Let the area of the region bounded by the curve y= max $${\sin x, \cos x}$$, lines x = O, $$x=\frac{3\pi}{2}$$, and the x-axis be A. Then, A+$$A^{2}$$ is equal to_____.

We need to find the area $$A$$ bounded by $$y = \max\{\sin x, \cos x\}$$, $$x = 0$$, $$x = \frac{3\pi}{2}$$, and the x-axis.

- $$[0, \frac{\pi}{4}]$$: $$\cos x \geq \sin x$$, so $$y = \cos x$$

- $$[\frac{\pi}{4}, \frac{5\pi}{4}]$$: $$\sin x \geq \cos x$$, so $$y = \sin x$$

- $$[\frac{5\pi}{4}, \frac{3\pi}{2}]$$: $$\cos x \geq \sin x$$, so $$y = \cos x$$

The area between the curve and x-axis counts absolute values. The function $$\max\{\sin x, \cos x\}$$:

- Is positive on $$[0, \pi]$$

- Becomes negative: $$\sin x < 0$$ for $$x \in (\pi, 2\pi)$$, and $$\cos x < 0$$ for $$x \in (\frac{\pi}{2}, \frac{3\pi}{2})$$

On $$[\frac{\pi}{4}, \pi]$$: $$y = \sin x \geq 0$$

On $$[\pi, \frac{5\pi}{4}]$$: $$y = \sin x \leq 0$$

On $$[\frac{5\pi}{4}, \frac{3\pi}{2}]$$: $$y = \cos x \leq 0$$

The area bounded by the curve and x-axis:

$$A = \int_0^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi} \sin x \, dx + \int_{\pi}^{5\pi/4} |\sin x| \, dx + \int_{5\pi/4}^{3\pi/2} |\cos x| \, dx$$

$$\int_0^{\pi/4} \cos x \, dx = \sin x \Big|_0^{\pi/4} = \frac{\sqrt{2}}{2}$$

$$\int_{\pi/4}^{\pi} \sin x \, dx = -\cos x \Big|_{\pi/4}^{\pi} = -(-1) - (-\frac{\sqrt{2}}{2}) = 1 + \frac{\sqrt{2}}{2}$$

$$\int_{\pi}^{5\pi/4} (-\sin x) \, dx = \cos x \Big|_{\pi}^{5\pi/4} = -\frac{\sqrt{2}}{2} - (-1) = 1 - \frac{\sqrt{2}}{2}$$

$$\int_{5\pi/4}^{3\pi/2} (-\cos x) \, dx = -\sin x \Big|_{5\pi/4}^{3\pi/2} = -(-1) - (-(-\frac{\sqrt{2}}{2})) = 1 - \frac{\sqrt{2}}{2}$$

$$A = \frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{2}}{2} + 1 - \frac{\sqrt{2}}{2} + 1 - \frac{\sqrt{2}}{2} = 3$$

$$A + A^2 = 3 + 9 = 12$$

The answer is $$12$$.