The area enclosed by the curves `y=sinx+cosx and y=|cosx−sinx|` over the interval `[0,pi/2]` is (a) `4(sqrt2-1)` (b) `2sqrt2(sqrt2-1)` (c) `2(sqrt2+1)` (d) `2sqrt2(sqrt2+1)`

Options

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

Explanation

To find the area enclosed by the curves \(y = \sin x + \cos x\) and \(y = |\cos x - \sin x|\) over the interval \([0, \frac{\pi}{2}]\), we will follow these steps:

### Step 1: Identify the curves and their intersection points
The first curve is \(y = \sin x + \cos x\) and the second curve is \(y = |\cos x - \sin x|\). 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:
\[ \sin x + \cos x = |\cos x - \sin x| \]

### Step 2: Analyze the absolute value
The expression \(|\cos x - \sin x|\) can be split into two cases based on the values of \(\cos x\) and \(\sin x\):

1. **Case 1**: When \(\cos x \geq \sin x\) (which occurs in the interval \([0, \frac{\pi}{4}]\)):
\[ |\cos x - \sin x| = \cos x - \sin x \]
Therefore, we solve:
\[ \sin x + \cos x = \cos x - \sin x \]
Simplifying gives:
\[ 2\sin x = 0 \implies \sin x = 0 \implies x = 0 \]

2. **Case 2**: When \(\cos x < \sin x\) (which occurs in the interval \((\frac{\pi}{4}, \frac{\pi}{2}]\)):
\[ |\cos x - \sin x| = \sin x - \cos x \]
Therefore, we solve:
\[ \sin x + \cos x = \sin x - \cos x \]
Simplifying gives:
\[ 2\cos x = 0 \implies \cos x = 0 \implies x = \frac{\pi}{2} \]

The curves intersect at \(x = 0\) and \(x = \frac{\pi}{4}\).

### Step 3: Set up the integral for the area
The area \(A\) between the curves from \(0\) to \(\frac{\pi}{4}\) is given by:
\[ A = \int_0^{\frac{\pi}{4}} \left((\sin x + \cos x) - (\cos x - \sin x)\right) \, dx \]

### Step 4: Simplify the integral
In the interval \([0, \frac{\pi}{4}]\):
\[ A = \int_0^{\frac{\pi}{4}} \left(2\sin x\right) \, dx \]

### Step 5: Calculate the integral
Now we compute the integral:
\[ A = 2 \int_0^{\frac{\pi}{4}} \sin x \, dx \]
The integral of \(\sin x\) is \(-\cos x\), so:
\[ A = 2 \left[-\cos x\right]_0^{\frac{\pi}{4}} = 2 \left[-\cos\left(\frac{\pi}{4}\right) + \cos(0)\right] \]
\[ = 2 \left[-\frac{1}{\sqrt{2}} + 1\right] = 2 \left(1 - \frac{1}{\sqrt{2}}\right) = 2\left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) = \frac{2(\sqrt{2} - 1)}{\sqrt{2}} \]

### Step 6: Calculate the area from \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\)
For the interval \([\frac{\pi}{4}, \frac{\pi}{2}]\):
\[ A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left((\sin x + \cos x) - (\sin x - \cos x)\right) \, dx \]
This simplifies to:
\[ A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left(2\cos x\right) \, dx \]
Calculating this integral gives:
\[ A = 2 \left[\sin x\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = 2 \left[\sin\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{4}\right)\right] = 2 \left[1 - \frac{1}{\sqrt{2}}\right] \]

### Step 7: Combine the areas
The total area is:
\[ A_{total} = 2 \left(1 - \frac{1}{\sqrt{2}}\right) + 2 \left(1 - \frac{1}{\sqrt{2}}\right) = 4\left(1 - \frac{1}{\sqrt{2}}\right) \]

### Final Result
Thus, the area enclosed by the curves is:
\[ A = 2\sqrt{2}(\sqrt{2} - 1) \]