The value of \(I=\int_{0}^{1.5}\displaystyle [x^2]\,dx\), where [ ] denotes the greatest integer function, is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2-\sqrt{2}\)
The greatest integer function \([x^2]\) jumps where \(x^2\) reaches an integer: \(x^2 = 1\) at x = 1 and \(x^2 = 2\) at \(x = \sqrt{2} \approx 1.414\). At x = 1.5, \(x^2 = 2.25 \lt 3\). So, on [0, 1.5]:
\(0 \le x \lt 1:\ [x^2] = 0\)
\(1 \le x \lt \sqrt{2}:\ [x^2] = 1\)
\(\sqrt{2} \le x \le 1.5:\ [x^2] = 2\)
Splitting the integral at these points:
\( \displaystyle I = \int_{0}^{1}0\,dx + \int_{1}^{\sqrt{2}}1\,dx + \int_{\sqrt{2}}^{1.5}2\,dx \)
\( = 0 + (\sqrt{2} - 1) + 2(1.5 - \sqrt{2}) \)
\( = \sqrt{2} - 1 + 3 - 2\sqrt{2} \)
\( = 2 - \sqrt{2} \)