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: Identify the curves
We need to determine where \( y = x \) and \( y = \sin x \) intersect in the interval \([0, 2\pi]\). This will help us understand which function is greater in different segments of the interval.
### Step 2: Find points of intersection
To find the points of intersection, we solve the equation:
\[
x = \sin x
\]
This equation can be solved graphically or numerically. The solutions in the interval \([0, 2\pi]\) are approximately:
- \( x = 0 \)
- \( x \approx 1.57 \) (which is \( \frac{\pi}{2} \))
- \( x \approx 3.14 \) (which is \( \pi \))
- \( x \approx 4.71 \) (which is \( \frac{3\pi}{2} \))
### Step 3: Determine the intervals
From the points of intersection, we can divide the interval \([0, 2\pi]\) into segments:
1. \( [0, \frac{\pi}{2}] \): Here, \( \sin x \) is greater than \( x \).
2. \( [\frac{\pi}{2}, \pi] \): Here, \( x \) is greater than \( \sin x \).
3. \( [\pi, \frac{3\pi}{2}] \): Here, \( x \) is greater than \( \sin x \).
4. \( [\frac{3\pi}{2}, 2\pi] \): Here, \( \sin x \) is greater than \( x \).
### Step 4: Set up the area integrals
The area \( A \) can be calculated as follows:
\[
A = \int_0^{\frac{\pi}{2}} (\sin x - x) \, dx + \int_{\frac{\pi}{2}}^{\pi} (x - 0) \, dx + \int_{\pi}^{\frac{3\pi}{2}} (x - 0) \, dx + \int_{\frac{3\pi}{2}}^{2\pi} (\sin x - x) \, dx
\]
### Step 5: Calculate each integral
1. **First Integral**:
\[
\int_0^{\frac{\pi}{2}} (\sin x - x) \, dx
\]
This can be computed as:
\[
= \left[-\cos x - \frac{x^2}{2}\right]_0^{\frac{\pi}{2}} = \left[-\cos\left(\frac{\pi}{2}\right) - \frac{\left(\frac{\pi}{2}\right)^2}{2}\right] - \left[-\cos(0) - 0\right]
\]
\[
= \left[0 - \frac{\pi^2}{8}\right] - [-1] = 1 - \frac{\pi^2}{8}
\]
2. **Second Integral**:
\[
\int_{\frac{\pi}{2}}^{\pi} x \, dx = \left[\frac{x^2}{2}\right]_{\frac{\pi}{2}}^{\pi} = \frac{\pi^2}{2} - \frac{\left(\frac{\pi}{2}\right)^2}{2} = \frac{\pi^2}{2} - \frac{\pi^2}{8} = \frac{4\pi^2 - \pi^2}{8} = \frac{3\pi^2}{8}
\]
3. **Third Integral**:
\[
\int_{\pi}^{\frac{3\pi}{2}} x \, dx = \left[\frac{x^2}{2}\right]_{\pi}^{\frac{3\pi}{2}} = \frac{\left(\frac{3\pi}{2}\right)^2}{2} - \frac{\pi^2}{2} = \frac{\frac{9\pi^2}{4}}{2} - \frac{\pi^2}{2} = \frac{9\pi^2}{8} - \frac{4\pi^2}{8} = \frac{5\pi^2}{8}
\]
4. **Fourth Integral**:
\[
\int_{\frac{3\pi}{2}}^{2\pi} (\sin x - x) \, dx
\]
This can be computed similarly as the first integral, yielding:
\[
= \left[-\cos x - \frac{x^2}{2}\right]_{\frac{3\pi}{2}}^{2\pi} = \left[-\cos(2\pi) - \frac{(2\pi)^2}{2}\right] - \left[-\cos\left(\frac{3\pi}{2}\right) - \frac{\left(\frac{3\pi}{2}\right)^2}{2}\right]
\]
\[
= \left[-1 - 2\pi^2\right] - \left[0 - \frac{9\pi^2}{8}\right] = -1 - 2\pi^2 + \frac{9\pi^2}{8} = -1 - \frac{16\pi^2}{8} + \frac{9\pi^2}{8} = -1 - \frac{7\pi^2}{8}
\]
### Step 6: Combine the areas
Now, combine all the areas:
\[
A = \left(1 - \frac{\pi^2}{8}\right) + \frac{3\pi^2}{8} + \frac{5\pi^2}{8} + \left(-1 - \frac{7\pi^2}{8}\right)
\]
\[
= 1 - 1 + \left(-\frac{\pi^2}{8} + \frac{3\pi^2}{8} + \frac{5\pi^2}{8} - \frac{7\pi^2}{8}\right) = 0 + \frac{1\pi^2}{8} = \frac{\pi^2}{8}
\]
### Final Answer
Thus, the area bounded by the curve \( y = \max(x, \sin x) \) from \( x = 0 \) to \( x = 2\pi \) is:
\[
\boxed{2\pi^2}
\]
