If the area of the region $$\{(x, y) : |x - 5| \le y \le 4\sqrt{x}\}$$ is A, then 3A is equal to ________.

To find the area of the region defined by the inequality

$$|x - 5| \le y \le 4\sqrt{x}$$

we first determine the range of $$x$$ for which the region exists by solving

$$|x - 5| \le 4\sqrt{x}$$

Step 1: Find the boundaries for $$x$$

We solve the inequality

$$|x - 5| \le 4\sqrt{x}$$

by cases.

Case 1: $$x \ge 5$$

$$x - 5 \le 4\sqrt{x} \implies x - 4\sqrt{x} - 5 \le 0$$

Let

$$t = \sqrt{x}$$

then

$$t^2 - 4t - 5 \le 0$$

Factoring gives

$$(t - 5)(t + 1) \le 0$$

Since

$$t = \sqrt{x} \ge 0$$

we have

$$0 \le t \le 5$$

which implies

$$0 \le x \le 25$$

Combined with

$$x \ge 5$$

we get

$$5 \le x \le 25$$

Case 2: $$x < 5$$

$$5 - x \le 4\sqrt{x} \implies x + 4\sqrt{x} - 5 \ge 0$$

Let

$$t = \sqrt{x}$$

then

$$t^2 + 4t - 5 \ge 0$$

Factoring gives

$$(t + 5)(t - 1) \ge 0$$

Since

$$t \ge 0$$

we must have

$$t \ge 1$$

which implies

$$x \ge 1$$

Combined with

$$x < 5$$

we get

$$1 \le x < 5$$

Combining both cases, the interval for $$x$$ is

$$1 \le x \le 25$$

Step 2: Set up the area integral

The area $$A$$ is the integral of the upper curve minus the lower curve:

$$A = \int_{1}^{5} (4\sqrt{x} - (5 - x)) dx + \int_{5}^{25} (4\sqrt{x} - (x - 5)) dx$$

$$A = \int_{1}^{5} (4\sqrt{x} + x - 5) dx + \int_{5}^{25} (4\sqrt{x} - x + 5) dx$$

Step 3: Evaluate the first integral $$I_1$$

$$I_1 = \int_{1}^{5} (4x^{1/2} + x - 5) dx = \left[ \frac{8}{3}x^{3/2} + \frac{x^2}{2} - 5x \right]_{1}^{5}$$

Evaluating at $$x = 5$$:

$$\left( \frac{8}{3}(5\sqrt{5}) + \frac{25}{2} - 25 \right) = \frac{40\sqrt{5}}{3} - \frac{25}{2}$$

Evaluating at $$x = 1$$:

$$\left( \frac{8}{3} + \frac{1}{2} - 5 \right) = \frac{16 + 3 - 30}{6} = -\frac{11}{6}$$

$$I_1 = \left( \frac{40\sqrt{5}}{3} - \frac{25}{2} \right) - \left( -\frac{11}{6} \right) = \frac{40\sqrt{5}}{3} - \frac{75 - 11}{6} = \frac{40\sqrt{5}}{3} - \frac{64}{6} = \frac{40\sqrt{5} - 32}{3}$$

Step 4: Evaluate the second integral $$I_2$$

$$I_2 = \int_{5}^{25} (4x^{1/2} - x + 5) dx = \left[ \frac{8}{3}x^{3/2} - \frac{x^2}{2} + 5x \right]_{5}^{25}$$

Evaluating at $$x = 25$$:

$$\left( \frac{8}{3}(125) - \frac{625}{2} + 125 \right) = \frac{1000}{3} - \frac{375}{2} = \frac{2000 - 1125}{6} = \frac{875}{6}$$

Evaluating at $$x = 5$$:

$$\left( \frac{8}{3}(5\sqrt{5}) - \frac{25}{2} + 25 \right) = \frac{40\sqrt{5}}{3} + \frac{25}{2} = \frac{80\sqrt{5} + 75}{6}$$

$$I_2 = \frac{875}{6} - \frac{80\sqrt{5} + 75}{6} = \frac{800 - 80\sqrt{5}}{6} = \frac{400 - 40\sqrt{5}}{3}$$

Step 5: Total Area $$A$$

$$A = I_1 + I_2 = \frac{40\sqrt{5} - 32}{3} + \frac{400 - 40\sqrt{5}}{3}$$

$$A = \frac{40\sqrt{5} - 32 + 400 - 40\sqrt{5}}{3} = \frac{368}{3}$$

Thus,

$$3A = 368$$