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 \( I = \int_{0}^{\pi} [|\sin x| + |\cos x|] \, dx \), where \([.]\) denotes the greatest integer function, we will follow these steps:

### Step 1: Analyze the functions involved
The functions \(|\sin x|\) and \(|\cos x|\) are both non-negative and periodic. Over the interval \([0, \pi]\):
- \(|\sin x|\) is non-negative and increases from \(0\) to \(1\) at \(x = \frac{\pi}{2}\), then decreases back to \(0\) at \(x = \pi\).
- \(|\cos x|\) starts at \(1\) at \(x = 0\), decreases to \(0\) at \(x = \frac{\pi}{2}\), and becomes negative but is taken as positive in absolute value, returning to \(0\) at \(x = \pi\).

### Step 2: Determine the range of \(|\sin x| + |\cos x|\)
- At \(x = 0\): \(|\sin 0| + |\cos 0| = 0 + 1 = 1\)
- At \(x = \frac{\pi}{2}\): \(|\sin \frac{\pi}{2}| + |\cos \frac{\pi}{2}| = 1 + 0 = 1\)
- At \(x = \pi\): \(|\sin \pi| + |\cos \pi| = 0 + 1 = 1\)

To find the maximum value of \(|\sin x| + |\cos x|\), we can analyze the function:
\[ |\sin x| + |\cos x| \leq \sqrt{(\sin^2 x + \cos^2 x)} \cdot \sqrt{2} = \sqrt{2} \]
The maximum occurs when both functions are equal, which happens at \(x = \frac{\pi}{4}\):
\[ |\sin \frac{\pi}{4}| + |\cos \frac{\pi}{4}| = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414 \]

### Step 3: Determine the greatest integer function
Since \(|\sin x| + |\cos x|\) ranges from \(1\) to \(\sqrt{2}\) over the interval \([0, \pi]\), we can conclude:
\[ 1 \leq |\sin x| + |\cos x| < \sqrt{2} \]
Thus, the greatest integer function \([|\sin x| + |\cos x|]\) will equal \(1\) for all \(x\) in \([0, \pi]\).

### Step 4: Evaluate the integral
Now we can evaluate the integral:
\[ I = \int_{0}^{\pi} [|\sin x| + |\cos x|] \, dx = \int_{0}^{\pi} 1 \, dx \]
Calculating this integral gives:
\[ I = \left[ x \right]_{0}^{\pi} = \pi - 0 = \pi \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\pi} \]