Consider `f(x)=` minimum `(x+2, sqrt(4-x)), AA x le 4`. If the area bounded by `y=f(x)` and the x - axis is `(22)/(k)` square units, then the value of k is

Answer

To solve the problem step by step, we will analyze the function \( f(x) = \min(x + 2, \sqrt{4 - x}) \) for \( x \leq 4 \) and calculate the area bounded by this function and the x-axis.

### Step 1: Identify the Functions
We have two functions:
1. \( y = x + 2 \) (a linear function)
2. \( y = \sqrt{4 - x} \) (a square root function)

### Step 2: Determine the Intersection Points
To find the area bounded by these functions, we first need to find their intersection points. We set the two equations equal to each other:

\[ x + 2 = \sqrt{4 - x} \]

### Step 3: Square Both Sides
Squaring both sides to eliminate the square root gives:

\[ (x + 2)^2 = 4 - x \]

Expanding the left side:

\[ x^2 + 4x + 4 = 4 - x \]

### Step 4: Rearranging the Equation
Rearranging the equation leads to:

\[ x^2 + 5x = 0 \]

### Step 5: Factor the Equation
Factoring gives:

\[ x(x + 5) = 0 \]

Thus, the solutions are:

\[ x = 0 \quad \text{and} \quad x = -5 \]

Since we are only interested in \( x \leq 4 \), we take \( x = 0 \).

### Step 6: Find the Other Boundaries
Next, we find where each function intersects the x-axis:
- For \( y = x + 2 \), set \( y = 0 \):

\[ x + 2 = 0 \implies x = -2 \]

- For \( y = \sqrt{4 - x} \), set \( y = 0 \):

\[ \sqrt{4 - x} = 0 \implies 4 - x = 0 \implies x = 4 \]

### Step 7: Determine the Area Bounded
The area bounded by the curve and the x-axis is from \( x = -2 \) to \( x = 0 \) for \( y = x + 2 \) and from \( x = 0 \) to \( x = 4 \) for \( y = \sqrt{4 - x} \).

### Step 8: Calculate the Area
The total area \( A \) can be calculated as:

\[ A = \int_{-2}^{0} (x + 2) \, dx + \int_{0}^{4} \sqrt{4 - x} \, dx \]

Calculating the first integral:

\[ \int (x + 2) \, dx = \frac{x^2}{2} + 2x \]

Evaluating from \( -2 \) to \( 0 \):

\[ \left[ \frac{0^2}{2} + 2 \cdot 0 \right] - \left[ \frac{(-2)^2}{2} + 2(-2) \right] = 0 - \left[ 2 - 4 \right] = 0 + 2 = 2 \]

Calculating the second integral:

Using the substitution \( u = 4 - x \), \( du = -dx \):

\[ \int \sqrt{u} (-du) = -\frac{2}{3} u^{3/2} \]

Evaluating from \( u = 4 \) to \( u = 0 \):

\[ -\frac{2}{3} \left[ 0 - 4^{3/2} \right] = -\frac{2}{3} \left[ 0 - 8 \right] = \frac{16}{3} \]

### Step 9: Total Area
Adding both areas:

\[ A = 2 + \frac{16}{3} = \frac{6}{3} + \frac{16}{3} = \frac{22}{3} \]

### Step 10: Relate Area to k
According to the problem, the area is given as \( \frac{22}{k} \). Therefore, we have:

\[ \frac{22}{k} = \frac{22}{3} \implies k = 3 \]

### Final Answer
Thus, the value of \( k \) is:

\[ \boxed{3} \]