Let f(x) = minimum `(x+1, sqrt(1-x))" for all "x le 1.` Then the area bounded by y=f(x) and the x-axis is

Explanation

To find the area bounded by the function \( f(x) = \min(x + 1, \sqrt{1 - x}) \) and the x-axis for \( x \leq 1 \), we will follow these steps:

### Step 1: Determine the intervals for \( f(x) \)

We need to find where \( x + 1 \) and \( \sqrt{1 - x} \) intersect to determine the intervals for \( f(x) \).

Set:
\[ x + 1 = \sqrt{1 - x} \]

Squaring both sides gives:
\[ (x + 1)^2 = 1 - x \]
\[ x^2 + 2x + 1 = 1 - x \]
\[ x^2 + 3x = 0 \]
Factoring out \( x \):
\[ x(x + 3) = 0 \]
Thus, \( x = 0 \) or \( x = -3 \).

Since we are interested in the interval \( x \leq 1 \), we will consider \( x = -1 \) and \( x = 0 \).

### Step 2: Evaluate \( f(x) \) in the intervals

1. For \( x \in [-1, 0] \):
- \( f(x) = x + 1 \)

2. For \( x \in [0, 1] \):
- \( f(x) = \sqrt{1 - x} \)

### Step 3: Calculate the area under the curve

The area \( A \) can be calculated as the sum of the areas under \( f(x) \) in both intervals:

\[ A = \int_{-1}^{0} (x + 1) \, dx + \int_{0}^{1} \sqrt{1 - x} \, dx \]

### Step 4: Compute the first integral

\[ \int_{-1}^{0} (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_{-1}^{0} \]
Calculating the limits:
\[ = \left( \frac{0^2}{2} + 0 \right) - \left( \frac{(-1)^2}{2} - 1 \right) \]
\[ = 0 - \left( \frac{1}{2} - 1 \right) = 0 - \left( -\frac{1}{2} \right) = \frac{1}{2} \]

### Step 5: Compute the second integral

Using the substitution \( u = 1 - x \), then \( du = -dx \):
- When \( x = 0 \), \( u = 1 \)
- When \( x = 1 \), \( u = 0 \)

Thus, the integral becomes:
\[ \int_{0}^{1} \sqrt{1 - x} \, dx = \int_{1}^{0} \sqrt{u} (-du) = \int_{0}^{1} u^{1/2} \, du \]
Calculating this integral:
\[ = \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{1} = \left[ \frac{2}{3} u^{3/2} \right]_{0}^{1} = \frac{2}{3} (1) - 0 = \frac{2}{3} \]

### Step 6: Combine the areas

Now, adding both areas:
\[ A = \frac{1}{2} + \frac{2}{3} \]
Finding a common denominator (which is 6):
\[ = \frac{3}{6} + \frac{4}{6} = \frac{7}{6} \]

### Final Answer

The area bounded by \( y = f(x) \) and the x-axis is:
\[ \boxed{\frac{7}{6}} \]