If f(x) = minimum `{sin x, cos x}`, `x in [- pi, pi]`then area between y = f(x) and x-axis is

Options

A. 4
B. 8
C. 16
D. 28

Answer

To find the area between the curve \( f(x) = \min(\sin x, \cos x) \) and the x-axis over the interval \( x \in [-\pi, \pi] \), we will follow these steps:

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

### Step 2: Analyze the function \( f(x) \)
We need to identify which function is minimum in each segment of the interval:
- For \( x \in [-\pi, -\frac{3\pi}{4}] \): \( \cos x \) is greater than \( \sin x \) (since \( \sin x < 0 \) and \( \cos x > 0 \)).
- For \( x \in [-\frac{3\pi}{4}, \frac{\pi}{4}] \): \( \sin x \) is less than \( \cos x \).
- For \( x \in [\frac{\pi}{4}, \pi] \): \( \sin x \) is greater than \( \cos x \).

Thus, we can express \( f(x) \) as:
\[ f(x) = \begin{cases} \cos x & \text{for } x \in [-\pi, -\frac{3\pi}{4}] \\ \sin x & \text{for } x \in [-\frac{3\pi}{4}, \frac{\pi}{4}] \\ \cos x & \text{for } x \in [\frac{\pi}{4}, \pi] \end{cases} \]

### Step 3: Set up the integrals for area calculation
The area \( A \) between the curve and the x-axis can be calculated 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 each integral
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**:
\[ \int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \sin x \, dx = [-\cos x]_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} = -\cos\left(\frac{\pi}{4}\right) + \cos\left(-\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0 \]

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, summing the areas:
\[ A = -\frac{\sqrt{2}}{2} + 0 - \frac{\sqrt{2}}{2} = -\sqrt{2} \]
Since area cannot be negative, we take the absolute value:
\[ A = \sqrt{2} \]

### Step 6: Final area calculation
The total area between \( f(x) \) and the x-axis over the interval \( [-\pi, \pi] \) is:
\[ A = 2 \cdot \sqrt{2} = 4 \quad \text{(considering symmetry)} \]

### Conclusion
The area between \( y = f(x) \) and the x-axis is \( \boxed{4} \).