Key Concepts and Formulas

• Area between curves: The area between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$, where $f(x) \geq g(x)$ on $[a, b]$, is given by $\int_a^b (f(x) - g(x)) dx$.
• Absolute value: $|x^2 - 2| = x^2 - 2$ when $x^2 \geq 2$ and $|x^2 - 2| = 2 - x^2$ when $x^2 < 2$.
• Solving inequalities: Understanding how to solve inequalities to determine the intervals of integration.

Step-by-Step Solution

Step 1: Analyze the inequalities and find intersection points

We are given the region defined by $|x^2 - 2| \leq y \leq x$. We need to find the intersection points of the curves $y = |x^2 - 2|$ and $y = x$. First, consider $x^2 - 2 \geq 0$, which means $x \leq -\sqrt{2}$ or $x \geq \sqrt{2}$. In this case, $|x^2 - 2| = x^2 - 2$. So we solve $x^2 - 2 = x$, which gives $x^2 - x - 2 = 0$, or $(x-2)(x+1) = 0$. The solutions are $x = 2$ and $x = -1$. Since we require $x \leq -\sqrt{2}$ or $x \geq \sqrt{2}$, only $x = 2$ is a valid solution. Next, consider $x^2 - 2 < 0$, which means $-\sqrt{2} < x < \sqrt{2}$. In this case, $|x^2 - 2| = 2 - x^2$. So we solve $2 - x^2 = x$, which gives $x^2 + x - 2 = 0$, or $(x+2)(x-1) = 0$. The solutions are $x = -2$ and $x = 1$. Since we require $-\sqrt{2} < x < \sqrt{2}$, only $x = 1$ is a valid solution. So the intersection points are at $x = 1$ and $x = 2$.

Step 2: Determine the intervals and the integrand

The area $A$ can be divided into two parts: from $x=1$ to $x=\sqrt{2}$, and from $x=\sqrt{2}$ to $x=2$. For $1 \leq x \leq \sqrt{2}$, we have $|x^2 - 2| = 2 - x^2$. Also, $x \geq 2-x^2$ in this interval. Thus, the area is $\int_1^{\sqrt{2}} (x - (2 - x^2)) dx = \int_1^{\sqrt{2}} (x^2 + x - 2) dx$. For $\sqrt{2} \leq x \leq 2$, we have $|x^2 - 2| = x^2 - 2$. Also, $x \geq x^2 - 2$ in this interval. Thus, the area is $\int_{\sqrt{2}}^2 (x - (x^2 - 2)) dx = \int_{\sqrt{2}}^2 (-x^2 + x + 2) dx$.

Step 3: Calculate the integrals

The first integral is $\int_1^{\sqrt{2}} (x^2 + x - 2) dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} - 2x \right]_1^{\sqrt{2}} = \left( \frac{2\sqrt{2}}{3} + 1 - 2\sqrt{2} \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right) = \frac{2\sqrt{2}}{3} + 1 - 2\sqrt{2} - \frac{1}{3} - \frac{1}{2} + 2 = 3 + \frac{2\sqrt{2}}{3} - 2\sqrt{2} - \frac{5}{6} = \frac{13}{6} - \frac{4\sqrt{2}}{3}.$ The second integral is $\int_{\sqrt{2}}^2 (-x^2 + x + 2) dx = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{\sqrt{2}}^2 = \left( -\frac{8}{3} + 2 + 4 \right) - \left( -\frac{2\sqrt{2}}{3} + 1 + 2\sqrt{2} \right) = -\frac{8}{3} + 6 + \frac{2\sqrt{2}}{3} - 1 - 2\sqrt{2} = 5 - \frac{8}{3} + \frac{2\sqrt{2}}{3} - 2\sqrt{2} = \frac{7}{3} - \frac{4\sqrt{2}}{3}.$

Step 4: Calculate the total area

The total area $A$ is the sum of the two integrals: $A = \left( \frac{13}{6} - \frac{4\sqrt{2}}{3} \right) + \left( \frac{7}{3} - \frac{4\sqrt{2}}{3} \right) = \frac{13}{6} + \frac{14}{6} - \frac{8\sqrt{2}}{3} = \frac{27}{6} - \frac{8\sqrt{2}}{3} = \frac{9}{2} - \frac{8\sqrt{2}}{3}.$

Step 5: Calculate the expression 6A + 16√2

Now we need to calculate $6A + 16\sqrt{2}$: $6A + 16\sqrt{2} = 6 \left( \frac{9}{2} - \frac{8\sqrt{2}}{3} \right) + 16\sqrt{2} = 27 - 16\sqrt{2} + 16\sqrt{2} = 27.$

Step 6: Sanity Check The provided answer is 2. There appears to be an error in the original question. It should have been: If the area of the region $\left\{(x, \mathrm{y}):\left|x^{2}-2\right| \leq y \leq x\right\}$ is $\mathrm{A}$, then $6 \mathrm{A} = 27 - 16\sqrt{2}$. Then we require $6A + 16\sqrt{2}$, which is $27 - 16\sqrt{2} + 16\sqrt{2} = 27$. The answer provided in the question is incorrect. Let's work backwards.

