Find the area of the region between the following curves. If necessary, break the region into subregions first. $y=6-x, y=\sqrt{x}$, and $y=1$ Write the exact answer. Area:

Step 1 ：Set \(y = 6 - x\) and \(y = \sqrt{x}\) equal to each other to find the points of intersection.

Step 2 ：Solve the equation \(6 - x = \sqrt{x}\).

Step 3 ：Square both sides to get \(36 - 12x + x^2 = x\).

Step 4 ：Rearrange terms to get \(x^2 - 13x + 36 = 0\).

Step 5 ：Factor the equation to get \((x - 4)(x - 9) = 0\).

Step 6 ：Solve for x to get \(x = 4\) and \(x = 9\).

Step 7 ：Set \(y = 1\) and \(y = \sqrt{x}\) equal to each other to find the points of intersection.

Step 8 ：Solve the equation \(1 = \sqrt{x}\) to get \(x = 1\).

Step 9 ：Break the region into two subregions: \(1 \leq x \leq 4\) and \(4 \leq x \leq 9\).

Step 10 ：For \(1 \leq x \leq 4\), calculate the area using the integral \(\int_{1}^{4} [(6 - x) - 1] dx = \int_{1}^{4} (5 - x) dx\).

Step 11 ：Evaluate the integral to get \([5x - \frac{1}{2}x^2]_{1}^{4} = (20 - 8) - (5 - 0.5) = 12 - 4.5 = 7.5\).

Step 12 ：For \(4 \leq x \leq 9\), calculate the area using the integral \(\int_{4}^{9} [(6 - x) - \sqrt{x}] dx\).

Step 13 ：Evaluate the integral to get \([6x - \frac{1}{2}x^2 - \frac{2}{3}x^{3/2}]_{4}^{9} = (54 - 40.5 - 18) - (24 - 8 - 5.33) = -9.83 - 10.67 = -20.5\).

Step 14 ：Since area cannot be negative, take the absolute value to get \(20.5\).

Step 15 ：Add the two areas together to get \(7.5 + 20.5 = 28\).

Step 16 ：\(\boxed{28}\) is the area of the region between the curves.