The area (in sq. units) bounded by the curve `y=max(x, sinx), AA x in [0, 2pi]` is

Options

A. `pi^(2)`
B. `2pi^(2)`
C. 2
D. 4

Answer

To find the area bounded by the curve \( y = \max(x, \sin x) \) for \( x \) in the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Determine the curves
We need to analyze the functions \( y = x \) and \( y = \sin x \) over the interval \([0, 2\pi]\).

### Step 2: Find the intersection points
To find where \( \sin x \) intersects \( x \), we need to solve the equation:
\[ \sin x = x \]
This equation has a solution at \( x = 0 \). To find other points, we can analyze the behavior of both functions.

- At \( x = 0 \), \( \sin 0 = 0 \) and \( y = 0 \).
- At \( x = \frac{\pi}{2} \), \( \sin \frac{\pi}{2} = 1 \) and \( y = \frac{\pi}{2} \) (since \( \frac{\pi}{2} > 1 \)).
- At \( x = \pi \), \( \sin \pi = 0 \) and \( y = \pi \) (since \( \pi > 0 \)).
- At \( x = \frac{3\pi}{2} \), \( \sin \frac{3\pi}{2} = -1 \) and \( y = \frac{3\pi}{2} \) (since \( \frac{3\pi}{2} > -1 \)).
- At \( x = 2\pi \), \( \sin 2\pi = 0 \) and \( y = 2\pi \) (since \( 2\pi > 0 \)).

From this analysis, we can conclude that \( \sin x \) is above \( x \) only in the interval \( [0, \frac{\pi}{2}] \) and below \( x \) for the rest of the interval.

### Step 3: Set up the area calculation
The area can be split into two parts:
1. From \( x = 0 \) to \( x = \frac{\pi}{2} \), where \( y = \sin x \).
2. From \( x = \frac{\pi}{2} \) to \( x = 2\pi \), where \( y = x \).

The area \( A \) can be expressed as:
\[ A = \int_0^{\frac{\pi}{2}} (\sin x - 0) \, dx + \int_{\frac{\pi}{2}}^{2\pi} (x - 0) \, dx \]

### Step 4: Calculate the integrals
1. For the first integral:
\[ \int_0^{\frac{\pi}{2}} \sin x \, dx = -\cos x \bigg|_0^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1 \]

2. For the second integral:
\[ \int_{\frac{\pi}{2}}^{2\pi} x \, dx = \frac{x^2}{2} \bigg|_{\frac{\pi}{2}}^{2\pi} = \frac{(2\pi)^2}{2} - \frac{\left(\frac{\pi}{2}\right)^2}{2} = \frac{4\pi^2}{2} - \frac{\frac{\pi^2}{4}}{2} = 2\pi^2 - \frac{\pi^2}{8} = 2\pi^2 - \frac{\pi^2}{8} = \frac{16\pi^2}{8} - \frac{\pi^2}{8} = \frac{15\pi^2}{8} \]

### Step 5: Combine the areas
Adding both areas together:
\[ A = 1 + \frac{15\pi^2}{8} \]

### Final Answer
The total area bounded by the curve \( y = \max(x, \sin x) \) from \( x = 0 \) to \( x = 2\pi \) is:
\[ A = 1 + \frac{15\pi^2}{8} \]