The area of the region $$A = \{(x,y) : |\cos x - \sin x| \leq y \leq \sin x, 0 \leq x \leq \frac{\pi}{2}\}$$

We need to find the area of the region $$A = \{(x,y) : |\cos x - \sin x| \leq y \leq \sin x, 0 \leq x \leq \pi/2\}$$.

First we analyze $$|\cos x - \sin x|$$ on $$[0, \pi/2]$$. On $$[0, \pi/4]$$: $$\cos x \geq \sin x$$, so $$|\cos x - \sin x| = \cos x - \sin x$$. On $$[\pi/4, \pi/2]$$: $$\sin x \geq \cos x$$, so $$|\cos x - \sin x| = \sin x - \cos x$$.

Next we determine where $$|\cos x - \sin x| \leq \sin x$$. On $$[0, \pi/4]$$: $$\cos x - \sin x \leq \sin x \implies \cos x \leq 2\sin x \implies \tan x \geq 1/2$$, so $$x \geq \arctan(1/2)$$. On $$[\pi/4, \pi/2]$$: $$\sin x - \cos x \leq \sin x \implies -\cos x \leq 0 \implies \cos x \geq 0$$, which holds throughout the interval.

Let $$\alpha = \arctan(1/2)$$. The area is given by

$$\text{Area} = \int_\alpha^{\pi/4}[\sin x - (\cos x - \sin x)]dx + \int_{\pi/4}^{\pi/2}[\sin x - (\sin x - \cos x)]dx$$

$$= \int_\alpha^{\pi/4}(2\sin x - \cos x)dx + \int_{\pi/4}^{\pi/2}\cos x\,dx$$

For the second integral, $$\int_{\pi/4}^{\pi/2}\cos x\,dx = [\sin x]_{\pi/4}^{\pi/2} = 1 - \frac{1}{\sqrt{2}}$$.

For the first integral, $$\int_\alpha^{\pi/4}(2\sin x - \cos x)dx = [-2\cos x - \sin x]_\alpha^{\pi/4}$$. At $$x = \pi/4$$: $$-2\cdot\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{3}{\sqrt{2}}$$. At $$x = \alpha$$, where $$\tan\alpha = 1/2$$, we have $$\sin\alpha = \frac{1}{\sqrt{5}}$$ and $$\cos\alpha = \frac{2}{\sqrt{5}}$$, giving $$-2\cdot\frac{2}{\sqrt{5}} - \frac{1}{\sqrt{5}} = -\frac{5}{\sqrt{5}} = -\sqrt{5}$$. Hence the first integral equals $$-\frac{3}{\sqrt{2}} - (-\sqrt{5}) = \sqrt{5} - \frac{3}{\sqrt{2}}$$.

Combining these results, the total area is

$$\text{Area} = \sqrt{5} - \frac{3}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} = \sqrt{5} - \frac{4}{\sqrt{2}} + 1 = \sqrt{5} - 2\sqrt{2} + 1$$.

The correct answer is Option 4: $$\sqrt{5} - 2\sqrt{2} + 1$$.