`int_(pi)^(10n) |sin x|dx` is equla to

Options

A. 20
B. 8
C. 10
D. 18

Answer

To solve the integral \( I = \int_{\pi}^{10\pi} |\sin x| \, dx \), we can break it down into manageable parts. Here’s a step-by-step solution:

### Step 1: Analyze the Periodicity of \( |\sin x| \)

The function \( \sin x \) is periodic with a period of \( 2\pi \). Therefore, \( |\sin x| \) is also periodic with the same period. This means that we can break the integral into intervals of \( 2\pi \).

### Step 2: Break Down the Integral

We can express the integral from \( \pi \) to \( 10\pi \) as follows:

\[ I = \int_{\pi}^{2\pi} |\sin x| \, dx + \int_{2\pi}^{10\pi} |\sin x| \, dx \]

### Step 3: Evaluate the First Integral

In the interval \( [\pi, 2\pi] \), \( \sin x \) is negative. Therefore, \( |\sin x| = -\sin x \) in this interval:

\[ \int_{\pi}^{2\pi} |\sin x| \, dx = \int_{\pi}^{2\pi} -\sin x \, dx \]

Calculating this integral:

\[ \int -\sin x \, dx = \cos x + C \]

Now, evaluating from \( \pi \) to \( 2\pi \):

\[ \left[ \cos x \right]_{\pi}^{2\pi} = \cos(2\pi) - \cos(\pi) = 1 - (-1) = 2 \]

### Step 4: Evaluate the Second Integral

For the interval \( [2\pi, 10\pi] \), we can break it down further into smaller intervals of \( 2\pi \):

\[ \int_{2\pi}^{10\pi} |\sin x| \, dx = \sum_{n=1}^{8} \int_{(2n-2)\pi}^{(2n)\pi} |\sin x| \, dx \]

Each of these integrals from \( (2n-2)\pi \) to \( (2n)\pi \) will contribute the same value as the integral from \( 0 \) to \( 2\pi \) because of the periodicity of \( |\sin x| \).

Calculating \( \int_{0}^{2\pi} |\sin x| \, dx \):

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

Calculating both parts:

1. From \( 0 \) to \( \pi \):

\[ \int_{0}^{\pi} \sin x \, dx = \left[-\cos x\right]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2 \]

2. From \( \pi \) to \( 2\pi \):

\[ \int_{\pi}^{2\pi} -\sin x \, dx = 2 \quad \text{(as calculated earlier)} \]

Thus,

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

### Step 5: Combine the Results

Now, we have:

\[ I = \int_{\pi}^{2\pi} |\sin x| \, dx + \int_{2\pi}^{10\pi} |\sin x| \, dx \]

Substituting the values we found:

\[ I = 2 + 4 \times 4 = 2 + 16 = 18 \]

### Final Answer

Thus, the value of the integral \( \int_{\pi}^{10\pi} |\sin x| \, dx \) is:

\[ \boxed{18} \]