The area of the region bounded by the `Y-"axis"` `y = "cos" x` and `y = "sin" x` Where `0lexlepi/2,` is

Options

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

Explanation

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

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

Setting \( \cos x = \sin x \):
\[ \tan x = 1 \implies x = \frac{\pi}{4} \]
Thus, the curves intersect at \( x = \frac{\pi}{4} \).

### Step 2: Determine the area between the curves
The area \( A \) between the curves from \( x = 0 \) to \( x = \frac{\pi}{4} \) can be calculated using the integral:
\[ A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx \]

### Step 3: Evaluate the integral
We can split the integral into two parts:
\[ A = \int_{0}^{\frac{\pi}{4}} \cos x \, dx - \int_{0}^{\frac{\pi}{4}} \sin x \, dx \]

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 \( \frac{\pi}{4} \):
\[ \left[-\cos x\right]_{0}^{\frac{\pi}{4}} = -\cos\left(\frac{\pi}{4}\right) - (-\cos(0)) = -\frac{1}{\sqrt{2}} + 1 = 1 - \frac{1}{\sqrt{2}} \]

### Step 4: Combine the results
Now, substituting back into the area formula:
\[ A = \frac{1}{\sqrt{2}} - \left(1 - \frac{1}{\sqrt{2}}\right) \]
This simplifies to:
\[ A = \frac{1}{\sqrt{2}} - 1 + \frac{1}{\sqrt{2}} = 2 \cdot \frac{1}{\sqrt{2}} - 1 = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1 \]

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