Step-by-Step Solution

1. Define the functions and the interval: We are asked to find the area enclosed by the curves:

• $y_1 = \sin x + \cos x$
• $y_2 = |\cos x - \sin x|$ over the interval $[0, \pi/2]$.

2. Analyze the absolute value function: The function $y_2 = |\cos x - \sin x|$ is a piecewise function. We need to determine where the expression inside the absolute value, $\cos x - \sin x$, is positive or negative within the interval $[0, \pi/2]$.

• The sign of $\cos x - \sin x$ depends on whether $\cos x$ is greater or less than $\sin x$. In the interval $[0, \pi/2]$, we know that $\cos x = \sin x$ at $x = \pi/4$.

• For $x \in [0, \pi/4]$, we have $\cos x \ge \sin x$, so $\cos x - \sin x \ge 0$. Therefore, $y_2 = \cos x - \sin x$ for $x \in [0, \pi/4]$.

• For $x \in [\pi/4, \pi/2]$, we have $\cos x \le \sin x$, so $\cos x - \sin x \le 0$. Therefore, $y_2 = -(\cos x - \sin x) = \sin x - \cos x$ for $x \in [\pi/4, \pi/2]$.

3. Set up the area integral: The required area $A$ is given by the integral of the difference between the upper curve and the lower curve over the given interval. $A = \int_{0}^{\pi/2} (y_{upper} - y_{lower}) dx$ We need to compare $y_1$ and $y_2$ to find the upper curve.

• Interval 1: $x \in [0, \pi/4]$ $y_1 = \sin x + \cos x$ $y_2 = \cos x - \sin x$ The difference is $y_1 - y_2 = (\sin x + \cos x) - (\cos x - \sin x) = 2\sin x$. For $x \in [0, \pi/4]$, $\sin x \ge 0$, so $y_1 - y_2 \ge 0$, which means $y_1 \ge y_2$. The integrand is $2\sin x$.

• Interval 2: $x \in [\pi/4, \pi/2]$ $y_1 = \sin x + \cos x$ $y_2 = \sin x - \cos x$ The difference is $y_1 - y_2 = (\sin x + \cos x) - (\sin x - \cos x) = 2\cos x$. For $x \in [\pi/4, \pi/2]$, $\cos x \ge 0$, so $y_1 - y_2 \ge 0$, which means $y_1 \ge y_2$. The integrand is $2\cos x$.

Since the integrand function changes at $x = \pi/4$, we must split the integral into two parts: $A = \int_{0}^{\pi/4} 2\sin x \,dx + \int_{\pi/4}^{\pi/2} 2\cos x \,dx$

4. Evaluate the integrals:

• First integral: $\int_{0}^{\pi/4} 2\sin x \,dx = 2[-\cos x]_{0}^{\pi/4}$ $= 2(-\cos(\pi/4) - (-\cos(0)))$ $= 2\left(-\frac{1}{\sqrt{2}} + 1\right) = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2}$

• Second integral: $\int_{\pi/4}^{\pi/2} 2\cos x \,dx = 2[\sin x]_{\pi/4}^{\pi/2}$ $= 2(\sin(\pi/2) - \sin(\pi/4))$ $= 2\left(1 - \frac{1}{\sqrt{2}}\right) = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2}$

5. Calculate the total area: The total area is the sum of the areas from the two intervals. $A = (2 - \sqrt{2}) + (2 - \sqrt{2}) = 4 - 2\sqrt{2}$

6. Match the result with the given options: We need to see which option matches our result $A = 4 - 2\sqrt{2}$.

• A: $4(\sqrt 2 - 1) = 4\sqrt 2 - 4$
• B: $2\sqrt 2 (\sqrt 2 - 1) = 2\sqrt 2 \cdot \sqrt 2 - 2\sqrt 2 \cdot 1 = 4 - 2\sqrt 2$
• C: $2(\sqrt 2 + 1) = 2\sqrt 2 + 2$
• D: $2\sqrt 2 (\sqrt 2 + 1) = 4 + 2\sqrt 2$

Our calculated area $4 - 2\sqrt{2}$ matches option B.

Final Answer: The area enclosed by the curves is $4 - 2\sqrt{2}$, which is equivalent to $2\sqrt{2}(\sqrt{2} - 1)$.