We know $6A + 16\sqrt{2} = c$, where $c = 2$. So, $6A = 2 - 16\sqrt{2}$. $A = \frac{1}{3} - \frac{8}{3}\sqrt{2}$.

The area we calculated was $A = \frac{9}{2} - \frac{8\sqrt{2}}{3}$. Something is very wrong with this problem.

Let's recalculate the integrals:

$\int_1^{\sqrt{2}} (x^2 + x - 2) dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} - 2x \right]_1^{\sqrt{2}} = \left( \frac{2\sqrt{2}}{3} + 1 - 2\sqrt{2} \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right) = \frac{2\sqrt{2}}{3} + 1 - 2\sqrt{2} - \frac{1}{3} - \frac{1}{2} + 2 = 3 + \frac{2\sqrt{2}}{3} - 2\sqrt{2} - \frac{5}{6} = \frac{13}{6} - \frac{4\sqrt{2}}{3}.$ $\int_{\sqrt{2}}^2 (-x^2 + x + 2) dx = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{\sqrt{2}}^2 = \left( -\frac{8}{3} + 2 + 4 \right) - \left( -\frac{2\sqrt{2}}{3} + 1 + 2\sqrt{2} \right) = -\frac{8}{3} + 6 + \frac{2\sqrt{2}}{3} - 1 - 2\sqrt{2} = 5 - \frac{8}{3} + \frac{2\sqrt{2}}{3} - 2\sqrt{2} = \frac{7}{3} - \frac{4\sqrt{2}}{3}.$

The total area $A$ is the sum of the two integrals: $A = \left( \frac{13}{6} - \frac{4\sqrt{2}}{3} \right) + \left( \frac{7}{3} - \frac{4\sqrt{2}}{3} \right) = \frac{13}{6} + \frac{14}{6} - \frac{8\sqrt{2}}{3} = \frac{27}{6} - \frac{8\sqrt{2}}{3} = \frac{9}{2} - \frac{8\sqrt{2}}{3}.$

$6A + 16\sqrt{2} = 6 \left( \frac{9}{2} - \frac{8\sqrt{2}}{3} \right) + 16\sqrt{2} = 27 - 16\sqrt{2} + 16\sqrt{2} = 27.$

Let's assume the answer is 27 and work backward to determine the correct question.

If the area of the region $\left\{(x, \mathrm{y}):\left|x^{2}-2\right| \leq y \leq x\right\}$ is $\mathrm{A}$, then $6 \mathrm{A}+16 \sqrt{2}$ is equal to __________.

We calculated $6A + 16\sqrt{2} = 27$.

If the question was: If the area of the region $\left\{(x, \mathrm{y}):\left|x^{2}-2\right| \leq y \leq x\right\}$ is $\mathrm{A}$, then $6 \mathrm{A}+16 \sqrt{2}$ is equal to 27.

Let's make an assumption that the intended question was to find the closest integer to 6A. $A = \frac{9}{2} - \frac{8\sqrt{2}}{3} = 4.5 - \frac{8(1.414)}{3} \approx 4.5 - 3.77 = 0.73.$ $6A \approx 6(0.73) = 4.38.$ The closest integer is 4.

Common Mistakes & Tips

• Be careful with absolute values. Always consider different cases.
• Sketching the region is helpful for visualizing the problem.
• Double-check the limits of integration.

Summary

We found the intersection points of the curves $y = |x^2 - 2|$ and $y = x$. Then we split the area into two integrals based on the absolute value function. We calculated the integrals and summed them to find the total area $A = \frac{9}{2} - \frac{8\sqrt{2}}{3}$. Finally, we calculated the expression $6A + 16\sqrt{2}$ which simplifies to 27. However, the question states that the correct answer is 2. There appears to be an error in the original question.

Final Answer

The question is incorrect. If the question was to find the value of $6A + 16\sqrt{2}$, the final answer is \boxed{27}.