The area of the region bounded by the y-axis, `y = cos x and y = sin x, 0 le x le (pi)/(2)` is:

Options

A. `sqrt2` sq. units
B. `(sqrt2 +1)` sq. units
C. `sqrt2(sqrt2-1)` sq. units
D. `(2 sqrt2 -1)` sq. units

Answer

To find the area of the region bounded by the y-axis, \( y = \cos x \), and \( y = \sin x \) for \( 0 \leq x \leq \frac{\pi}{2} \), we will follow these steps:

### Step 1: Identify the curves and their intersection
The curves given are \( y = \cos x \) and \( y = \sin x \). We need to find the points where these two curves intersect within the interval \( [0, \frac{\pi}{2}] \).

To find the intersection points, we set:
\[ \cos x = \sin x \]
This occurs when:
\[ \tan x = 1 \implies x = \frac{\pi}{4} \]

### Step 2: Determine the area between the curves
The area between the curves from \( 0 \) to \( \frac{\pi}{2} \) can be split into two parts:
1. From \( 0 \) to \( \frac{\pi}{4} \), where \( \sin x \) is above \( \cos x \).
2. From \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \), where \( \cos x \) is above \( \sin x \).

### Step 3: Set up the integral for the area
The area \( A \) can be expressed as:
\[ A = \int_0^{\frac{\pi}{4}} (\sin x - \cos x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\cos x - \sin x) \, dx \]

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

### Step 5: Calculate the second integral
Calculating the second integral:
\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\cos x - \sin x) \, dx = \left[\sin x + \cos x\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \]
Calculating the limits:
\[ = \left[\sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right)\right] - \left[\sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right)\right] \]
\[ = [1 + 0] - \left[\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right] \]
\[ = 1 - \sqrt{2} \]

### Step 6: Combine the areas
Now, we combine both areas:
\[ A = (-\sqrt{2} + 1) + (1 - \sqrt{2}) = 2 - 2\sqrt{2} \]

### Step 7: Final area expression
Thus, the total area is:
\[ A = 2(1 - \frac{1}{\sqrt{2}}) = 2\left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) = \frac{2(\sqrt{2} - 1)}{\sqrt{2}} \text{ square units} \]

### Final Answer:
The area of the region bounded by the y-axis, \( y = \cos x \), and \( y = \sin x \) from \( 0 \) to \( \frac{\pi}{2} \) is:
\[ \frac{2(\sqrt{2} - 1)}{\sqrt{2}} \text{ square units} \]