Key Concepts and Formulas

• Area Under a Curve: The area between a curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by $\int_a^b |f(x)| \, dx$.
• Absolute Value: $|x| = \begin{cases} x, & \text{if } x \ge 0 \\ -x, & \text{if } x < 0 \end{cases}$
• Maximum Function: $\max\{a, b\} = \begin{cases} a, & \text{if } a \ge b \\ b, & \text{if } a < b \end{cases}$

Step-by-Step Solution

Step 1: Define the function y = max{|x|, x|x-2|} piecewise.

We need to determine when $|x| \ge x|x-2|$ and when $|x| < x|x-2|$ to express the function as a piecewise function.

Case 1: $x \ge 0$. Then $|x| = x$, and $y = \max\{x, x|x-2|\}$.

• If $0 \le x \le 2$, then $|x-2| = 2-x$, so $x|x-2| = x(2-x) = 2x - x^2$. We need to determine when $x \ge 2x - x^2$, which simplifies to $x^2 - x \ge 0$, or $x(x-1) \ge 0$. This is true when $x \le 0$ or $x \ge 1$. Since $0 \le x \le 2$, we have $1 \le x \le 2$. Thus, $y = \begin{cases} x, & \text{if } 0 \le x \le 1 \\ 2x - x^2, & \text{if } 1 < x \le 2 \end{cases}$

• If $x > 2$, then $|x-2| = x-2$, so $x|x-2| = x(x-2) = x^2 - 2x$. We need to determine when $x \ge x^2 - 2x$, which simplifies to $x^2 - 3x \le 0$, or $x(x-3) \le 0$. This is true when $0 \le x \le 3$. Since $x > 2$, we have $2 < x \le 3$. Thus, $y = \begin{cases} x, & \text{if } 2 < x \le 3 \\ x^2 - 2x, & \text{if } x > 3 \end{cases}$

Case 2: $x < 0$. Then $|x| = -x$, and $y = \max\{-x, x|x-2|\}$. Since $x < 0$, $|x-2| = 2-x$, so $x|x-2| = x(2-x) = 2x - x^2$. We need to determine when $-x \ge 2x - x^2$, which simplifies to $x^2 - 3x \ge 0$, or $x(x-3) \ge 0$. This is true when $x \le 0$ or $x \ge 3$. Since $x < 0$, we have $x < 0$. Thus, $y = -x$.

Combining the cases, we have $y = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } 0 \le x \le 1 \\ 2x - x^2, & \text{if } 1 < x \le 2 \\ x, & \text{if } 2 < x \le 3 \\ x^2 - 2x, & \text{if } x > 3 \end{cases}$

Step 2: Calculate the area.

The area is given by $\int_{-2}^4 y \, dx = \int_{-2}^4 \max\{|x|, x|x-2|\} \, dx$. Using the piecewise definition: $\text{Area} = \int_{-2}^0 -x \, dx + \int_0^1 x \, dx + \int_1^2 (2x - x^2) \, dx + \int_2^3 x \, dx + \int_3^4 (x^2 - 2x) \, dx$

$\text{Area} = \left[ -\frac{x^2}{2} \right]_{-2}^0 + \left[ \frac{x^2}{2} \right]_0^1 + \left[ x^2 - \frac{x^3}{3} \right]_1^2 + \left[ \frac{x^2}{2} \right]_2^3 + \left[ \frac{x^3}{3} - x^2 \right]_3^4$

$\text{Area} = \left[ 0 - (-\frac{4}{2}) \right] + \left[ \frac{1}{2} - 0 \right] + \left[ (4 - \frac{8}{3}) - (1 - \frac{1}{3}) \right] + \left[ \frac{9}{2} - \frac{4}{2} \right] + \left[ (\frac{64}{3} - 16) - (\frac{27}{3} - 9) \right]$

$\text{Area} = 2 + \frac{1}{2} + \left[ \frac{4}{3} - \frac{2}{3} \right] + \frac{5}{2} + \left[ \frac{16}{3} - 0 \right] = 2 + \frac{1}{2} + \frac{2}{3} + \frac{5}{2} + \frac{16}{3}$

$\text{Area} = 2 + \frac{6}{2} + \frac{18}{3} = 2 + 3 + 6 = 11$

Step 3: Re-evaluate the solution, given the answer is 1.

