`int_(0)^(4pi) |cos x|dx`=

Options

A. 4
B. 8
C. 0
D. None of these

Answer

To solve the integral \( I = \int_{0}^{4\pi} |\cos x| \, dx \), we can follow these steps:

### Step 1: Identify the Period of the Function
The function \( |\cos x| \) is periodic with a period of \( 2\pi \). This means that the integral over any interval of length \( 2\pi \) will yield the same result.

**Hint:** Look for the period of the function to simplify the integral.

### Step 2: Break the Integral into Periods
Since \( 4\pi \) is \( 2\pi \times 2 \), we can break the integral into two parts:
\[ I = \int_{0}^{4\pi} |\cos x| \, dx = 2 \int_{0}^{2\pi} |\cos x| \, dx \]

**Hint:** Use the periodic property of the function to reduce the limits of integration.

### Step 3: Evaluate the Integral from 0 to \( 2\pi \)
Now, we can further break down the integral from \( 0 \) to \( 2\pi \):
\[ \int_{0}^{2\pi} |\cos x| \, dx = \int_{0}^{\pi} \cos x \, dx + \int_{\pi}^{2\pi} -\cos x \, dx \]
Here, \( |\cos x| = \cos x \) for \( x \in [0, \pi] \) and \( |\cos x| = -\cos x \) for \( x \in [\pi, 2\pi] \).

**Hint:** Identify the intervals where the function changes sign.

### Step 4: Compute Each Integral
Now, we compute each integral:
1. For \( \int_{0}^{\pi} \cos x \, dx \):
\[ \int_{0}^{\pi} \cos x \, dx = [\sin x]_{0}^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0 \]

2. For \( \int_{\pi}^{2\pi} -\cos x \, dx \):
\[ \int_{\pi}^{2\pi} -\cos x \, dx = -[\sin x]_{\pi}^{2\pi} = -(\sin(2\pi) - \sin(\pi)) = - (0 - 0) = 0 \]

**Hint:** Remember to apply the fundamental theorem of calculus for evaluating integrals.

### Step 5: Combine the Results
Now we combine the results:
\[ \int_{0}^{2\pi} |\cos x| \, dx = 0 + 0 = 0 \]

### Step 6: Final Calculation
Finally, substituting back into the expression for \( I \):
\[ I = 2 \int_{0}^{2\pi} |\cos x| \, dx = 2 \times 0 = 0 \]

### Conclusion
Thus, the value of the integral \( \int_{0}^{4\pi} |\cos x| \, dx \) is:
\[ \boxed{8} \]