`int_(0)^(1) |sin 2pi x|dx` id equal to

Options

A. 0
B. `-(1)/(pi)`
C. `(1)/(pi)`
D. `(2)/(pi)`

Answer

To solve the integral \( \int_{0}^{1} |\sin(2\pi x)| \, dx \), we will follow these steps:

### Step 1: Identify the behavior of \( \sin(2\pi x) \)
The function \( \sin(2\pi x) \) oscillates between -1 and 1 as \( x \) varies from 0 to 1. Specifically, it completes one full cycle from 0 to 1. The sine function is positive in the intervals where \( 2\pi x \) is in the first and second quadrants, which corresponds to \( x \) in the intervals \( [0, \frac{1}{4}] \) and \( [\frac{1}{2}, \frac{3}{4}] \). It is negative in the intervals \( [\frac{1}{4}, \frac{1}{2}] \) and \( [\frac{3}{4}, 1] \).

### Step 2: Break the integral into intervals
We can break the integral into parts where the sine function is either positive or negative:
\[ \int_{0}^{1} |\sin(2\pi x)| \, dx = \int_{0}^{\frac{1}{4}} \sin(2\pi x) \, dx + \int_{\frac{1}{4}}^{\frac{1}{2}} -\sin(2\pi x) \, dx + \int_{\frac{1}{2}}^{\frac{3}{4}} \sin(2\pi x) \, dx + \int_{\frac{3}{4}}^{1} -\sin(2\pi x) \, dx \]

### Step 3: Evaluate each integral
1. **For \( \int_{0}^{\frac{1}{4}} \sin(2\pi x) \, dx \)**:
\[ = \left[-\frac{1}{2\pi} \cos(2\pi x)\right]_{0}^{\frac{1}{4}} = -\frac{1}{2\pi} \left( \cos\left(\frac{\pi}{2}\right) - \cos(0) \right) = -\frac{1}{2\pi} (0 - 1) = \frac{1}{2\pi} \]

2. **For \( \int_{\frac{1}{4}}^{\frac{1}{2}} -\sin(2\pi x) \, dx \)**:
\[ = -\left[-\frac{1}{2\pi} \cos(2\pi x)\right]_{\frac{1}{4}}^{\frac{1}{2}} = \frac{1}{2\pi} \left( \cos(\pi) - \cos\left(\frac{\pi}{2}\right) \right) = \frac{1}{2\pi} (-1 - 0) = -\frac{1}{2\pi} \]

3. **For \( \int_{\frac{1}{2}}^{\frac{3}{4}} \sin(2\pi x) \, dx \)**:
\[ = \left[-\frac{1}{2\pi} \cos(2\pi x)\right]_{\frac{1}{2}}^{\frac{3}{4}} = -\frac{1}{2\pi} \left( \cos\left(\frac{3\pi}{2}\right) - \cos(\pi) \right) = -\frac{1}{2\pi} (0 - (-1)) = \frac{1}{2\pi} \]

4. **For \( \int_{\frac{3}{4}}^{1} -\sin(2\pi x) \, dx \)**:
\[ = -\left[-\frac{1}{2\pi} \cos(2\pi x)\right]_{\frac{3}{4}}^{1} = \frac{1}{2\pi} \left( \cos(2\pi) - \cos\left(\frac{3\pi}{2}\right) \right) = \frac{1}{2\pi} (1 - 0) = \frac{1}{2\pi} \]

### Step 4: Combine the results
Now, we combine all the evaluated integrals:
\[ \int_{0}^{1} |\sin(2\pi x)| \, dx = \frac{1}{2\pi} - \frac{1}{2\pi} + \frac{1}{2\pi} + \frac{1}{2\pi} = \frac{2}{2\pi} = \frac{1}{\pi} \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{1} |\sin(2\pi x)| \, dx \) is:
\[ \frac{1}{\pi} \]