The value of $I=\int_{0}^{1.5} [x^2]dx$, where [] denotes the greatest integer function, is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $2-\sqrt{2}$
$I=\int_{0}^{1.5} [x^2]dx$
$=\int_{0}^10\,dx+\int_{1}^{\sqrt{2}}1\,dx+\int_{\sqrt{2}}^{1.5}2\,dx$
$=0+(\sqrt{2}-1)+2(1.5-\sqrt{2})$
$=\sqrt{2}-1+3-2\sqrt{2}$
$=2-\sqrt{2}$