Question

If \( \left\lfloor x^2 \right\rfloor \) is the greatest integer less than or equal to \( x^2 \), then \[ \int_0^{\sqrt{2}} \left\lfloor x^2 \right\rfloor \, dx = \]

• A. \( \sqrt{2} \)
• B. 2
• C. \( \sqrt{2} - 1 \)
• D. \( \sqrt{2} + 1 \)
• E. \( 2\sqrt{2} + 1 \)

Hint:

When dealing with integrals involving the greatest integer function, divide the integral based on the intervals where the greatest integer function remains constant. In this case, it was split at \( x = 1 \), where the value of \( \left\lfloor x^2 \right\rfloor \) changed.

Answer 1

The Correct Option is C

We are given the integral: \[ I = \int_0^{\sqrt{2}} \left\lfloor x^2 \right\rfloor \, dx \] where \( \left\lfloor x^2 \right\rfloor \) represents the greatest integer less than or equal to \( x^2 \).
Step 1: Analyze the function \( \left\lfloor x^2 \right\rfloor \) for \( x \in [0, \sqrt{2}] \):
- For \( x \in [0, 1) \), \( x^2 \in [0, 1) \), so \( \left\lfloor x^2 \right\rfloor = 0 \).
- For \( x \in [1, \sqrt{2}) \), \( x^2 \in [1, 2) \), so \( \left\lfloor x^2 \right\rfloor = 1 \).
- At \( x = \sqrt{2} \), \( x^2 = 2 \), and \( \left\lfloor 2 \right\rfloor = 2 \), but the integral does not consider this single point.
Step 2: Break the integral into two parts: \[ I = \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx \] Step 3: Evaluate the integrals: \[ I = 0 + \int_1^{\sqrt{2}} 1 \, dx = \left[ x \right]_1^{\sqrt{2}} = \sqrt{2} - 1 \]
Thus, the value of the integral is \( \sqrt{2} - 1 \).
Thus, the correct answer is option (C).