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, we need to find the area bounded by the function \( f(x) = \min(x + 2, \sqrt{4 - x}) \) and the x-axis for \( x \leq 4 \). Let's break down the steps:

### Step 1: Identify the Functions
We have two functions:
1. \( y_1 = x + 2 \)
2. \( y_2 = \sqrt{4 - x} \)

### Step 2: Find the Intersection Points
To find the area, we first need to determine where these two functions intersect. We set them 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: Rearrange the Equation
Rearranging the equation results in:

\[ x^2 + 4x + x + 4 - 4 = 0 \]
\[ x^2 + 5x = 0 \]

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

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

Thus, the solutions are:

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

### Step 6: Determine the Relevant Interval
Since we are considering \( x \leq 4 \), the relevant intersection points are \( x = 0 \) and \( x = -5 \).

### Step 7: Determine the Area
The area bounded by the curve and the x-axis can be calculated by integrating the function \( f(x) \) from \( x = -5 \) to \( x = 0 \).

Since \( f(x) = \sqrt{4 - x} \) for \( x \in [-5, 0] \):

\[ \text{Area} = \int_{-5}^{0} \sqrt{4 - x} \, dx \]

### Step 8: Compute the Integral
To compute the integral, we can use the substitution \( u = 4 - x \), which gives \( du = -dx \) and changes the limits of integration:

When \( x = -5 \), \( u = 9 \) and when \( x = 0 \), \( u = 4 \).

Thus, we have:

\[ \text{Area} = \int_{9}^{4} \sqrt{u} (-du) = \int_{4}^{9} \sqrt{u} \, du \]

### Step 9: Evaluate the Integral
The integral of \( \sqrt{u} \) is:

\[ \int \sqrt{u} \, du = \frac{2}{3} u^{3/2} \]

Evaluating from 4 to 9:

\[ \text{Area} = \left[ \frac{2}{3} u^{3/2} \right]_{4}^{9} = \frac{2}{3} (9^{3/2} - 4^{3/2}) = \frac{2}{3} (27 - 8) = \frac{2}{3} \cdot 19 = \frac{38}{3} \]

### Step 10: Relate to the Given Area
The problem states that the area is \( \frac{22}{k} \). Setting this equal to our calculated area:

\[ \frac{38}{3} = \frac{22}{k} \]

### Step 11: Solve for \( k \)
Cross-multiplying gives:

\[ 38k = 66 \implies k = \frac{66}{38} = \frac{33}{19} \]

Thus, the value of \( k \) is:

\[ k = 3 \]

### Final Answer
The value of \( k \) is \( \boxed{3} \).