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: Function Definition: \( \left\lfloor x^2 \right\rfloor \).
\nThe floor function \( \left\lfloor x^2 \right\rfloor \) returns the largest integer less than or equal to \( x^2 \).\n\nWe need to compute:\n\[\n\int_0^{1.5} \left\lfloor x^2 \right\rfloor \, dx\n\]\n\n
Step 2: Analyze \( x^2 \) on \( [0, 1.5] \).\n\n For \( 0 \leq x<1 \):\n \[\n 0 \leq x^2<1\n \quad \Rightarrow \quad\n \left\lfloor x^2 \right\rfloor = 0\n \]\n\n For \( 1 \leq x<\sqrt{2} \) (approximately \( 1.414 \)):\n \[\n 1 \leq x^2<2\n \quad \Rightarrow \quad\n \left\lfloor x^2 \right\rfloor = 1\n \]\n\n For \( \sqrt{2} \leq x \leq 1.5 \):\n \[\n 2 \leq x^2 \leq 2.25\n \quad \Rightarrow \quad\n \left\lfloor x^2 \right\rfloor = 2\n \]\n\n
Step 3: Integral Decomposition.
\n\nTherefore:\n\[\n\int_0^{1.5} \left\lfloor x^2 \right\rfloor \, dx = \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{1.5} 2 \, dx\n\]\n\n
Step 4: Evaluate the integral parts.
\n\n \[\n\int_0^1 0 \, dx = 0\n\]\n \[\n\int_1^{\sqrt{2}} 1 \, dx = \sqrt{2} - 1\n\]\n \[\n\int_{\sqrt{2}}^{1.5} 2 \, dx = 2(1.5 - \sqrt{2})\n\]\n\nSumming the results:\n\[\n0 + (\sqrt{2} - 1) + 2(1.5 - \sqrt{2})\n\]\nExpand:\n\[\n= \sqrt{2} - 1 + 3 - 2\sqrt{2}\n\]\n\[\n= (3 - 1) + (\sqrt{2} - 2\sqrt{2})\n\]\n\[\n= 2 - \sqrt{2}\n\]\n\nThus, the answer is \( 2 - \sqrt{2} \).