Question

If $[\cdot]$ denotes the greatest integer function, then $\int_1^2 [x^2] dx =$

• A. $5+\sqrt{2}+\sqrt{3}$
• B. $5+\sqrt{2}-\sqrt{3}$
• C. $5-\sqrt{2}-\sqrt{3}$
• D. $5-\sqrt{2}+\sqrt{3}$

Hint:

When integrating the greatest integer function $[f(x)]$, the key is to break the interval of integration at the points where $f(x)$ takes on integer values. For each subinterval created, $[f(x)]$ will be a constant, making the integration straightforward.

Answer 1

The Correct Option is C

Step 1: Identify intervals The function \( [x^2] \) is constant between values of \( x \) where \( x^2 \) is an integer. For \( x \in [1, 2] \), \( x^2 \in [1, 4] \). Integers for \( x^2 \) are 1, 2, 3. Break points: \( x = 1, \sqrt{2}, \sqrt{3}, 2 \).
Step 2: Split the integral \[ I = \int_1^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] dx + \int_{\sqrt{3}}^2 [x^2] dx \] - In \( [1, \sqrt{2}) \), \( 1 \le x^2<2 \implies [x^2] = 1 \). - In \( [\sqrt{2}, \sqrt{3}) \), \( 2 \le x^2<3 \implies [x^2] = 2 \). - In \( [\sqrt{3}, 2) \), \( 3 \le x^2<4 \implies [x^2] = 3 \).
Step 3: Calculate values \[ I = \int_1^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^2 3 dx \] \[ I = 1(\sqrt{2}-1) + 2(\sqrt{3}-\sqrt{2}) + 3(2-\sqrt{3}) \] \[ I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} \] \[ I = (6-1) + (\sqrt{2}-2\sqrt{2}) + (2\sqrt{3}-3\sqrt{3}) \] \[ I = 5 - \sqrt{2} - \sqrt{3} \]