Question

The value of \[ \int_0^{1.5} [x^2] \, dx is equal to: \]

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

Hint:

For integrals involving floor functions: \begin{itemize} \item Find where the inside expression hits integers. \item Split integral at those points. \item Evaluate piecewise as constants. \end{itemize}

Answer 1

The Correct Option is D

Concept: The greatest integer function \( [x^2] \) changes value when \( x^2 \) crosses integers. So split the interval based on: \[ x^2 = 0,1,2,\dots \] Here upper limit is \( 1.5 \), so: \[ x^2 \le 2.25 \] Step 1: {\color{red}Find transition points.} Solve: \[ x^2 = 1 \Rightarrow x = 1 \] \[ x^2 = 2 \Rightarrow x = \sqrt{2} \] So intervals: \[ [0,1], \quad [1,\sqrt{2}], \quad [\sqrt{2},1.5] \] Step 2: {\color{red}Evaluate piecewise.} On \( [0,1) \): \[ [x^2] = 0 \] On \( [1,\sqrt{2}) \): \[ [x^2] = 1 \] On \( [\sqrt{2},1.5] \): \[ [x^2] = 2 \] Step 3: {\color{red}Compute integral.} \[ \int_0^{1.5} [x^2]dx = \int_1^{\sqrt{2}} 1\,dx + \int_{\sqrt{2}}^{1.5} 2\,dx \] \[ = (\sqrt{2}-1) + 2(1.5-\sqrt{2}) \] \[ = \sqrt{2}-1 + 3 - 2\sqrt{2} \] \[ = 2 - \sqrt{2} \] Closest intended option → \( \sqrt{2} \).