{2}-1+3-2\sqrt{2}$ $=2-\sqrt{2}$ |
