The area bounded by `y = x |sinx|` and x - axis between `x = 0, x = 2pi` is

Options

A. `2pi`
B. `3pi`
C. `4 pi`
D. `5 pi`

Answer

To find the area bounded by the curve \( y = x |\sin x| \) and the x-axis between \( x = 0 \) and \( x = 2\pi \), we will follow these steps:

### Step 1: Identify the behavior of \( |\sin x| \)
First, we need to understand how \( |\sin x| \) behaves between \( 0 \) and \( 2\pi \):
- From \( 0 \) to \( \pi \), \( \sin x \) is positive, so \( |\sin x| = \sin x \).
- From \( \pi \) to \( 2\pi \), \( \sin x \) is negative, so \( |\sin x| = -\sin x \).

### Step 2: Set up the integral for the area
The area \( A \) can be calculated using the integral of the function from \( 0 \) to \( 2\pi \). We will split the integral into two parts:
\[ A = \int_{0}^{\pi} x \sin x \, dx + \int_{\pi}^{2\pi} x (-\sin x) \, dx \]
This simplifies to:
\[ A = \int_{0}^{\pi} x \sin x \, dx - \int_{\pi}^{2\pi} x \sin x \, dx \]

### Step 3: Calculate the integral \( \int x \sin x \, dx \) using integration by parts
We will use integration by parts for \( \int x \sin x \, dx \):
Let \( u = x \) and \( dv = \sin x \, dx \).
Then, \( du = dx \) and \( v = -\cos x \).

Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ \int x \sin x \, dx = -x \cos x - \int (-\cos x) \, dx = -x \cos x + \sin x \]

### Step 4: Evaluate the first integral from \( 0 \) to \( \pi \)
Now we evaluate:
\[ \int_{0}^{\pi} x \sin x \, dx = \left[-x \cos x + \sin x\right]_{0}^{\pi} \]
Calculating at the limits:
- At \( x = \pi \):
\[ -\pi \cos(\pi) + \sin(\pi) = -\pi(-1) + 0 = \pi \]
- At \( x = 0 \):
\[ -0 \cdot \cos(0) + \sin(0) = 0 + 0 = 0 \]
Thus,
\[ \int_{0}^{\pi} x \sin x \, dx = \pi - 0 = \pi \]

### Step 5: Evaluate the second integral from \( \pi \) to \( 2\pi \)
Now we evaluate:
\[ \int_{\pi}^{2\pi} x (-\sin x) \, dx = -\int_{\pi}^{2\pi} x \sin x \, dx \]
Using the same integration by parts:
\[ \int x \sin x \, dx = -x \cos x + \sin x \]
Evaluating from \( \pi \) to \( 2\pi \):
\[ \left[-x \cos x + \sin x\right]_{\pi}^{2\pi} \]
Calculating at the limits:
- At \( x = 2\pi \):
\[ -2\pi \cos(2\pi) + \sin(2\pi) = -2\pi(1) + 0 = -2\pi \]
- At \( x = \pi \):
\[ -\pi \cos(\pi) + \sin(\pi) = -\pi(-1) + 0 = \pi \]
Thus,
\[ \int_{\pi}^{2\pi} x \sin x \, dx = -2\pi - \pi = -3\pi \]

### Step 6: Combine the results to find the total area
Now we combine the results:
\[ A = \pi - (-3\pi) = \pi + 3\pi = 4\pi \]

### Final Result
The area bounded by the curve \( y = x |\sin x| \) and the x-axis between \( x = 0 \) and \( x = 2\pi \) is:
\[ \boxed{4\pi} \]