To solve the integral \( \int_{-1}^{1} f(x) \, dx \) where \( f(x) = \min[x + 1, \sqrt{1 - x}] \), we need to analyze the function \( f(x) \) over the interval \([-1, 1]\).
### Step 1: Analyze the function \( f(x) \)
We need to determine where \( x + 1 \) is less than or equal to \( \sqrt{1 - x} \) and where it is greater.
1. Set the two expressions equal to find the points of intersection:
\[
x + 1 = \sqrt{1 - x}
\]
2. Square both sides:
\[
(x + 1)^2 = 1 - x
\]
\[
x^2 + 2x + 1 = 1 - x
\]
\[
x^2 + 3x = 0
\]
\[
x(x + 3) = 0
\]
This gives us \( x = 0 \) and \( x = -3 \). However, since we are only interested in the interval \([-1, 1]\), we will only consider \( x = 0 \).
### Step 2: Determine the behavior of \( f(x) \)
- For \( x \in [-1, 0] \):
- At \( x = -1 \): \( f(-1) = \min[-1 + 1, \sqrt{1 - (-1)}] = \min[0, \sqrt{2}] = 0 \)
- At \( x = 0 \): \( f(0) = \min[0 + 1, \sqrt{1 - 0}] = \min[1, 1] = 1 \)
Since \( x + 1 \) increases from 0 to 1 and \( \sqrt{1 - x} \) decreases from \( \sqrt{2} \) to 1, we can conclude that:
\[
f(x) = x + 1 \quad \text{for } x \in [-1, 0]
\]
- For \( x \in [0, 1] \):
- At \( x = 0 \): \( f(0) = 1 \)
- At \( x = 1 \): \( f(1) = \min[1 + 1, \sqrt{1 - 1}] = \min[2, 0] = 0 \)
Here, \( \sqrt{1 - x} \) decreases from 1 to 0 while \( x + 1 \) increases from 1 to 2. Thus:
\[
f(x) = \sqrt{1 - x} \quad \text{for } x \in [0, 1]
\]
### Step 3: Set up the integral
Now we can break the integral into two parts:
\[
\int_{-1}^{1} f(x) \, dx = \int_{-1}^{0} (x + 1) \, dx + \int_{0}^{1} \sqrt{1 - x} \, dx
\]
### Step 4: Calculate the first integral
Calculate \( \int_{-1}^{0} (x + 1) \, dx \):
\[
\int (x + 1) \, dx = \frac{x^2}{2} + x
\]
Evaluating from \(-1\) to \(0\):
\[
\left[ \frac{0^2}{2} + 0 \right] - \left[ \frac{(-1)^2}{2} + (-1) \right] = 0 - \left[ \frac{1}{2} - 1 \right] = 0 - \left[ -\frac{1}{2} \right] = \frac{1}{2}
\]
### Step 5: Calculate the second integral
Calculate \( \int_{0}^{1} \sqrt{1 - x} \, dx \):
Using the substitution \( u = 1 - x \), \( du = -dx \):
\[
\int_{0}^{1} \sqrt{1 - x} \, dx = \int_{1}^{0} \sqrt{u} (-du) = \int_{0}^{1} u^{1/2} \, du
\]
\[
= \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{1} = \left[ \frac{2}{3} u^{3/2} \right]_{0}^{1} = \frac{2}{3} (1) - 0 = \frac{2}{3}
\]
### Step 6: Combine the results
Now, combine both integrals:
\[
\int_{-1}^{1} f(x) \, dx = \frac{1}{2} + \frac{2}{3}
\]
Finding a common denominator (which is 6):
\[
= \frac{3}{6} + \frac{4}{6} = \frac{7}{6}
\]
### Final Answer
Thus, the value of the integral \( \int_{-1}^{1} f(x) \, dx \) is:
\[
\boxed{\frac{7}{6}}
\]
