Question

The value of \( \displaystyle \int_0^{1.5} \left\lfloor x^2 \right\rfloor \, dx \) is:

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

Hint:

When integrating a floor function, split the integral at points where the floor value changes (discontinuities), and compute each interval separately.

Answer 1

The Correct Option is B

Step 1: Understand the function \( \left\lfloor x^2 \right\rfloor \).
The floor function \( \left\lfloor x^2 \right\rfloor \) gives the greatest integer less than or equal to \( x^2 \). We need to evaluate: \[ \int_0^{1.5} \left\lfloor x^2 \right\rfloor \, dx \]
Step 2: Analyze \( x^2 \) over the range \( [0, 1.5] \). For \( 0 \leq x<1 \): \[ 0 \leq x^2<1 \quad \Rightarrow \quad \left\lfloor x^2 \right\rfloor = 0 \] For \( 1 \leq x<\sqrt{2} \) (approximately \( 1.414 \)): \[ 1 \leq x^2<2 \quad \Rightarrow \quad \left\lfloor x^2 \right\rfloor = 1 \] For \( \sqrt{2} \leq x \leq 1.5 \): \[ 2 \leq x^2 \leq 2.25 \quad \Rightarrow \quad \left\lfloor x^2 \right\rfloor = 2 \]
Step 3: Break the integral accordingly.
Thus: \[ \int_0^{1.5} \left\lfloor x^2 \right\rfloor \, dx = \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{1.5} 2 \, dx \]
Step 4: Evaluate each part.
\[ \int_0^1 0 \, dx = 0 \] \[ \int_1^{\sqrt{2}} 1 \, dx = \sqrt{2} - 1 \] \[ \int_{\sqrt{2}}^{1.5} 2 \, dx = 2(1.5 - \sqrt{2}) \] Adding all parts: \[ 0 + (\sqrt{2} - 1) + 2(1.5 - \sqrt{2}) \] Expand: \[ = \sqrt{2} - 1 + 3 - 2\sqrt{2} \] \[ = (3 - 1) + (\sqrt{2} - 2\sqrt{2}) \] \[ = 2 - \sqrt{2} \] Thus, the final value is \( 2 - \sqrt{2} \).