Let the area of the region enclosed by the curve `y = min {sin x, cos x}` and the x-axis between `x = - pi` to `x = pi` be `A`. Then `A` is equal to _____________.

Answer

To find the area \( A \) enclosed by the curve \( y = \min\{\sin x, \cos x\} \) and the x-axis between \( x = -\pi \) and \( x = \pi \), we will follow these steps:

### Step 1: Identify the points of intersection
The functions \( \sin x \) and \( \cos x \) intersect where \( \sin x = \cos x \). This occurs at:
\[ x = \frac{\pi}{4} + n\frac{\pi}{2} \quad (n \in \mathbb{Z}) \]
Within the interval \([- \pi, \pi]\), the points of intersection are:
- \( x = -\frac{3\pi}{4} \)
- \( x = \frac{\pi}{4} \)

### Step 2: Determine the regions
We need to analyze the behavior of \( \sin x \) and \( \cos x \) in the intervals:
1. From \( -\pi \) to \( -\frac{3\pi}{4} \)
2. From \( -\frac{3\pi}{4} \) to \( \frac{\pi}{4} \)
3. From \( \frac{\pi}{4} \) to \( \pi \)

- In the interval \( [-\pi, -\frac{3\pi}{4}] \), \( \sin x \) is less than \( \cos x \).
- In the interval \( [-\frac{3\pi}{4}, \frac{\pi}{4}] \), \( \sin x \) is greater than \( \cos x \).
- In the interval \( [\frac{\pi}{4}, \pi] \), \( \sin x \) is less than \( \cos x \).

### Step 3: Set up the integrals
The area \( A \) can be computed as:
\[ A = \int_{-\pi}^{-\frac{3\pi}{4}} \cos x \, dx + \int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \sin x \, dx + \int_{\frac{\pi}{4}}^{\pi} \cos x \, dx \]

### Step 4: Calculate the integrals
1. **First Integral**:
\[ \int_{-\pi}^{-\frac{3\pi}{4}} \cos x \, dx = [\sin x]_{-\pi}^{-\frac{3\pi}{4}} = \sin\left(-\frac{3\pi}{4}\right) - \sin(-\pi) = -\frac{\sqrt{2}}{2} - 0 = -\frac{\sqrt{2}}{2} \]

2. **Second Integral**:
\[ \int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \sin x \, dx = [-\cos x]_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} = -\cos\left(\frac{\pi}{4}\right) + \cos\left(-\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} + \left(-\frac{\sqrt{2}}{2}\right) = -\sqrt{2} \]

3. **Third Integral**:
\[ \int_{\frac{\pi}{4}}^{\pi} \cos x \, dx = [\sin x]_{\frac{\pi}{4}}^{\pi} = \sin(\pi) - \sin\left(\frac{\pi}{4}\right) = 0 - \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} \]

### Step 5: Combine the areas
Now, we combine the absolute values of the areas to find the total area \( A \):
\[ A = \left| -\frac{\sqrt{2}}{2} \right| + \left| -\sqrt{2} \right| + \left| -\frac{\sqrt{2}}{2} \right| = \frac{\sqrt{2}}{2} + \sqrt{2} + \frac{\sqrt{2}}{2} = 2\sqrt{2} \]

### Final Answer
Thus, the area \( A \) is equal to:
\[ \boxed{2\sqrt{2}} \]