Question

The value of \( I = \int_{0}^{1.5} \left\lfloor x^2 \right\rfloor dx \), where [ ] denotes the greatest integer function, is:

• A. \( 2 - \sqrt{2} \)
• B. \( \sqrt{2} \)
• C. \( 5 \sqrt{2} \)
• D. \( 3 - 2 \sqrt{2} \)

Answer 1

The Correct Option is A

The greatest integer function \([x^2]\) takes different values depending on \(x^2\).

Over \(0 \leq x \leq \sqrt{2}\), we split the integral into ranges where \([x^2]\) is constant:

For \(0 \leq x < 1\), \(x^2 < 1\) and \([x^2] = 0\). The contribution to the integral is:

\[ \int_{0}^{1} [x^2]dx = \int_{0}^{1} 0 \, dx = 0. \]

For \(1 \leq x \leq \sqrt{2}\), \(1 \leq x^2 < 2\) and \([x^2] = 1\). The contribution to the integral is:

\[ \int_{1}^{\sqrt{2}} [x^2]dx = \int_{1}^{\sqrt{2}} 1 \, dx = (\sqrt{2} - 1). \]

Adding these results:

\[ I = 0 + (\sqrt{2} - 1) = \sqrt{2} - 1. \]

Thus, the value of \(I\) is \(2 - \sqrt{2}\).