Area enclosed by the curves `y=sinx+cosx and y=abs(cosx-sinx)` and the lines `x=0 and x= pi/2`

Answer

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

### Step 1: Determine the intersection points
First, we need to find the points where the curves intersect. We set the equations equal to each other:

\[ \sin x + \cos x = |\cos x - \sin x| \]

We need to consider two cases for the absolute value.

**Case 1:** \( \cos x - \sin x \geq 0 \) (which occurs when \( x \leq \frac{\pi}{4} \))

In this case, we have:

\[ \sin x + \cos x = \cos x - \sin x \]

Simplifying this gives:

\[ 2\sin x = 0 \implies \sin x = 0 \implies x = 0 \]

**Case 2:** \( \cos x - \sin x < 0 \) (which occurs when \( x > \frac{\pi}{4} \))

In this case, we have:

\[ \sin x + \cos x = -(\cos x - \sin x) \]

This simplifies to:

\[ \sin x + \cos x = -\cos x + \sin x \implies 2\cos x = 0 \implies \cos x = 0 \implies x = \frac{\pi}{2} \]

Thus, the points of intersection are \( x = 0 \) and \( x = \frac{\pi}{4} \).

### Step 2: Set up the area integral
The area \( A \) enclosed by the curves from \( x = 0 \) to \( x = \frac{\pi}{2} \) can be calculated by integrating the difference between the upper and lower curves.

From \( x = 0 \) to \( x = \frac{\pi}{4} \), \( y = \sin x + \cos x \) is above \( y = \cos x - \sin x \). From \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \), \( y = \sin x + \cos x \) is above \( y = \sin x - \cos x \).

Thus, we can express the area as:

\[ A = \int_0^{\frac{\pi}{4}} \left( (\sin x + \cos x) - (\cos x - \sin x) \right) dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left( (\sin x + \cos x) - (\sin x - \cos x) \right) dx \]

### Step 3: Simplify the integrals
Calculating the first integral:

\[ \int_0^{\frac{\pi}{4}} \left( \sin x + \cos x - \cos x + \sin x \right) dx = \int_0^{\frac{\pi}{4}} 2\sin x \, dx \]

Calculating the second integral:

\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left( \sin x + \cos x - \sin x + \cos x \right) dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2\cos x \, dx \]

### Step 4: Evaluate the integrals
Now we evaluate both integrals:

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

2. **For the second integral:**
\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2\cos x \, dx = 2\sin x \bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = 2\left(\sin\frac{\pi}{2} - \sin\frac{\pi}{4}\right) = 2\left(1 - \frac{1}{\sqrt{2}}\right) \]

### Step 5: Combine the areas
Now, we combine the areas:

\[ A = 2\left(1 - \frac{1}{\sqrt{2}}\right) + 2\left(1 - \frac{1}{\sqrt{2}}\right) = 4\left(1 - \frac{1}{\sqrt{2}}\right) \]

### Final Area Calculation
Thus, the total area enclosed by the curves is:

\[ A = 4 - 4\frac{1}{\sqrt{2}} = 4 - 2\sqrt{2} \]

### Final Answer
The area enclosed by the curves is:

\[ \boxed{4 - 2\sqrt{2}} \text{ square units} \]