`int_(0)^((14pi)/3) |sinx|dx`

Answer

To solve the integral \( \int_{0}^{\frac{14\pi}{3}} |\sin x| \, dx \), we will first analyze the behavior of the function \( |\sin x| \) over the given interval.

### Step 1: Determine the periodicity and behavior of \( |\sin x| \)

The function \( \sin x \) has a period of \( 2\pi \). The absolute value \( |\sin x| \) will repeat every \( 2\pi \) and will have the following behavior:
- From \( 0 \) to \( \pi \), \( |\sin x| = \sin x \) (positive).
- From \( \pi \) to \( 2\pi \), \( |\sin x| = -\sin x \) (negative).
- From \( 2\pi \) to \( 3\pi \), \( |\sin x| = \sin x \) (positive).
- From \( 3\pi \) to \( 4\pi \), \( |\sin x| = -\sin x \) (negative).

### Step 2: Break down the integral into segments

We need to evaluate the integral from \( 0 \) to \( \frac{14\pi}{3} \). First, we find how many complete periods of \( 2\pi \) fit into \( \frac{14\pi}{3} \):

\[ \frac{14\pi/3}{2\pi} = \frac{14}{6} = \frac{7}{3} \]

This means \( \frac{14\pi}{3} \) is \( 2\pi \) plus an additional \( \frac{2\pi}{3} \). Thus, we can break the integral into the following parts:

\[ \int_{0}^{\frac{14\pi}{3}} |\sin x| \, dx = \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} -\sin x \, dx + \int_{2\pi}^{3\pi} \sin x \, dx + \int_{3\pi}^{4\pi} -\sin x \, dx + \int_{4\pi}^{\frac{14\pi}{3}} \sin x \, dx \]

### Step 3: Calculate each integral

1. **From \( 0 \) to \( \pi \)**:
\[ \int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2 \]

2. **From \( \pi \) to \( 2\pi \)**:
\[ \int_{\pi}^{2\pi} -\sin x \, dx = [\cos x]_{\pi}^{2\pi} = \cos(2\pi) - \cos(\pi) = 1 - (-1) = 2 \]

3. **From \( 2\pi \) to \( 3\pi \)**:
\[ \int_{2\pi}^{3\pi} \sin x \, dx = [-\cos x]_{2\pi}^{3\pi} = -\cos(3\pi) - (-\cos(2\pi)) = -(-1) - (-1) = 2 \]

4. **From \( 3\pi \) to \( 4\pi \)**:
\[ \int_{3\pi}^{4\pi} -\sin x \, dx = [\cos x]_{3\pi}^{4\pi} = \cos(4\pi) - \cos(3\pi) = 1 - (-1) = 2 \]

5. **From \( 4\pi \) to \( \frac{14\pi}{3} \)**:
\[ 4\pi = \frac{12\pi}{3}, \quad \text{and } \frac{14\pi}{3} - 4\pi = \frac{14\pi}{3} - \frac{12\pi}{3} = \frac{2\pi}{3} \]
\[ \int_{4\pi}^{\frac{14\pi}{3}} \sin x \, dx = [-\cos x]_{4\pi}^{\frac{14\pi}{3}} = -\cos\left(\frac{14\pi}{3}\right) + \cos(4\pi) \]
Since \( \frac{14\pi}{3} = 4\pi + \frac{2\pi}{3} \), we have:
\[ \cos\left(\frac{14\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2 \]
Thus,
\[ -\left(-\frac{1}{2}\right) + 1 = \frac{1}{2} + 1 = \frac{3}{2} \]

### Step 4: Combine all parts

Now we sum all the parts:
\[ 2 + 2 + 2 + 2 + \frac{3}{2} = 8 + \frac{3}{2} = \frac{16}{2} + \frac{3}{2} = \frac{19}{2} \]

### Final Answer

Thus, the value of the integral \( \int_{0}^{\frac{14\pi}{3}} |\sin x| \, dx \) is:

\[ \boxed{\frac{19}{2}} \]