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

$f(x) = ax^{b} \implies f'(x) = abx^{b - 1} \implies f''(x) = a(b - 1)(b)x^{b - 2}$

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

For $x^{b - 2} = x^{\dfrac{1}{b}} \implies b - 2 = \dfrac{1}{b} \implies b^2 - 2b - 1 = 0$

For $b > 1 \implies b = 1 + \sqrt{2}$

For $a(b - 1)b = \dfrac{1}{a^{(\dfrac{1}{b})}} = \dfrac{1}{a^{b - 2}} \implies a = (\dfrac{1}{b(b - 1)})^{\dfrac{1}{b - 1}}$

$f''(x) = (b(b - 1))^{(\dfrac{b - 2}{b - 1})} x^{b - 2} = f^{-1}(x)$

and

$f(x) = (\dfrac{1}{b(b - 1)})^{\dfrac{1}{b - 1}} x^{b}$ and $f(x) = f^{-1}(x) \implies$

$x^{b - 2}((\dfrac{1}{b(b - 1)})^{\dfrac{1}{b - 1}} x^2 - (b(b - 1))^{\dfrac{b - 2}{b - 1}}) = 0$

$\implies x = 0 \:\, x = \sqrt{b(b - 1)} = \sqrt{2 + \sqrt{2}}$ for $x \geq 0$ .

The area $A = \displaystyle\int_{0}^{\sqrt{b(b - 1)}} f^{-1}(x) - f(x) \:\ dx$ .

$\displaystyle\int_{0}^{\sqrt{b(b - 1)}} f(x) \:\ dx =$ $(\dfrac{1}{b + 1})(\dfrac{1}{b(b - 1)})^{\dfrac{1}{b - 1}} (b(b - 1))^{\dfrac{b + 1}{2}} =$

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

$(\dfrac{1}{b(b - 1)})^{\dfrac{b}{b - 1}} (b(b - 1))^{\dfrac{b + 1}{2}} =$ $(b(b - 1))^{\dfrac{b^2 - 2b - 1}{2(b - 1)}} = {(b(b - 1))}^{0} = \boxed{1}$

and

$\displaystyle\int_{0}^{\sqrt{b(b - 1)}} f^{-1}(x) \:\ dx =$ $\dfrac{(b(b - 1))^{\dfrac{b - 2}{b - 1}}}{b - 1} (b(b - 1))^{\dfrac{b - 1}{2}} =$

$(b - 1)^{\dfrac{-1}{b - 1}} (b - 1)^{\dfrac{b - 1}{2}} * b^{(\dfrac{b^2 - 3}{2(b - 1)})} =$

$(b - 1)^{\dfrac{b^2 - 2b - 1}{2(b - 1)}} * b^{(\dfrac{2 b + 1 -3}{2b - 2})} =(b - 1)^{0} * b = \boxed{b}$

$\implies A = b - 1 = 1 + \sqrt{2} - 1 = \boxed{\sqrt{2}} \approx \boxed{1.41421}$ .