Key Concepts and Formulas

• Area Between Curves: The area between two curves $f(x)$ and $g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$, is given by $A = \int_a^b (f(x) - g(x)) \, dx$.
• Absolute Value Function: $|x-a| = \begin{cases} x-a, & \text{if } x \ge a \\ a-x, & \text{if } x < a \end{cases}$.
• Power Rule for Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$.

Step-by-Step Solution

Step 1: Identify the upper and lower curves and the intervals of integration.

The region is defined by $|x-5| \leq y \leq 4\sqrt{x}$. This means $y = 4\sqrt{x}$ is the upper curve and $y = |x-5|$ is the lower curve. The absolute value function changes its definition at $x=5$. Therefore, we need to consider two cases:

• For $x \geq 5$, $|x-5| = x-5$.
• For $x < 5$, $|x-5| = 5-x$.

We need to find the intersection points of $y = 4\sqrt{x}$ and $y = |x-5|$ to determine the limits of integration.

Step 2: Find the intersection points.

Case 1: $x \geq 5$, so $4\sqrt{x} = x-5$. Squaring both sides gives $16x = (x-5)^2 = x^2 - 10x + 25$. Rearranging, we get $x^2 - 26x + 25 = 0$, which factors to $(x-1)(x-25) = 0$. Thus, $x=1$ or $x=25$. Since we assumed $x \geq 5$, we discard $x=1$ and keep $x=25$.

Case 2: $x < 5$, so $4\sqrt{x} = 5-x$. Squaring both sides gives $16x = (5-x)^2 = x^2 - 10x + 25$. Rearranging, we get $x^2 - 26x + 25 = 0$, which factors to $(x-1)(x-25) = 0$. Thus, $x=1$ or $x=25$. Since we assumed $x < 5$, we discard $x=25$ and keep $x=1$.

Therefore, the intersection points occur at $x=1$ and $x=25$.

Step 3: Set up the integral for the area.

Since the definition of $|x-5|$ changes at $x=5$, we split the integral into two parts: $A = \int_1^5 (4\sqrt{x} - (5-x)) \, dx + \int_5^{25} (4\sqrt{x} - (x-5)) \, dx$ $A = \int_1^5 (4\sqrt{x} - 5 + x) \, dx + \int_5^{25} (4\sqrt{x} - x + 5) \, dx$

Step 4: Evaluate the integrals.

$A = \left[ 4 \cdot \frac{2}{3} x^{3/2} - 5x + \frac{1}{2}x^2 \right]_1^5 + \left[ 4 \cdot \frac{2}{3} x^{3/2} - \frac{1}{2}x^2 + 5x \right]_5^{25}$ $A = \left[ \frac{8}{3} x^{3/2} - 5x + \frac{1}{2}x^2 \right]_1^5 + \left[ \frac{8}{3} x^{3/2} - \frac{1}{2}x^2 + 5x \right]_5^{25}$ $A = \left( \frac{8}{3} (5)^{3/2} - 25 + \frac{25}{2} \right) - \left( \frac{8}{3} - 5 + \frac{1}{2} \right) + \left( \frac{8}{3} (25)^{3/2} - \frac{625}{2} + 125 \right) - \left( \frac{8}{3} (5)^{3/2} - \frac{25}{2} + 25 \right)$ $A = \frac{8}{3} (5\sqrt{5}) - 25 + \frac{25}{2} - \frac{8}{3} + 5 - \frac{1}{2} + \frac{8}{3} (125) - \frac{625}{2} + 125 - \frac{8}{3} (5\sqrt{5}) + \frac{25}{2} - 25$ $A = -25 + \frac{25}{2} - \frac{8}{3} + 5 - \frac{1}{2} + \frac{1000}{3} - \frac{625}{2} + 125 + \frac{25}{2} - 25$ $A = \frac{992}{3} - \frac{626}{2} + 80 = \frac{992}{3} - 313 + 80 = \frac{992}{3} - 233 = \frac{992 - 699}{3} = \frac{293}{3}$

Step 5: Calculate 3A.

We are asked to find $3A$. $3A = 3 \cdot \frac{293}{3} = 293$

Common Mistakes & Tips

• Extraneous Solutions: Remember to check for extraneous solutions when squaring equations, especially when dealing with square roots.
• Splitting the Integral: When dealing with absolute values, remember to split the integral into multiple parts based on where the expression inside the absolute value changes sign.
• Visualization: Sketching a graph of the functions can help you visualize the region and identify the upper and lower curves correctly.

Summary

We found the area of the region by identifying the upper and lower curves, finding their intersection points, and setting up and evaluating the definite integrals. Due to the absolute value in the definition of the region, we split the integral into two parts. Finally, we multiplied the area by 3 to obtain the desired result.

The final answer is \boxed{293}.