The area of the region $$\{(x, y) : |x - 1| \leq y \leq \sqrt{5 - x^2}\}$$ is equal to

We need to find the area of the region $$\{(x, y) : |x - 1| \leq y \leq \sqrt{5 - x^2}\}$$. The upper boundary is the semicircle $$y = \sqrt{5 - x^2}$$, which is the upper half of the circle $$x^2 + y^2 = 5$$ centred at the origin with radius $$\sqrt{5}$$. The lower boundary is $$y = |x - 1|$$.

We first find where these curves intersect. We need $$|x - 1|^2 = 5 - x^2$$, i.e., $$(x-1)^2 + x^2 = 5$$, giving $$2x^2 - 2x + 1 = 5$$, so $$2x^2 - 2x - 4 = 0$$, hence $$x^2 - x - 2 = 0$$, which factors as $$(x-2)(x+1) = 0$$. Thus $$x = -1$$ or $$x = 2$$.

At $$x = -1$$: $$y = |{-1}-1| = 2$$, giving the point $$(-1, 2)$$. At $$x = 2$$: $$y = |2-1| = 1$$, giving the point $$(2, 1)$$. We can verify both lie on the circle: $$1 + 4 = 5$$ and $$4 + 1 = 5$$.

Since $$y = |x-1|$$ has a corner at $$x = 1$$, we split the integral at $$x = 1$$:

$$A = \int_{-1}^{1}\left[\sqrt{5-x^2} - (1-x)\right]dx + \int_1^2\left[\sqrt{5-x^2} - (x-1)\right]dx$$

We compute the integral of $$\sqrt{5-x^2}$$ first. Using the standard formula $$\int\sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$$ with $$a = \sqrt{5}$$:

$$\int_{-1}^{2}\sqrt{5-x^2}\,dx = \left[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}\sin^{-1}\frac{x}{\sqrt{5}}\right]_{-1}^{2}$$

At $$x = 2$$: $$\frac{2}{2}\sqrt{1}+\frac{5}{2}\sin^{-1}\frac{2}{\sqrt{5}} = 1 + \frac{5}{2}\sin^{-1}\frac{2}{\sqrt{5}}$$.

At $$x = -1$$: $$\frac{-1}{2}\sqrt{4}+\frac{5}{2}\sin^{-1}\frac{-1}{\sqrt{5}} = -1 - \frac{5}{2}\sin^{-1}\frac{1}{\sqrt{5}}$$.

So the integral equals $$2 + \frac{5}{2}\left(\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\frac{1}{\sqrt{5}}\right)$$. Now, since $$\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\frac{1}{\sqrt{5}} = \frac{\pi}{2}$$ (because if $$\sin\alpha = 2/\sqrt{5}$$ then $$\cos\alpha = 1/\sqrt{5}$$, so $$\sin^{-1}(1/\sqrt{5}) = \pi/2 - \alpha$$), this becomes $$2 + \frac{5}{2}\cdot\frac{\pi}{2} = 2 + \frac{5\pi}{4}$$.

Now we compute the remaining parts. We have:

$$\int_{-1}^{1}(1-x)\,dx = \left[x - \frac{x^2}{2}\right]_{-1}^{1} = \left(1-\frac{1}{2}\right)-\left(-1-\frac{1}{2}\right) = \frac{1}{2}+\frac{3}{2} = 2$$

$$\int_1^2(x-1)\,dx = \left[\frac{x^2}{2}-x\right]_1^2 = (2-2)-\left(\frac{1}{2}-1\right) = 0+\frac{1}{2} = \frac{1}{2}$$

Therefore the total area is:

$$A = \left(2+\frac{5\pi}{4}\right) - 2 - \frac{1}{2} = \frac{5\pi}{4} - \frac{1}{2}$$

Hence, the correct answer is Option D.