Evaluate `int_-1^1 [x[1+sinpix]+1]dx`, [.] is the greatest integer function.

Options

A. 2
B. 0
C. 1
D. None of these

Answer

To evaluate the integral

\[ I = \int_{-1}^{1} \left\lfloor x(1 + \sin(\pi x)) + 1 \right\rfloor \, dx, \]

where \(\lfloor . \rfloor\) is the greatest integer function, we can break this integral into two parts: from \(-1\) to \(0\) and from \(0\) to \(1\).

### Step 1: Split the Integral
We can write the integral as:

\[ I = \int_{-1}^{0} \left\lfloor x(1 + \sin(\pi x)) + 1 \right\rfloor \, dx + \int_{0}^{1} \left\lfloor x(1 + \sin(\pi x)) + 1 \right\rfloor \, dx. \]

### Step 2: Evaluate the Integral from \(-1\) to \(0\)
For \(x\) in the interval \([-1, 0]\):

- The sine function \(\sin(\pi x)\) varies from \(-1\) to \(0\).
- Therefore, \(1 + \sin(\pi x)\) varies from \(0\) to \(1\).
- Thus, \(x(1 + \sin(\pi x))\) will be between \(0\) and \(x\) (which is negative).
- Therefore, \(x(1 + \sin(\pi x)) + 1\) will be between \(1\) and \(1 + x\) (which is between \(0\) and \(1\)).
- The greatest integer function \(\lfloor x(1 + \sin(\pi x)) + 1 \rfloor\) will be \(0\) in this interval.

So,

\[ \int_{-1}^{0} \left\lfloor x(1 + \sin(\pi x)) + 1 \right\rfloor \, dx = \int_{-1}^{0} 0 \, dx = 0. \]

### Step 3: Evaluate the Integral from \(0\) to \(1\)
For \(x\) in the interval \([0, 1]\):

- The sine function \(\sin(\pi x)\) varies from \(0\) to \(1\).
- Therefore, \(1 + \sin(\pi x)\) varies from \(1\) to \(2\).
- Thus, \(x(1 + \sin(\pi x))\) will be between \(0\) and \(2\).
- Therefore, \(x(1 + \sin(\pi x)) + 1\) will be between \(1\) and \(3\).
- The greatest integer function \(\lfloor x(1 + \sin(\pi x)) + 1 \rfloor\) will be \(1\) in this interval.

So,

\[ \int_{0}^{1} \left\lfloor x(1 + \sin(\pi x)) + 1 \right\rfloor \, dx = \int_{0}^{1} 1 \, dx = 1. \]

### Step 4: Combine the Results
Now we can combine the results from both intervals:

\[ I = 0 + 1 = 1. \]

### Final Answer
Thus, the value of the integral is

\[ \boxed{1}. \]