The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and `x = (3pi)/(2)`, is

Options

A. `(4sqrt(2)-2)` sq units
B. `(4sqrt(2)+2)` sq units
C. `(4sqrt(2)-1)` sq units
D. `(4sqrt(2)+1)` sq units

Explanation

To find the area bounded by the curves \( y = \cos x \) and \( y = \sin x \) between the ordinates \( x = 0 \) and \( x = \frac{3\pi}{2} \), we will follow these steps:

### Step 1: Identify the points of intersection
To find the area between the curves, we first need to determine where the curves intersect. This occurs when \( \cos x = \sin x \).

Setting the equations equal to each other:
\[ \cos x = \sin x \]
This can be solved by dividing both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)):
\[ 1 = \tan x \quad \Rightarrow \quad x = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z} \]

Within the interval \( [0, \frac{3\pi}{2}] \), the relevant points of intersection are:
- \( x = \frac{\pi}{4} \)
- \( x = \frac{5\pi}{4} \)

### Step 2: Determine the area segments
Next, we will find the area between the curves in three segments:
1. From \( x = 0 \) to \( x = \frac{\pi}{4} \)
2. From \( x = \frac{\pi}{4} \) to \( x = \frac{5\pi}{4} \)
3. From \( x = \frac{5\pi}{4} \) to \( x = \frac{3\pi}{2} \)

### Step 3: Calculate the area for each segment
1. **From \( x = 0 \) to \( x = \frac{\pi}{4} \)**:
Here, \( \cos x \) is above \( \sin x \):
\[ A_1 = \int_0^{\frac{\pi}{4}} (\cos x - \sin x) \, dx \]

2. **From \( x = \frac{\pi}{4} \) to \( x = \frac{5\pi}{4} \)**:
Here, \( \sin x \) is above \( \cos x \):
\[ A_2 = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) \, dx \]

3. **From \( x = \frac{5\pi}{4} \) to \( x = \frac{3\pi}{2} \)**:
Here, \( \cos x \) is above \( \sin x \):
\[ A_3 = \int_{\frac{5\pi}{4}}^{\frac{3\pi}{2}} (\cos x - \sin x) \, dx \]

### Step 4: Compute the integrals
Now we compute each integral:

1. **For \( A_1 \)**:
\[ A_1 = \int_0^{\frac{\pi}{4}} (\cos x - \sin x) \, dx = \left[ \sin x + \cos x \right]_0^{\frac{\pi}{4}} = \left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \right) - \left( \sin 0 + \cos 0 \right) \]
\[ = \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - (0 + 1) = \sqrt{2} - 1 \]

2. **For \( A_2 \)**:
\[ A_2 = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) \, dx = \left[ -\cos x - \sin x \right]_{\frac{\pi}{4}}^{\frac{5\pi}{4}} = \left( -\cos \frac{5\pi}{4} - \sin \frac{5\pi}{4} \right) - \left( -\cos \frac{\pi}{4} - \sin \frac{\pi}{4} \right) \]
\[ = \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) = \sqrt{2} + \sqrt{2} = 2\sqrt{2} \]

3. **For \( A_3 \)**:
\[ A_3 = \int_{\frac{5\pi}{4}}^{\frac{3\pi}{2}} (\cos x - \sin x) \, dx = \left[ \sin x + \cos x \right]_{\frac{5\pi}{4}}^{\frac{3\pi}{2}} = \left( \sin \frac{3\pi}{2} + \cos \frac{3\pi}{2} \right) - \left( \sin \frac{5\pi}{4} + \cos \frac{5\pi}{4} \right) \]
\[ = (-1 + 0) - \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) = -1 + \sqrt{2} = \sqrt{2} - 1 \]

### Step 5: Total Area
Now, we can sum the areas:
\[ A = A_1 + A_2 + A_3 = (\sqrt{2} - 1) + (2\sqrt{2}) + (\sqrt{2} - 1) = 4\sqrt{2} - 2 \]

Thus, the area bounded by the curves \( y = \cos x \) and \( y = \sin x \) between \( x = 0 \) and \( x = \frac{3\pi}{2} \) is:
\[ \boxed{4\sqrt{2} - 2} \]