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. `(sqrt2-1)` sq. uints
D. `(2sqrt2-1)` 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 can follow these steps:

### Step 1: Find the points of intersection
We need to find where \( \cos x = \sin x \).

\[ \cos x = \sin x \]

Dividing both sides by \( \cos x \) (where \( \cos x \neq 0 \)) gives:

\[ 1 = \tan x \]

This implies:

\[ x = \frac{\pi}{4} \]

### Step 2: Determine the area between the curves
The area between the curves from \( x = 0 \) to \( x = \frac{\pi}{4} \) can be calculated as:

\[ \text{Area} = \int_0^{\frac{\pi}{4}} (\cos x - \sin x) \, dx \]

### Step 3: Compute the integral
We can split the integral into two parts:

\[ \text{Area} = \int_0^{\frac{\pi}{4}} \cos x \, dx - \int_0^{\frac{\pi}{4}} \sin x \, dx \]

Calculating each integral separately:

1. **Integral of \( \cos x \)**:
\[ \int \cos x \, dx = \sin x \]
Evaluating from \( 0 \) to \( \frac{\pi}{4} \):
\[ \left[ \sin x \right]_0^{\frac{\pi}{4}} = \sin\left(\frac{\pi}{4}\right) - \sin(0) = \frac{1}{\sqrt{2}} - 0 = \frac{1}{\sqrt{2}} \]

2. **Integral of \( \sin x \)**:
\[ \int \sin x \, dx = -\cos x \]
Evaluating from \( 0 \) to \( \frac{\pi}{4} \):
\[ \left[ -\cos x \right]_0^{\frac{\pi}{4}} = -\cos\left(\frac{\pi}{4}\right) + \cos(0) = -\frac{1}{\sqrt{2}} + 1 \]

### Step 4: Combine the results
Now we can combine the results:

\[ \text{Area} = \left(\frac{1}{\sqrt{2}}\right) - \left(-\frac{1}{\sqrt{2}} + 1\right) \]

This simplifies to:

\[ \text{Area} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 1 = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1 \]

### Final Answer
Thus, the area of the region bounded by the y-axis, \( y = \cos x \), and \( y = \sin x \) for \( 0 \leq x \leq \frac{\pi}{2} \) is:

\[ \text{Area} = \sqrt{2} - 1 \]