Let e be the base of natural logarithm and let f: \ 1, 2, 3, 4\...

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

Arranging the domain and codomain in increasing order, we get $f(1) = e^3$, $f(2) = e^2$, $f(3) = e$, and $f(4) = 1$.

Given $g: \{1, e, e^2, e^3\} \rightarrow \left\{1, \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}\right\}$ is a strictly increasing bijective function.

Arranging the domain and codomain in increasing order, we get $g(1) = \dfrac{1}{4}$, $g(e) = \dfrac{1}{3}$, $g(e^2) = \dfrac{1}{2}$, and $g(e^3) = 1$.

Evaluating $\phi(x) = \left[f^{-1}\left\{g^{-1}\left(\dfrac{1}{2}\right)\right\}\right]^x$:

From the mapping of $g$, $g^{-1}\left(\dfrac{1}{2}\right) = e^2$.

From the mapping of $f$, $f^{-1}(e^2) = 2$.

Substituting these values, we get $\phi(x) = 2^x$.

The region $R$ is given by $x^2 \leq y \leq 2^x$ for $0 \leq x \leq 1$.

The area of the region $R$ is:

$\int_{0}^{1} (2^x - x^2) dx = \left[ \dfrac{2^x}{\log_e(2)} - \dfrac{x^3}{3} \right]_{0}^{1}$

$= \left( \dfrac{2}{\log_e(2)} - \dfrac{1}{3} \right) - \left( \dfrac{1}{\log_e(2)} - 0 \right)$

$= \dfrac{1}{\log_e(2)} - \dfrac{1}{3}$

$= \dfrac{3 - \log_e(2)}{3\log_e(2)}$

Answer: $\dfrac{3 - \log_e(2)}{3\log_e(2)}$