Area Enclosed Between Sine Cosine Curves in First Quadrant Interval

To find the area bounded by the curves $y = \sin x + \cos x$ and $y = |\cos x - \sin x|$ over the interval $\left[0, \frac{\pi}{2}\right]$, we set up the definite integral:

$A = \int_{0}^{\pi/2} \left( (\sin x + \cos x) - |\cos x - \sin x| \right) dx$

We split the interval $\left[0, \frac{\pi}{2}\right]$ at $x = \frac{\pi}{4}$ based on the definition of the absolute value function:

1. For $x \in \left[0, \frac{\pi}{4}\right]$: $\cos x \ge \sin x \implies |\cos x - \sin x| = \cos x - \sin x$ The integrand becomes: $(\sin x + \cos x) - (\cos x - \sin x) = 2\sin x$

2. For $x \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right]$: $\sin x \ge \cos x \implies |\cos x - \sin x| = \sin x - \cos x$ The integrand becomes: $(\sin x + \cos x) - (\sin x - \cos x) = 2\cos x$

Now, we evaluate the integral over both sub-intervals:

$A = \int_{0}^{\pi/4} 2\sin x \, dx + \int_{\pi/4}^{\pi/2} 2\cos x \, dx$

Evaluating each integral separately: $\int_{0}^{\pi/4} 2\sin x \, dx = \left[ -2\cos x \right]_0^{\pi/4} = -2\left(\frac{1}{\sqrt{2}}\right) - (-2(1)) = 2 - \sqrt{2}$

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

Adding the two results together gives the total area: $A = (2 - \sqrt{2}) + (2 - \sqrt{2}) = 4 - 2\sqrt{2} = 2\sqrt{2}(\sqrt{2} - 1)$

Therefore, the correct option is B.