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: Understanding the Concept
The integral involves the greatest integer function (floor function). The value of $[x^2]$ is a step function that changes its value only at points where $x^2$ becomes an integer. To evaluate the definite integral, we must split the interval of integration $[1, 2]$ into subintervals based on where the value of $[x^2]$ is constant.
Step 2: Key Formula or Approach
1. Identify the integer values that $x^2$ takes as $x$ ranges from 1 to 2. As $x$ goes from 1 to 2, $x^2$ goes from $1^2=1$ to $2^2=4$. The integers are 1, 2, 3. 2. Find the values of $x$ where $x^2$ equals these integers: $x^2=1, x^2=2, x^2=3, x^2=4$. This gives $x=1, \sqrt{2}, \sqrt{3}, 2$. 3. Split the integral into subintervals: $[1, \sqrt{2}]$, $[\sqrt{2}, \sqrt{3}]$, and $[\sqrt{3}, 2]$. 4. In each subinterval, the value of $[x^2]$ is constant. Evaluate the integral over each subinterval and sum the results. \[ \int_1^2 [x^2] dx = \int_1^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] dx + \int_{\sqrt{3}}^{2} [x^2] dx \] Step 3: Detailed Explanation
We split the integral based on the integer values of $x^2$. - For $1 \le x<\sqrt{2}$, we have $1 \le x^2<2$, so $[x^2] = 1$. - For $\sqrt{2} \le x<\sqrt{3}$, we have $2 \le x^2<3$, so $[x^2] = 2$. - For $\sqrt{3} \le x<2$, we have $3 \le x^2<4$, so $[x^2] = 3$. - At $x=2$, $[x^2]=[4]=4$, but this single point does not affect the value of the definite integral. Now we can write the integral as a sum of integrals over these subintervals: \[ I = \int_1^{\sqrt{2}} 1 \,dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \,dx + \int_{\sqrt{3}}^{2} 3 \,dx \] Evaluate each integral: \[ \int_1^{\sqrt{2}} 1 \,dx = [x]_1^{\sqrt{2}} = \sqrt{2} - 1 \] \[ \int_{\sqrt{2}}^{\sqrt{3}} 2 \,dx = [2x]_{\sqrt{2}}^{\sqrt{3}} = 2\sqrt{3} - 2\sqrt{2} \] \[ \int_{\sqrt{3}}^{2} 3 \,dx = [3x]_{\sqrt{3}}^{2} = 3(2) - 3\sqrt{3} = 6 - 3\sqrt{3} \] Now, sum these results: \[ I = (\sqrt{2} - 1) + (2\sqrt{3} - 2\sqrt{2}) + (6 - 3\sqrt{3}) \] \[ I = (-1 + 6) + (\sqrt{2} - 2\sqrt{2}) + (2\sqrt{3} - 3\sqrt{3}) \] \[ I = 5 - \sqrt{2} - \sqrt{3} \] Step 4: Final Answer
The value of the integral is $5 - \sqrt{2} - \sqrt{3}$.