To find the area bounded by the curve \( y = f(x) \), the x-axis, and the line \( x = 0 \) from \( y = 0 \) to \( x = 1 \), we need to analyze the function \( f(x) = \min(x + 1, \sqrt{1 - x}) \) for \( x \leq 1 \).
### Step 1: Identify the Functions
We have two functions to consider:
1. \( y = x + 1 \)
2. \( y = \sqrt{1 - x} \)
### Step 2: Find Points of Intersection
To find the area, we need to determine where these two functions intersect. We set them equal to each other:
\[
x + 1 = \sqrt{1 - x}
\]
Squaring both sides gives:
\[
(x + 1)^2 = 1 - x
\]
Expanding and rearranging:
\[
x^2 + 2x + 1 = 1 - x \\
x^2 + 3x = 0 \\
x(x + 3) = 0
\]
This gives us \( x = 0 \) and \( x = -3 \). Since we are only considering \( x \leq 1 \), the relevant intersection point is \( x = 0 \).
### Step 3: Determine Which Function is Minimum
Next, we need to determine which function is the minimum in the interval \( [0, 1] \):
- At \( x = 0 \): \( f(0) = \min(0 + 1, \sqrt{1 - 0}) = \min(1, 1) = 1 \)
- At \( x = 1 \): \( f(1) = \min(1 + 1, \sqrt{1 - 1}) = \min(2, 0) = 0 \)
To find the minimum function in the interval \( [0, 1] \), we can evaluate both functions at a point in between, say \( x = 0.5 \):
- \( f(0.5) = 0.5 + 1 = 1.5 \)
- \( f(0.5) = \sqrt{1 - 0.5} = \sqrt{0.5} \approx 0.707 \)
Thus, \( \sqrt{1 - x} \) is the minimum function in the interval \( [0, 1] \).
### Step 4: Set Up the Integral for Area
The area \( A \) bounded by \( y = f(x) \), the x-axis, and the line \( x = 0 \) from \( x = 0 \) to \( x = 1 \) is given by the integral:
\[
A = \int_0^1 \sqrt{1 - x} \, dx
\]
### Step 5: Calculate the Integral
To solve the integral:
\[
A = \int_0^1 (1 - x)^{1/2} \, dx
\]
Using the substitution \( u = 1 - x \), we have \( du = -dx \). Changing the limits accordingly:
- When \( x = 0 \), \( u = 1 \)
- When \( x = 1 \), \( u = 0 \)
Thus, the integral becomes:
\[
A = \int_1^0 u^{1/2} (-du) = \int_0^1 u^{1/2} \, du
\]
Now, integrating:
\[
A = \left[ \frac{u^{3/2}}{3/2} \right]_0^1 = \left[ \frac{2}{3} u^{3/2} \right]_0^1 = \frac{2}{3} (1^{3/2} - 0^{3/2}) = \frac{2}{3}
\]
### Final Answer
Thus, the area bounded by the curves is:
\[
\boxed{\frac{2}{3}} \text{ square units}
\]
