The value of `int_(0)^(2pi)[|sin x |+|cos x |] d x ` is equal to

Options

A. `(pi)/(2)`
B. `pi`
C. `(3pi)/(2)`
D. `2pi`

Answer

To solve the integral \( \int_{0}^{2\pi} (|\sin x| + |\cos x|) \, dx \), we can follow these steps:

### Step 1: Analyze the Function
The function \( |\sin x| + |\cos x| \) is periodic with a period of \( \pi \). This is because both \( |\sin x| \) and \( |\cos x| \) have a period of \( 2\pi \), but their sum will repeat every \( \pi \).

### Step 2: Determine the Integral Over One Period
To find the integral over the interval \( [0, 2\pi] \), we can calculate it over \( [0, \pi] \) and then double the result:
\[ \int_{0}^{2\pi} (|\sin x| + |\cos x|) \, dx = 2 \int_{0}^{\pi} (|\sin x| + |\cos x|) \, dx \]

### Step 3: Evaluate the Integral from 0 to π
In the interval \( [0, \pi] \):
- \( |\sin x| = \sin x \) (since \( \sin x \) is non-negative)
- \( |\cos x| = \cos x \) for \( x \in [0, \frac{\pi}{2}] \) and \( |\cos x| = -\cos x \) for \( x \in [\frac{\pi}{2}, \pi] \)

Thus, we can split the integral:
\[ \int_{0}^{\pi} (|\sin x| + |\cos x|) \, dx = \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) \, dx + \int_{\frac{\pi}{2}}^{\pi} (\sin x - \cos x) \, dx \]

### Step 4: Calculate Each Integral
1. **First Integral**:
\[ \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) \, dx = \int_{0}^{\frac{\pi}{2}} \sin x \, dx + \int_{0}^{\frac{\pi}{2}} \cos x \, dx \]
- \( \int \sin x \, dx = -\cos x \) and \( \int \cos x \, dx = \sin x \)
- Evaluating:
\[ \int_{0}^{\frac{\pi}{2}} \sin x \, dx = [-\cos x]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1 \]
\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]
- Thus, the first integral is:
\[ \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) \, dx = 1 + 1 = 2 \]

2. **Second Integral**:
\[ \int_{\frac{\pi}{2}}^{\pi} (\sin x - \cos x) \, dx = \int_{\frac{\pi}{2}}^{\pi} \sin x \, dx - \int_{\frac{\pi}{2}}^{\pi} \cos x \, dx \]
- Evaluating:
\[ \int_{\frac{\pi}{2}}^{\pi} \sin x \, dx = [-\cos x]_{\frac{\pi}{2}}^{\pi} = -\cos(\pi) + \cos\left(\frac{\pi}{2}\right) = 1 + 0 = 1 \]
\[ \int_{\frac{\pi}{2}}^{\pi} \cos x \, dx = [\sin x]_{\frac{\pi}{2}}^{\pi} = \sin(\pi) - \sin\left(\frac{\pi}{2}\right) = 0 - 1 = -1 \]
- Thus, the second integral is:
\[ \int_{\frac{\pi}{2}}^{\pi} (\sin x - \cos x) \, dx = 1 - (-1) = 1 + 1 = 2 \]

### Step 5: Combine the Results
Now, we combine the results of the two integrals:
\[ \int_{0}^{\pi} (|\sin x| + |\cos x|) \, dx = 2 + 2 = 4 \]

### Step 6: Final Calculation
Finally, we multiply by 2 to account for the interval \( [0, 2\pi] \):
\[ \int_{0}^{2\pi} (|\sin x| + |\cos x|) \, dx = 2 \times 4 = 8 \]

Thus, the value of the integral is:
\[ \boxed{8} \]