If `I=int_(-20pi)^(20pi)|sinx|[sinx]dx` (where [.] denotes the greatest integer function) then the value of `I` is

Options

A. `-40`
B. `40`
C. `20`
D. `-20`

Explanation

To solve the integral \( I = \int_{-20\pi}^{20\pi} |\sin x| \lfloor \sin x \rfloor \, dx \), where \( \lfloor . \rfloor \) denotes the greatest integer function, we can follow these steps:

### Step 1: Change of Variable
We can use the property of definite integrals that states:
\[ \int_{-a}^{a} f(x) \, dx = \int_{-a}^{a} f(-x) \, dx \]
Let’s perform the substitution \( x = -t \). Then \( dx = -dt \) and the limits change as follows:
- When \( x = -20\pi \), \( t = 20\pi \)
- When \( x = 20\pi \), \( t = -20\pi \)

Thus, we can rewrite the integral:
\[ I = \int_{20\pi}^{-20\pi} |\sin(-t)| \lfloor \sin(-t) \rfloor (-dt) = \int_{-20\pi}^{20\pi} |\sin t| \lfloor -\sin t \rfloor \, dt \]
Since \( |\sin(-t)| = |\sin t| \) and \( \lfloor -\sin t \rfloor = -\lfloor \sin t \rfloor - 1 \) when \( \sin t \) is not an integer, we can express \( I \) as:
\[ I = \int_{-20\pi}^{20\pi} |\sin t| (-\lfloor \sin t \rfloor - 1) \, dt \]

### Step 2: Combine the Integrals
Now we have:
\[ I = -\int_{-20\pi}^{20\pi} |\sin t| \lfloor \sin t \rfloor \, dt - \int_{-20\pi}^{20\pi} |\sin t| \, dt \]
Let’s denote the first integral as \( I \):
\[ I = -I - \int_{-20\pi}^{20\pi} |\sin t| \, dt \]
This leads to:
\[ 2I = -\int_{-20\pi}^{20\pi} |\sin t| \, dt \]
Thus:
\[ I = -\frac{1}{2} \int_{-20\pi}^{20\pi} |\sin t| \, dt \]

### Step 3: Evaluate the Integral of \( |\sin t| \)
The function \( |\sin t| \) has a period of \( 2\pi \). The integral over one period is:
\[ \int_0^{2\pi} |\sin t| \, dt = 2 \]
Since there are \( 20\pi/(2\pi) = 10 \) complete periods from \( -20\pi \) to \( 20\pi \):
\[ \int_{-20\pi}^{20\pi} |\sin t| \, dt = 10 \times 2 = 20 \]

### Step 4: Substitute Back to Find \( I \)
Now substituting back, we have:
\[ I = -\frac{1}{2} \times 20 = -10 \]

### Final Answer
Thus, the value of the integral \( I \) is:
\[ \boxed{-10} \]