The value of `overset(16pi//3)underset(0)int |sinx|dx` is

Options

A. `17//2`
B. `19//2`
C. `21//2`
D. None of these

Answer

To find the value of the integral \( \int_0^{\frac{16\pi}{3}} |\sin x| \, dx \), we will break it down into intervals where the sine function is positive and negative.

### Step 1: Identify the intervals
The sine function, \( \sin x \), is periodic with a period of \( 2\pi \). The integral from \( 0 \) to \( \frac{16\pi}{3} \) can be divided into segments where \( \sin x \) changes its sign:
- From \( 0 \) to \( \pi \): \( \sin x \) is positive.
- From \( \pi \) to \( 2\pi \): \( \sin x \) is negative.
- From \( 2\pi \) to \( 3\pi \): \( \sin x \) is positive.
- From \( 3\pi \) to \( 4\pi \): \( \sin x \) is negative.
- From \( 4\pi \) to \( 5\pi \): \( \sin x \) is positive.
- From \( 5\pi \) to \( \frac{16\pi}{3} \): \( \sin x \) is negative.

### Step 2: Rewrite the integral
Using the properties of absolute value, we can rewrite the integral as:
\[ \int_0^{\frac{16\pi}{3}} |\sin x| \, dx = \int_0^{\pi} \sin x \, dx - \int_{\pi}^{2\pi} \sin x \, dx + \int_{2\pi}^{3\pi} \sin x \, dx - \int_{3\pi}^{4\pi} \sin x \, dx + \int_{4\pi}^{5\pi} \sin x \, dx - \int_{5\pi}^{\frac{16\pi}{3}} \sin x \, dx \]

### 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 \( 5\pi \)**:
\[ \int_{4\pi}^{5\pi} \sin x \, dx = [-\cos x]_{4\pi}^{5\pi} = -\cos(5\pi) - (-\cos(4\pi)) = -(-1) - (-1) = 2 \]

6. **From \( 5\pi \) to \( \frac{16\pi}{3} \)**:
\[ \int_{5\pi}^{\frac{16\pi}{3}} \sin x \, dx = [-\cos x]_{5\pi}^{\frac{16\pi}{3}} = -\cos\left(\frac{16\pi}{3}\right) - (-\cos(5\pi)) \]
To find \( \cos\left(\frac{16\pi}{3}\right) \):
\[ \frac{16\pi}{3} = 5\pi + \frac{\pi}{3} \quad \text{(since } 5\pi = \frac{15\pi}{3}\text{)} \]
Thus, \( \cos\left(\frac{16\pi}{3}\right) = \cos\left(5\pi + \frac{\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2} \).
Therefore,
\[ -\cos\left(\frac{16\pi}{3}\right) - (-\cos(5\pi)) = -\left(-\frac{1}{2}\right) - (-(-1)) = \frac{1}{2} + 1 = \frac{3}{2} \]

### Step 4: Combine all parts
Now, we combine all the parts:
\[ \int_0^{\frac{16\pi}{3}} |\sin x| \, dx = 2 + 2 + 2 + 2 + 2 + \frac{3}{2} = 10 + \frac{3}{2} = \frac{20}{2} + \frac{3}{2} = \frac{23}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_0^{\frac{16\pi}{3}} |\sin x| \, dx = \frac{23}{2} \]