The area of the region A=`{(x,y):|cosx-sinx|leylesinx,0lexlepi/2}` is

Options

A. `sqrt5-2sqrt2+1`
B. `1-3/sqrt2+4/sqrt5`
C. `3/sqrt5-3/sqrt2+1`
D. `sqrt5+2sqrt2-4.5`

Answer

To find the area of the region defined by the inequalities \( | \cos x - \sin x | \leq y \leq \sin x \) for \( 0 \leq x \leq \frac{\pi}{2} \), we can break down the problem step by step.

### Step 1: Analyze the inequalities
The inequality \( | \cos x - \sin x | \leq y \) can be split into two cases:
1. \( \cos x - \sin x \leq y \)
2. \( -(\cos x - \sin x) \leq y \) or \( \sin x - \cos x \leq y \)

Thus, we have:
- \( \sin x - \cos x \leq y \leq \sin x \) for \( \frac{\pi}{4} < x \leq \frac{\pi}{2} \)
- \( \cos x - \sin x \leq y \leq \sin x \) for \( 0 \leq x \leq \frac{\pi}{4} \)

### Step 2: Find intersection points
To find the intersection points of \( \cos x = \sin x \):
- This occurs when \( x = \frac{\pi}{4} \).

### Step 3: Set up the area calculation
The area \( A \) can be calculated as the integral of the upper function minus the lower function over the specified intervals.

#### For \( 0 \leq x \leq \frac{\pi}{4} \):
The upper function is \( \sin x \) and the lower function is \( \cos x - \sin x \).

\[ A_1 = \int_0^{\frac{\pi}{4}} \left( \sin x - (\cos x - \sin x) \right) dx = \int_0^{\frac{\pi}{4}} (2 \sin x - \cos x) dx \]

#### For \( \frac{\pi}{4} < x \leq \frac{\pi}{2} \):
The upper function is \( \sin x \) and the lower function is \( \sin x - \cos x \).

\[ A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left( \sin x - (\sin x - \cos x) \right) dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x dx \]

### Step 4: Calculate the integrals

#### Integral \( A_1 \):
\[ A_1 = \int_0^{\frac{\pi}{4}} (2 \sin x - \cos x) dx \]
Calculating this integral:
\[ = \left[-2 \cos x - \sin x \right]_0^{\frac{\pi}{4}} = \left[-2 \cdot \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right] - \left[-2 \cdot 1 - 0 \right] \]
\[ = \left[-\frac{3\sqrt{2}}{2}\right] - (-2) = 2 - \frac{3\sqrt{2}}{2} \]

#### Integral \( A_2 \):
\[ A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x dx = [\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \left[1 - \frac{1}{\sqrt{2}}\right] \]

### Step 5: Combine the areas
The total area \( A \) is:
\[ A = A_1 + A_2 = \left(2 - \frac{3\sqrt{2}}{2}\right) + \left(1 - \frac{1}{\sqrt{2}}\right) \]
\[ = 3 - \frac{3\sqrt{2}}{2} - \frac{1}{\sqrt{2}} = 3 - \frac{5\sqrt{2}}{2} \]

### Final Answer
Thus, the area of the region \( A \) is:
\[ A = 3 - \frac{5\sqrt{2}}{2} \]