$f^{-1}(x) = f^{n}(x)$

$f(x) = ax^{b} \implies f^{n}(x) = am x^{b - n}$ where $m = b(b - 1)(b - 2) *** (b - n + 1)$

and $f^{-1}(x) = \frac{1}{a^{\frac{1}{b}}} x^{\frac{1}{b}}$

For $x^{b - n} = x^{\frac{1}{b}} \implies b - n = \dfrac{1}{b} \implies b = \dfrac{n + \sqrt{n^2 + 4}}{2}$ for $b > n - 1$ .

For $am = \dfrac{1}{a^{\dfrac{1}{b}}} = \dfrac{1}{a^{b - n}} \implies$ $a^{b - n + 1} m = 1 \implies a = (\dfrac{1}{m})^{\dfrac{1}{b - n + 1}}$

$\implies f(x) = (\dfrac{1}{m})^{\dfrac{1}{b - n + 1}} * x^{b}$

and

$f^{n}(x) = m^{\frac{b - n}{b - n + 1}} x^{b - n} = f^{-1}(x)$

$f(x) = f^{-1}(x) \implies x^{b - n}(m^{\frac{b - n}{b - n + 1}} - (\dfrac{1}{m})^{\frac{1}{b - n + 1}} x^{n})$

$x = 0$ and for $x \geq 0 \implies x = m^{\frac{1}{n}}$ for case $n$ even.

$\implies$ The area $A = \displaystyle\int_{0}^{m^{\frac{1}{n}}} f^{-1}(x) - f(x) \:\ dx =$ $(\dfrac{1}{b - n + 1})m^{\frac{2}{n}} - (\dfrac{1}{b(b - n + 1)})m^{\frac{2}{n}}$

$= (\dfrac{1}{b - n + 1})(m^{\frac{2}{n}})(\dfrac{b - 1}{b}) =$

$\boxed{(\dfrac{1}{b - n + 1})(b(b - 1)(b - 2) * * * (b - n + 1))^{\dfrac{2}{n}}(\dfrac{b - 1}{b})}$

Note: You could have used the Gamma function if so desired.

For $n = 3$ we have:

$b = \dfrac{3 + \sqrt{13}}{2} \implies$ The area $A =$ $(\dfrac{2}{\sqrt{13} - 1}) (\dfrac{9 + 3\sqrt{13}}{2})^{\dfrac{2}{3}} (\dfrac{\sqrt{13} + 1}{\sqrt{13} + 3})$ $\approx \boxed{2.46889797}$ .