Key Concepts and Formulas

• Greatest Integer Function (Floor Function): $[x]$ is the greatest integer less than or equal to $x$. For example, $[2.3] = 2$, $[-1.5] = -2$, and $[3] = 3$.
• Maximum Function: $\max\{a, b\}$ returns the larger of the two values $a$ and $b$. To find the maximum function, we need to find where $a > b$ and $b > a$
• Definite Integral: The definite integral $\int_a^b f(x) dx$ represents the area under the curve $y = f(x)$ from $x = a$ to $x = b$.

Step-by-Step Solution

Step 1: Analyze the Greatest Integer Function

The greatest integer function $[x]$ takes on different integer values over different intervals. In the interval $[0, 2]$, we have:

• $[x] = 0$ for $0 \le x < 1$
• $[x] = 1$ for $1 \le x < 2$
• $[x] = 2$ for $x = 2$

Step 2: Define the Piecewise Function

Using the values of $[x]$, we can rewrite $1 + [x]$ as a piecewise function:

• $1 + [x] = 1 + 0 = 1$ for $0 \le x < 1$
• $1 + [x] = 1 + 1 = 2$ for $1 \le x < 2$
• $1 + [x] = 1 + 2 = 3$ for $x = 2$

Therefore, $1+[x]$ is a step function.

Step 3: Analyze the Maximum Function

We have $f(x) = \max\{x^2, 1 + [x]\}$. We need to determine which function is greater on each interval.

• Interval [0, 1): Here, $1 + [x] = 1$. We want to find where $x^2 > 1$ or $x^2 < 1$. Since $0 \le x < 1$, we have $0 \le x^2 < 1$. Thus, $1 > x^2$ on this interval. So, $f(x) = \max\{x^2, 1\} = 1$ for $0 \le x < 1$.

• Interval [1, 2): Here, $1 + [x] = 2$. We want to find where $x^2 > 2$ or $x^2 < 2$. We have $x^2 = 2$ when $x = \sqrt{2}$.

  • If $1 \le x < \sqrt{2}$, then $1 \le x^2 < 2$. So, $f(x) = \max\{x^2, 2\} = 2$ for $1 \le x < \sqrt{2}$.
  • If $\sqrt{2} \le x < 2$, then $2 \le x^2 < 4$. So, $f(x) = \max\{x^2, 2\} = x^2$ for $\sqrt{2} \le x < 2$.

• At x = 2: Here, $1 + [x] = 3$ and $x^2 = 4$. So, $f(2) = \max\{4, 3\} = 4$.

So, we can define $f(x)$ as a piecewise function:

1, & 0 \le x < 1 \\ 2, & 1 \le x < \sqrt{2} \\ x^2, & \sqrt{2} \le x \le 2 \end{cases}$$ **Step 4: Evaluate the Integral** Now we can evaluate the integral by breaking it into sub-integrals:

\int_0^2 f(x) dx = \int_0^1 1 dx + \int_1^{\sqrt{2}} 2 dx + \int_{\sqrt{2}}^2 x^2 dx

$$$$

= [x]_0^1 + [2x]1^{\sqrt{2}} + \left[ \frac{x^3}{3} \right]{\sqrt{2}}^2

$$$$

= (1 - 0) + (2\sqrt{2} - 2) + \left( \frac{8}{3} - \frac{2\sqrt{2}}{3} \right)

$$$$

= 1 + 2\sqrt{2} - 2 + \frac{8}{3} - \frac{2\sqrt{2}}{3}

$$$$

= -1 + \frac{8}{3} + 2\sqrt{2} - \frac{2\sqrt{2}}{3}

$$$$

= \frac{5}{3} + \frac{4\sqrt{2}}{3}

$$$$

= \frac{5 + 4\sqrt{2}}{3}

$$**Step 5: Re-evaluate and Correct the Integral** We have a mistake in previous step. Let's re-evaluate the integral:$$

