The area between the curve `y=xsin x ` and x-axis where `o le x le 2 pi` , is

Options

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

Answer

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

### Step 1: Identify the intervals
The function \( y = x \sin x \) oscillates above and below the x-axis between \( 0 \) and \( 2\pi \). We need to determine where the function is positive and where it is negative within this interval.

### Step 2: Find the points of intersection with the x-axis
To find the points where the curve intersects the x-axis, we set \( y = 0 \):
\[ x \sin x = 0 \]
This gives us the solutions:
- \( x = 0 \)
- \( \sin x = 0 \) at \( x = n\pi \) where \( n \) is an integer. Within \( [0, 2\pi] \), the points are \( x = 0, \pi, 2\pi \).

### Step 3: Set up the integral
The area \( A \) between the curve and the x-axis can be calculated as the sum of the areas from \( 0 \) to \( \pi \) (where the curve is above the x-axis) and from \( \pi \) to \( 2\pi \) (where the curve is below the x-axis):
\[ A = \int_0^\pi x \sin x \, dx + \left| \int_\pi^{2\pi} x \sin x \, dx \right| \]

### Step 4: Calculate the integral from \( 0 \) to \( \pi \)
We will first compute the integral \( \int_0^\pi x \sin x \, dx \) using integration by parts. Let:
- \( u = x \) (thus \( du = dx \))
- \( dv = \sin x \, dx \) (thus \( v = -\cos x \))

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ \int x \sin x \, dx = -x \cos x \bigg|_0^\pi + \int \cos x \, dx \]
Calculating the integral of \( \cos x \):
\[ \int \cos x \, dx = \sin x \bigg|_0^\pi \]
Now substituting back:
\[ \int_0^\pi x \sin x \, dx = -\left[ \pi \cos \pi - 0 \cdot \cos 0 \right] + \left[ \sin \pi - \sin 0 \right] \]
\[ = -(-\pi) + (0 - 0) = \pi \]

### Step 5: Calculate the integral from \( \pi \) to \( 2\pi \)
Next, we compute \( \int_\pi^{2\pi} x \sin x \, dx \) similarly:
\[ \int_\pi^{2\pi} x \sin x \, dx = -x \cos x \bigg|_\pi^{2\pi} + \int \cos x \, dx \bigg|_\pi^{2\pi} \]
Calculating:
\[ = -\left[ 2\pi \cos(2\pi) - \pi \cos(\pi) \right] + \left[ \sin(2\pi) - \sin(\pi) \right] \]
\[ = -\left[ 2\pi \cdot 1 - \pi \cdot (-1) \right] + (0 - 0) \]
\[ = -\left[ 2\pi + \pi \right] = -3\pi \]

### Step 6: Combine the areas
Now, substituting back into the area formula:
\[ A = \pi + | -3\pi | = \pi + 3\pi = 4\pi \]

### Final Answer:
The area between the curve \( y = x \sin x \) and the x-axis from \( x = 0 \) to \( x = 2\pi \) is \( 4\pi \).

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