`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 need to analyze the behavior of the function \( \sin(2\pi x) \) over the interval from 0 to 1.

### Step 1: Identify the points where \( \sin(2\pi x) \) changes sign
The function \( \sin(2\pi x) \) completes one full cycle from 0 to 1. It is zero at:
- \( x = 0 \)
- \( x = \frac{1}{2} \)
- \( x = 1 \)

In the interval \( [0, 1] \):
- \( \sin(2\pi x) \) is positive for \( x \in [0, \frac{1}{2}) \)
- \( \sin(2\pi x) \) is negative for \( x \in (\frac{1}{2}, 1] \)

### Step 2: Break the integral into two parts
We can split the integral at \( x = \frac{1}{2} \):
\[ \int_{0}^{1} |\sin(2\pi x)| \, dx = \int_{0}^{\frac{1}{2}} \sin(2\pi x) \, dx + \int_{\frac{1}{2}}^{1} -\sin(2\pi x) \, dx \]

### Step 3: Calculate the first integral
For the first integral, where \( \sin(2\pi x) \) is positive:
\[ \int_{0}^{\frac{1}{2}} \sin(2\pi x) \, dx \]
Using the integral formula \( \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C \):
\[ = -\frac{1}{2\pi} \cos(2\pi x) \bigg|_{0}^{\frac{1}{2}} = -\frac{1}{2\pi} \left[ \cos(\pi) - \cos(0) \right] \]
\[ = -\frac{1}{2\pi} \left[ -1 - 1 \right] = \frac{2}{2\pi} = \frac{1}{\pi} \]

### Step 4: Calculate the second integral
For the second integral, where \( \sin(2\pi x) \) is negative:
\[ \int_{\frac{1}{2}}^{1} -\sin(2\pi x) \, dx = -\left(-\frac{1}{2\pi} \cos(2\pi x) \bigg|_{\frac{1}{2}}^{1}\right) \]
\[ = \frac{1}{2\pi} \left[ \cos(2\pi) - \cos(\pi) \right] = \frac{1}{2\pi} \left[ 1 - (-1) \right] = \frac{1}{2\pi} \cdot 2 = \frac{1}{\pi} \]

### Step 5: Combine the results
Now, we combine the results of both integrals:
\[ \int_{0}^{1} |\sin(2\pi x)| \, dx = \frac{1}{\pi} + \frac{1}{\pi} = \frac{2}{\pi} \]

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