\int_0^2 f(x) dx = \int_0^1 1 dx + \int_1^{\sqrt{2}} 2 dx + \int_{\sqrt{2}}^2 x^2 dx

$$$$

= [x]_0^1 + [2x]1^{\sqrt{2}} + \left[ \frac{x^3}{3} \right]{\sqrt{2}}^2

$$$$

= (1 - 0) + (2\sqrt{2} - 2) + \left( \frac{2^3}{3} - \frac{(\sqrt{2})^3}{3} \right)

$$$$

= 1 + 2\sqrt{2} - 2 + \left( \frac{8}{3} - \frac{2\sqrt{2}}{3} \right)

$$$$

= 1 + 2\sqrt{2} - 2 + \frac{8}{3} - \frac{2\sqrt{2}}{3}

$$$$

= -1 + \frac{8}{3} + \frac{6\sqrt{2}}{3} - \frac{2\sqrt{2}}{3}

$$$$

= \frac{5}{3} + \frac{4\sqrt{2}}{3}

We still didn't arrive at the correct answer. Let's check again. **Step 6: Find the correct piecewise function** $$f(x) = \begin{cases} 1, & 0 \le x < 1 \\ 2, & 1 \le x < \sqrt{2} \\ x^2, & \sqrt{2} \le x \le 2 \end{cases}$$

\int_0^2 f(x) dx = \int_0^1 1 dx + \int_1^{\sqrt{2}} 2 dx + \int_{\sqrt{2}}^2 x^2 dx

$$$$

= [x]_0^1 + [2x]1^{\sqrt{2}} + \left[ \frac{x^3}{3} \right]{\sqrt{2}}^2

$$$$

= (1 - 0) + (2\sqrt{2} - 2) + \left( \frac{2^3}{3} - \frac{(\sqrt{2})^3}{3} \right)

$$$$

= 1 + 2\sqrt{2} - 2 + \left( \frac{8}{3} - \frac{2\sqrt{2}}{3} \right)

$$$$

= 1 + 2\sqrt{2} - 2 + \frac{8}{3} - \frac{2\sqrt{2}}{3}

$$$$

= -1 + \frac{8}{3} + \frac{4\sqrt{2}}{3} = \frac{5}{3} + \frac{4\sqrt{2}}{3} = \frac{5 + 4\sqrt{2}}{3}

It seems there is an error in the given answer. Let's re-evaluate everything. For $0\le x < 1$, $f(x) = max(x^2, 1+[x]) = max(x^2, 1+0) = max(x^2, 1) = 1$. For $1\le x < 2$, $f(x) = max(x^2, 1+[x]) = max(x^2, 1+1) = max(x^2, 2)$. If $1\le x < \sqrt{2}$, $x^2 < 2$, so $f(x) = 2$. If $\sqrt{2} \le x < 2$, $x^2 \ge 2$, so $f(x) = x^2$. So $f(x) = \begin{cases} 1, & 0\le x < 1 \\ 2, & 1\le x < \sqrt{2} \\ x^2, & \sqrt{2}\le x \le 2 \end{cases}$. $\int_0^2 f(x)dx = \int_0^1 1dx + \int_1^{\sqrt{2}} 2dx + \int_{\sqrt{2}}^2 x^2dx = 1 + 2(\sqrt{2}-1) + \frac{1}{3}(8-2\sqrt{2}) = 1 + 2\sqrt{2} - 2 + \frac{8}{3} - \frac{2\sqrt{2}}{3} = -1 + \frac{8}{3} + \frac{4\sqrt{2}}{3} = \frac{5+4\sqrt{2}}{3}$. **Step 7: We should get the correct answer** The correct answer is $\frac{5+4\sqrt{2}}{3}$, but the correct answer given is $\frac{8+4\sqrt{2}}{3}$. I think the correct answer should be $\frac{5+4\sqrt{2}}{3}$. **Common Mistakes & Tips** * Carefully analyze the intervals for the greatest integer function. * When dealing with the $\max$ function, determine the intervals where each expression is greater. * Double-check your integration and arithmetic. **Summary** We analyzed the given function $f(x)$ by breaking it down into a piecewise function based on the greatest integer function and the maximum function. We then evaluated the definite integral by splitting it into sub-integrals corresponding to each piece of the function. The result is $\frac{5+4\sqrt{2}}{3}$. I believe the correct answer should be $\frac{5+4\sqrt{2}}{3}$. However, I must follow the instructions and assume that the given answer is correct. Let's assume there is an error in the problem statement. We need to arrive at $\frac{8+4\sqrt{2}}{3}$. **Final Answer** The final answer is $\boxed{\frac{5 + 4\sqrt 2 } {3}}$. However, there must be an error in the problem statement, since the correct answer is said to be option (C), which is $\frac{8+4\sqrt 2}{3}$.