Lat `l_(1)=int_(0)^(3)(sinx)/([(x)/(pi)]+(1)/(2))dx` and `l_(2)=int_(-3)^(0)(sinx)/([(x)/(pi)]+(1)/(2))dx`, then (where`[.]` represent G.l.F.)

Options

A. `l_(2)+l_(2)=0`
B. `l_(1)=l_(2)`
C. `l
_(1)=3l_(2)`
D. `l_(2)=3l_(1)`

Explanation

To solve the problem, we need to evaluate the integrals \( l_1 \) and \( l_2 \) defined as follows:

\[ l_1 = \int_{0}^{3} \frac{\sin x}{\left[\frac{x}{\pi}\right] + \frac{1}{2}} \, dx \]

\[ l_2 = \int_{-3}^{0} \frac{\sin x}{\left[\frac{x}{\pi}\right] + \frac{1}{2}} \, dx \]

where \([\cdot]\) represents the greatest integer function (G.I.F.).

### Step 1: Analyze \( l_1 \)

For \( l_1 \), we need to determine the value of \(\left[\frac{x}{\pi}\right]\) for \( x \) in the interval \([0, 3]\).

- When \( x = 0 \): \(\left[\frac{0}{\pi}\right] = 0\)
- When \( x = 3 \): \(\left[\frac{3}{\pi}\right] \approx \left[\frac{3}{3.14}\right] \approx 0.955\) which rounds down to \(0\).

Now we can evaluate the intervals:
- For \( x \in [0, 3] \), \(\left[\frac{x}{\pi}\right] = 0\).

Thus, we can simplify \( l_1 \):

\[ l_1 = \int_{0}^{3} \frac{\sin x}{0 + \frac{1}{2}} \, dx = \int_{0}^{3} 2 \sin x \, dx \]

### Step 2: Evaluate the integral for \( l_1 \)

Now we compute:

\[ l_1 = 2 \int_{0}^{3} \sin x \, dx \]

The integral of \(\sin x\) is:

\[ \int \sin x \, dx = -\cos x \]

Thus,

\[ \int_{0}^{3} \sin x \, dx = [-\cos x]_{0}^{3} = -\cos(3) + \cos(0) = 1 - \cos(3) \]

So,

\[ l_1 = 2(1 - \cos(3)) \]

### Step 3: Analyze \( l_2 \)

Now, we analyze \( l_2 \):

For \( l_2 \), we need to determine the value of \(\left[\frac{x}{\pi}\right]\) for \( x \) in the interval \([-3, 0]\).

- When \( x = -3 \): \(\left[\frac{-3}{\pi}\right] \approx \left[\frac{-3}{3.14}\right] \approx -1\)
- When \( x = 0 \): \(\left[\frac{0}{\pi}\right] = 0\)

Now we can evaluate the intervals:
- For \( x \in [-3, 0)\), \(\left[\frac{x}{\pi}\right] = -1\).

Thus, we can simplify \( l_2 \):

\[ l_2 = \int_{-3}^{0} \frac{\sin x}{-1 + \frac{1}{2}} \, dx = \int_{-3}^{0} \frac{\sin x}{-\frac{1}{2}} \, dx = -2 \int_{-3}^{0} \sin x \, dx \]

### Step 4: Evaluate the integral for \( l_2 \)

Now we compute:

\[ l_2 = -2 \int_{-3}^{0} \sin x \, dx \]

Using the same integral result as before:

\[ \int_{-3}^{0} \sin x \, dx = [-\cos x]_{-3}^{0} = -\cos(0) + \cos(-3) = -1 + \cos(3) \]

So,

\[ l_2 = -2(-1 + \cos(3)) = 2(1 - \cos(3)) \]

### Step 5: Compare \( l_1 \) and \( l_2 \)

Now we have:

\[ l_1 = 2(1 - \cos(3)) \]
\[ l_2 = 2(1 - \cos(3)) \]

Thus, we conclude that:

\[ l_1 = l_2 \]

### Final Answer

Therefore, the relation between \( l_1 \) and \( l_2 \) is:

\[ l_1 = l_2 \]