Let $$e$$ be the base of natural logarithm and let $$f : \{1, 2, 3, 4\} \to \{1, e, e^2, e^3\}$$ and $$g : \{1, e, e^2, e^3\} \to \left\{1,\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\}$$ be two bijective functions such that $$f$$ is strictly decreasing and $$g$$ is strictly increasing. If $$\phi(x) = \left[f^{-1}\left\{g^{-1}\left(\frac{1}{2}\right)\right\}\right]^x$$, then the area of the region R = {(x, y): $$x^2 \leq y \leq \phi(x)$$, $$0 \leq x \leq 1$$} is:

The function $$g: \{1, e, e^2, e^3\} \rightarrow \{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\}$$ is a strictly increasing bijection

Domain elements: $$1 < e < e^2 < e^3$$

Range elements: $$\frac{1}{4} < \frac{1}{3} < \frac{1}{2} < 1$$

$$g(e^2) = \frac{1}{2}$$ $$\implies$$ $$\therefore g^{-1}\left(\frac{1}{2}\right) = e^2$$

The function $$f: \{1, 2, 3, 4\} \rightarrow \{1, e, e^2, e^3\}$$ is a strictly decreasing bijection.

Domain elements: $$1 < 2 < 3 < 4$$

Range elements: $$1 < e < e^2 < e^3$$

Since it is decreasing, the smallest domain element maps to the largest range element, and vice versa: 

$$\therefore f^{-1}(e^2) = 2$$

$$\phi(x) = 2^x$$

The area of the bounded region $$R = \{(x, y): x^2 \le y \le 2^x, \, 0 \le x \le 1\}$$ is:

$$\text{Area} = \int_{0}^{1} (2^x - x^2) \, dx = \left[ \frac{2^x}{\ln 2} - \frac{x^3}{3} \right]_{0}^{1}$$

$$\text{Area} = \left( \frac{2}{\ln 2} - \frac{1}{3} \right) - \left( \frac{1}{\ln 2} - 0 \right) = \frac{1}{\ln 2} - \frac{1}{3}$$

$$\text{Area} = \frac{3 - \ln 2}{3 \ln 2} = \frac{3 - \log_e(2)}{3 \log_e(2)}$$