There seems to be an error in the above calculation. We will re-evaluate. $y = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } 0 \le x \le 1 \\ 2x - x^2, & \text{if } 1 < x \le 2 \\ x, & \text{if } 2 < x \le 3 \\ x^2 - 2x, & \text{if } x > 3 \end{cases}$

$\text{Area} = \int_{-2}^0 -x \, dx + \int_0^1 x \, dx + \int_1^2 (2x - x^2) \, dx + \int_2^3 x \, dx + \int_3^4 (x^2 - 2x) \, dx$

$\text{Area} = \left[ -\frac{x^2}{2} \right]_{-2}^0 + \left[ \frac{x^2}{2} \right]_0^1 + \left[ x^2 - \frac{x^3}{3} \right]_1^2 + \left[ \frac{x^2}{2} \right]_2^3 + \left[ \frac{x^3}{3} - x^2 \right]_3^4$

$\text{Area} = \left[ 0 - (-\frac{4}{2}) \right] + \left[ \frac{1}{2} - 0 \right] + \left[ (4 - \frac{8}{3}) - (1 - \frac{1}{3}) \right] + \left[ \frac{9}{2} - \frac{4}{2} \right] + \left[ (\frac{64}{3} - 16) - (\frac{27}{3} - 9) \right]$

$\text{Area} = 2 + \frac{1}{2} + \left[ \frac{4}{3} - \frac{2}{3} \right] + \frac{5}{2} + \left[ \frac{16}{3} - 0 \right] = 2 + \frac{1}{2} + \frac{2}{3} + \frac{5}{2} + \frac{16}{3}$

$\text{Area} = 2 + 3 + \frac{18}{3} = 2+3+6 = 11$

The error must be in the piecewise definition. Let's re-evaluate that. $|x| \ge x|x-2|$.

If $x \ge 0$: $1 \ge |x-2|$. This means $-1 \le x-2 \le 1$, so $1 \le x \le 3$. If $x < 0$: $-1 \ge |x-2|$. This is never true, since $|x-2|$ is always non-negative.

So $y = \begin{cases} |x|, & \text{if } 1 \le x \le 3 \\ x|x-2|, & \text{otherwise} \end{cases}$

$y = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } 0 \le x < 1 \\ x, & \text{if } 1 \le x \le 3 \\ x^2-2x, & \text{if } 3 < x \le 4 \end{cases}$

$\text{Area} = \int_{-2}^0 -x \, dx + \int_0^1 x \, dx + \int_1^3 x \, dx + \int_3^4 (x^2 - 2x) \, dx$

$\text{Area} = \left[ -\frac{x^2}{2} \right]_{-2}^0 + \left[ \frac{x^2}{2} \right]_0^1 + \left[ \frac{x^2}{2} \right]_1^3 + \left[ \frac{x^3}{3} - x^2 \right]_3^4$ $= 2 + \frac{1}{2} + (\frac{9}{2} - \frac{1}{2}) + (\frac{64}{3} - 16 - (9-9)) = 2 + \frac{1}{2} + 4 + \frac{16}{3} = 6 + \frac{1}{2} + \frac{16}{3} = 6 + \frac{3 + 32}{6} = 6 + \frac{35}{6} = \frac{71}{6}$

The area is given by $\int_{-2}^4 \max\{|x|, x|x-2|\} \, dx$.

If $x \in [-2, 0]$, $\max\{|x|, x|x-2|\} = \max\{-x, x(2-x)\} = \max\{-x, 2x-x^2\}$. $-x \ge 2x-x^2$ when $x^2-3x \ge 0$, i.e. $x(x-3) \ge 0$. Since $x<0$, this is true.

If $x \in [0, 2]$, $\max\{|x|, x|x-2|\} = \max\{x, x(2-x)\} = \max\{x, 2x-x^2\}$. $x \ge 2x-x^2$ when $x^2-x \ge 0$, i.e. $x(x-1) \ge 0$. So $x \in [0,1]$.

If $x \in [2, 4]$, $\max\{|x|, x|x-2|\} = \max\{x, x(x-2)\} = \max\{x, x^2-2x\}$. $x \ge x^2-2x$ when $x^2-3x \le 0$, i.e. $x(x-3) \le 0$. So $x \in [2, 3]$.

$\int_{-2}^0 -x dx + \int_0^1 x dx + \int_1^2 (2x-x^2) dx + \int_2^3 x dx + \int_3^4 (x^2-2x) dx = 2 + \frac{1}{2} + \frac{2}{3} + \frac{5}{2} + \frac{16}{3} = 11$.

Let us plot the functions. The max is the x axis. Therefore the area is zero. The correct answer is not 1.

If the correct answer is 1, there is a mistake in the question.

The correct answer is 11.

Common Mistakes & Tips

• Carefully consider the absolute value and maximum functions to define the piecewise function correctly.
• Double-check the limits of integration for each piece of the integral.
• Be meticulous with the arithmetic when evaluating the definite integrals.

Summary

We first expressed the function $y = \max\{|x|, x|x-2|\}$ as a piecewise function. Then, we calculated the area under the curve by splitting the integral into intervals based on the piecewise definition. After evaluating the integrals, we obtained the area.

Final Answer

The final answer is \boxed{11}. There is no correct option.