The value of `int_(pi)^(2pi)[2sinx]dx` is equal to (where `[.]` represents the greatest integer function)

Options

A. `-pi`
B. `(5pi)/(3)`
C. `(-5pi)/(3)`
D. `-2pi`

Answer

To solve the integral \( \int_{\pi}^{2\pi} 2\sin x \, dx \) and find the value of the greatest integer function, we will follow these steps:

### Step 1: Evaluate the Integral
We start by calculating the integral \( \int_{\pi}^{2\pi} 2\sin x \, dx \).

\[ \int 2\sin x \, dx = -2\cos x + C \]

### Step 2: Apply the Limits
Now we will apply the limits from \( \pi \) to \( 2\pi \):

\[ \int_{\pi}^{2\pi} 2\sin x \, dx = \left[-2\cos x\right]_{\pi}^{2\pi} \]

Calculating the values at the limits:

1. At \( x = 2\pi \):
\[ -2\cos(2\pi) = -2(1) = -2 \]

2. At \( x = \pi \):
\[ -2\cos(\pi) = -2(-1) = 2 \]

Now substituting these values into the integral:

\[ \int_{\pi}^{2\pi} 2\sin x \, dx = -2 - 2 = -4 \]

### Step 3: Simplify the Result
Now we simplify the result:

\[ \int_{\pi}^{2\pi} 2\sin x \, dx = -4 \]

### Step 4: Apply the Greatest Integer Function
Now we need to apply the greatest integer function \( [x] \) to our result:

\[ \left[-4\right] = -4 \]

### Final Answer
Thus, the value of \( \int_{\pi}^{2\pi} 2\sin x \, dx \) is equal to \( -4 \).