The value of the integral `I=int_(0)^(pi)[|sinx|+|cosx|]dx,` (where `[.]` denotes the greatest integer function) is equal to

Options

A. 1
B. 2
C. `pi`
D. `2pi`

Answer

To solve the integral \( I = \int_{0}^{\pi} \lfloor |\sin x| + |\cos x| \rfloor \, dx \), we will follow these steps:

### Step 1: Analyze the expression \( |\sin x| + |\cos x| \)
The functions \( |\sin x| \) and \( |\cos x| \) are both non-negative and periodic. Over the interval \( [0, \pi] \), we can observe the behavior of these functions.

### Step 2: Determine the maximum value of \( |\sin x| + |\cos x| \)
- At \( x = 0 \), \( |\sin 0| + |\cos 0| = 0 + 1 = 1 \).
- At \( x = \frac{\pi}{4} \), \( |\sin \frac{\pi}{4}| + |\cos \frac{\pi}{4}| = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \).
- At \( x = \frac{\pi}{2} \), \( |\sin \frac{\pi}{2}| + |\cos \frac{\pi}{2}| = 1 + 0 = 1 \).
- At \( x = \pi \), \( |\sin \pi| + |\cos \pi| = 0 + 1 = 1 \).

Thus, the maximum value of \( |\sin x| + |\cos x| \) on the interval \( [0, \pi] \) is \( \sqrt{2} \).

### Step 3: Determine the range of \( |\sin x| + |\cos x| \)
From the analysis:
- The minimum value is \( 1 \) (at \( x = 0, \frac{\pi}{2}, \pi \)).
- The maximum value is \( \sqrt{2} \) (at \( x = \frac{\pi}{4} \)).

### Step 4: Apply the greatest integer function
Since \( 1 \leq |\sin x| + |\cos x| < \sqrt{2} \) and \( \sqrt{2} \approx 1.414 \), we can conclude:
\[ \lfloor |\sin x| + |\cos x| \rfloor = 1 \quad \text{for all } x \in [0, \pi]. \]

### Step 5: Evaluate the integral
Now we can substitute this result into the integral:
\[ I = \int_{0}^{\pi} \lfloor |\sin x| + |\cos x| \rfloor \, dx = \int_{0}^{\pi} 1 \, dx. \]
Calculating this integral gives:
\[ I = [x]_{0}^{\pi} = \pi - 0 = \pi. \]

### Final Answer
Thus, the value of the integral \( I \) is \( \pi \).