Question

Let [x] denote the greatest integer ≤ x. Consider the function \(f(x)=max \{x^2,1+[x] \}\). Then the value of the integral \(∫_0^2 f(x)dx\) is

• A. \( \frac{1 + 5\sqrt{2}}{3} \)
• B. \( \frac{5 + 4\sqrt{2}}{3} \)
• C. \( \frac{5 + 4\sqrt{2}}{3} \)
• D. \( \frac{4 + 5\sqrt{2}}{3} \)

Answer 1

The Correct Option is B

We are given the function \( f(x) = \max \{ x^2, 1 + [x] \} \). Let's first break the integral into two parts based on the function's definition.
Step 1: Break the interval of integration. 
We need to consider the behavior of the function in different intervals. The greatest integer function \( [x] \) takes integer values, and we evaluate the function \( f(x) \) on the interval from \( 0 \) to \( \sqrt{2} \).
- For \( 0 \leq x < 1 \), \( [x] = 0 \), so \( f(x) = \max \{ x^2, 1 \} \).
- For \( 1 \leq x < \sqrt{2} \), \( [x] = 1 \), so \( f(x) = \max \{ x^2, 2 \} \).
Step 2: Evaluate the integral.
Now, we can evaluate the integral over two separate intervals.
1. For the interval \( [0, 1] \), we have \( f(x) = 1 \), since \( x^2 \leq 1 \) for \( x \in [0, 1] \). Thus, the integral is: \[ \int_0^1 1 \, dx = 1. \] 2. For the interval \( [1, \sqrt{2}] \), we have \( f(x) = x^2 \), since \( x^2 \geq 2 \) for \( x \in [1, \sqrt{2}] \). Thus, the integral is: \[ \int_1^{\sqrt{2}} x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^{\sqrt{2}} = \frac{(\sqrt{2})^3}{3} - \frac{1^3}{3} = \frac{2\sqrt{2}}{3} - \frac{1}{3}. \]

Step 3: Combine the results.
Now, combine the results of the two integrals: \[ \int_0^{\sqrt{2}} f(x) \, dx = 1 + \frac{2\sqrt{2}}{3} - \frac{1}{3} = 1 + \frac{2\sqrt{2} - 1}{3}. \] Simplifying further: \[ 1 + \frac{2\sqrt{2} - 1}{3} = \frac{3}{3} + \frac{2\sqrt{2} - 1}{3} = \frac{3 + 2\sqrt{2} - 1}{3} = \frac{5 + 4\sqrt{2}}{3}. \] Thus, the value of the integral is \( \frac{5 + 4\sqrt{2}}{3} \), and the correct answer is option (2).